93、R语言教程详解
加载数据 > w<-read.table("test.prn",header = T) > w X.. X...1 1 A 2 2 B 3 3 C 5 4 D 5 > library(readxl) > dat<-read_excel("test.xlsx") > dat # A tibble: 4 x 2 `商品` `价格` <chr> <dbl> 1 A 2 2 B 3 3 C 5 4 D 5 > bank=read.table("bank-full.csv",header = TRUE,sep=",") 查看数据结构 > str(bank) 'data.frame': 41188 obs. of 21 variables: $ age : int 56 57 37 40 56 45 59 41 24 25 ... $ job : Factor w/ 12 levels "admin.","blue-collar",..: 4 8 8 1 8 8 1 2 10 8 ... $ marital : Factor w/ 4 levels "divorced","married",..: 2 2 2 2 2 2 2 2 3 3 ... $ education : Factor w/ 8 levels "basic.4y","basic.6y",..: 1 4 4 2 4 3 6 8 6 4 ... $ default : Factor w/ 3 levels "no","unknown",..: 1 2 1 1 1 2 1 2 1 1 ... $ housing : Factor w/ 3 levels "no","unknown",..: 1 1 3 1 1 1 1 1 3 3 ... $ loan : Factor w/ 3 levels "no","unknown",..: 1 1 1 1 3 1 1 1 1 1 ... $ contact : Factor w/ 2 levels "cellular","telephone": 2 2 2 2 2 2 2 2 2 2 ... $ month : Factor w/ 10 levels "apr","aug","dec",..: 7 7 7 7 7 7 7 7 7 7 ... $ day_of_week : Factor w/ 5 levels "fri","mon","thu",..: 2 2 2 2 2 2 2 2 2 2 ... $ duration : int 261 149 226 151 307 198 139 217 380 50 ... $ campaign : int 1 1 1 1 1 1 1 1 1 1 ... $ pdays : int 999 999 999 999 999 999 999 999 999 999 ... $ previous : int 0 0 0 0 0 0 0 0 0 0 ... $ poutcome : Factor w/ 3 levels "failure","nonexistent",..: 2 2 2 2 2 2 2 2 2 2 ... $ emp.var.rate : num 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 ... $ cons.price.idx: num 94 94 94 94 94 ... $ cons.conf.idx : num -36.4 -36.4 -36.4 -36.4 -36.4 -36.4 -36.4 -36.4 -36.4 -36.4 ... $ euribor3m : num 4.86 4.86 4.86 4.86 4.86 ... $ nr.employed : num 5191 5191 5191 5191 5191 ... $ y : Factor w/ 2 levels "no","yes": 1 1 1 1 1 1 1 1 1 1 ... 查看数据的最小值，最大值，中位数，平均数，分位数 > summary(bank) age job marital Min. :17.00 admin. :10422 divorced: 4612 1st Qu.:32.00 blue-collar: 9254 married :24928 Median :38.00 technician : 6743 single :11568 Mean :40.02 services : 3969 unknown : 80 3rd Qu.:47.00 management : 2924 Max. :98.00 retired : 1720 (Other) : 6156 education default housing university.degree :12168 no :32588 no :18622 high.school : 9515 unknown: 8597 unknown: 990 basic.9y : 6045 yes : 3 yes :21576 professional.course: 5243 basic.4y : 4176 basic.6y : 2292 (Other) : 1749 loan contact month day_of_week no :33950 cellular :26144 may :13769 fri:7827 unknown: 990 telephone:15044 jul : 7174 mon:8514 yes : 6248 aug : 6178 thu:8623 jun : 5318 tue:8090 nov : 4101 wed:8134 apr : 2632 (Other): 2016 duration campaign pdays Min. : 0.0 Min. : 1.000 Min. : 0.0 1st Qu.: 102.0 1st Qu.: 1.000 1st Qu.:999.0 Median : 180.0 Median : 2.000 Median :999.0 Mean : 258.3 Mean : 2.568 Mean :962.5 3rd Qu.: 319.0 3rd Qu.: 3.000 3rd Qu.:999.0 Max. :4918.0 Max. :56.000 Max. :999.0 previous poutcome emp.var.rate Min. :0.000 failure : 4252 Min. :-3.40000 1st Qu.:0.000 nonexistent:35563 1st Qu.:-1.80000 Median :0.000 success : 1373 Median : 1.10000 Mean :0.173 Mean : 0.08189 3rd Qu.:0.000 3rd Qu.: 1.40000 Max. :7.000 Max. : 1.40000 cons.price.idx cons.conf.idx euribor3m Min. :92.20 Min. :-50.8 Min. :0.634 1st Qu.:93.08 1st Qu.:-42.7 1st Qu.:1.344 Median :93.75 Median :-41.8 Median :4.857 Mean :93.58 Mean :-40.5 Mean :3.621 3rd Qu.:93.99 3rd Qu.:-36.4 3rd Qu.:4.961 Max. :94.77 Max. :-26.9 Max. :5.045 nr.employed y Min. :4964 no :36548 1st Qu.:5099 yes: 4640 Median :5191 Mean :5167 3rd Qu.:5228 Max. :5228 > psych::describe(bank) 方差 个数 平均值 标准差 均值 去掉最大 中位数 最小值 最大值 极差 偏差 峰度 绝对偏差 最小值 之后 的平均数 vars n mean sd median trimmed mad min max range skew kurtosis age 1 41188 40.02 10.42 38.00 39.30 10.38 17.00 98.00 81.00 0.78 0.79 job* 2 41188 4.72 3.59 3.00 4.48 2.97 1.00 12.00 11.00 0.45 -1.39 marital* 3 41188 2.17 0.61 2.00 2.21 0.00 1.00 4.00 3.00 -0.06 -0.34 education* 4 41188 4.75 2.14 4.00 4.88 2.97 1.00 8.00 7.00 -0.24 -1.21 default* 5 41188 1.21 0.41 1.00 1.14 0.00 1.00 3.00 2.00 1.44 0.07 housing* 6 41188 2.07 0.99 3.00 2.09 0.00 1.00 3.00 2.00 -0.14 -1.95 loan* 7 41188 1.33 0.72 1.00 1.16 0.00 1.00 3.00 2.00 1.82 1.38 contact* 8 41188 1.37 0.48 1.00 1.33 0.00 1.00 2.00 1.00 0.56 -1.69 month* 9 41188 5.23 2.32 5.00 5.31 2.97 1.00 10.00 9.00 -0.31 -1.03 day_of_week* 10 41188 3.00 1.40 3.00 3.01 1.48 1.00 5.00 4.00 0.01 -1.27 duration 11 41188 258.29 259.28 180.00 210.61 139.36 0.00 4918.00 4918.00 3.26 20.24 campaign 12 41188 2.57 2.77 2.00 1.99 1.48 1.00 56.00 55.00 4.76 36.97 pdays 13 41188 962.48 186.91 999.00 999.00 0.00 0.00 999.00 999.00 -4.92 22.23 previous 14 41188 0.17 0.49 0.00 0.05 0.00 0.00 7.00 7.00 3.83 20.11 poutcome* 15 41188 1.93 0.36 2.00 2.00 0.00 1.00 3.00 2.00 -0.88 3.98 emp.var.rate 16 41188 0.08 1.57 1.10 0.27 0.44 -3.40 1.40 4.80 -0.72 -1.06 cons.price.idx 17 41188 93.58 0.58 93.75 93.58 0.56 92.20 94.77 2.57 -0.23 -0.83 cons.conf.idx 18 41188 -40.50 4.63 -41.80 -40.60 6.52 -50.80 -26.90 23.90 0.30 -0.36 euribor3m 19 41188 3.62 1.73 4.86 3.81 0.16 0.63 5.04 4.41 -0.71 -1.41 nr.employed 20 41188 5167.04 72.25 5191.00 5178.43 55.00 4963.60 5228.10 264.50 -1.04 0.00 y* 21 41188 1.11 0.32 1.00 1.02 0.00 1.00 2.00 1.00 2.45 4.00 se age 0.05 job* 0.02 marital* 0.00 education* 0.01 default* 0.00 housing* 0.00 loan* 0.00 contact* 0.00 month* 0.01 day_of_week* 0.01 duration 1.28 campaign 0.01 pdays 0.92 previous 0.00 poutcome* 0.00 emp.var.rate 0.01 cons.price.idx 0.00 cons.conf.idx 0.02 euribor3m 0.01 nr.employed 0.36 y* 0.00 查看数据是否有缺失值 > sapply(bank,anyNA) age job marital education FALSE FALSE FALSE FALSE default housing loan contact FALSE FALSE FALSE FALSE month day_of_week duration campaign FALSE FALSE FALSE FALSE pdays previous poutcome emp.var.rate FALSE FALSE FALSE FALSE cons.price.idx cons.conf.idx euribor3m nr.employed FALSE FALSE FALSE FALSE y FALSE 成功与不成功的个数 > table(bank$y) no yes 36548 4640 在是否结婚这个属性的取值与 是否成功的数量比较 > table(bank$y,bank$marital) divorced married single unknown no 4136 22396 9948 68 yes 476 2532 1620 12 > xtabs(~y+marital,data=bank) marital y divorced married single unknown no 4136 22396 9948 68 yes 476 2532 1620 12 > tab=table(bank$y,bank$marital) > tab divorced married single unknown no 4136 22396 9948 68 yes 476 2532 1620 12 在是否结婚这个属性上的取值 > margin.table(tab,2) divorced married single unknown 4612 24928 11568 80 > margin.table(tab,1) no yes 36548 4640 在是否结婚这个属性上横向看概率 > prop.table(tab,1) divorced married single unknown no 0.113166247 0.612783189 0.272189997 0.001860567 yes 0.102586207 0.545689655 0.349137931 0.002586207 在是否结婚这个属性上纵向看概率 > prop.table(tab,2) divorced married single unknown no 0.8967910 0.8984275 0.8599585 0.8500000 yes 0.1032090 0.1015725 0.1400415 0.1500000 平的列联表 以第一列和第二列，展开分类group by 1,2 以col.vars 的取值 进行次数统计 > ftable(bank[,c(3,4,21)],row.vars = 1:2,col.vars = "y") y no yes marital education divorced basic.4y 406 83 basic.6y 169 13 basic.9y 534 31 high.school 1086 107 illiterate 1 1 professional.course 596 61 university.degree 1177 160 unknown 167 20 married basic.4y 2915 313 basic.6y 1628 139 basic.9y 3858 298 high.school 4683 475 illiterate 12 3 professional.course 2799 357 university.degree 5573 821 unknown 928 126 single basic.4y 422 31 basic.6y 301 36 basic.9y 1174 142 high.school 2702 448 illiterate 1 0 professional.course 1247 177 university.degree 3723 683 unknown 378 103 unknown basic.4y 5 1 basic.6y 6 0 basic.9y 6 2 high.school 13 1 illiterate 0 0 professional.course 6 0 university.degree 25 6 unknown 7 2 卡方检验，在p值小于2.2e-16时，拒绝原假设，认为数据不服从卡方分布 > chisq.test(tab) Pearson's Chi-squared test data: tab X-squared = 122.66, df = 3, p-value < 2.2e-16 画直方图 > hist(bank$age) > library(lattice) 画连续变量的分布，就是把直方图的中位数连接起来 以年龄为横轴，y为纵轴，数据是bank,画图，auto.key是否有图例 > densityplot(~age,groups = y,data=bank,plot.point=FALSE,auto.key = TRUE) 画Box图 > boxplot(age~y,data=bank) 双样本t分布检验，p值小于0.05时拒绝原假设 这里的原假设是两个样本没有相关性 得到的结果是p值为1.805e-06,拒绝两个样本没有相关性的假设 这里认为两个样本有相关性 > t.test(age~y,data=bank,alternative="two.sided",var.equal=FALSE) Welch Two Sample t-test data: age by y t = -4.7795, df = 5258.5, p-value = 1.805e-06 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -1.4129336 -0.5909889 sample estimates: mean in group no mean in group yes 39.91119 40.91315 数据可视化 画饼图 > tab=table(bank$marital) > pie(tab) 画直方图 > tab=table(bank$marital) > barplot(tab) 画下面这个图 > tab=table(bank$marital,bank$y) > plot(tab) 画层叠直方图 > tab=table(bank$marital,bank$y) > lattice::barchart(tab,auto.key=TRUE) 加载这个包，准备画图 > library(dplyr) > data=group_by(bank,marital,y) > data=tally(data) ！！！！！！！！！！！！！ > ggplot2::ggplot(data=data,mapping=aes(marital,n))+geom_bar(mapping=aes(fill=y),position="dodge",stat="identity") 数据预处理 分组之后再画图 > labels=c('青年','中年','老年') > bank$age_group=cut(bank$age,breaks = c(0,35,55,100),right = FALSE,labels = labels) > library(ggplot2) > ggplot(data=bank,mapping = aes(age_group))+geom_bar(mapping = aes(fill=y),position="dodge",stat="count") 衍生变量 直接使用$符向原数据框添加新的变量 > bank$log.cons.price.idx=log(bank$cons.price.idx) 使用transform函数向原数据框添加变量 > bank<-transform(bank,log.cons.price.idx=log(cons.price.idx),log.nr.employed=log(nr.employed)) 使用dplyr包里的mutate函数增加变量 > bank<-dplyr::mutate(bank,log.cons.price.idx=log(cons.price.idx)) 使用dplyr包里的transmute函数只保留新生成的变量 > bank2<-dplyr::transmute(bank,log.cons.price.idx=log(cons.price.idx),log.nr.employed=log(nr.employed)) 中心化 > v=1:10 > v1=v-mean(v) > v2=scale(v,center=TRUE,scale = FALSE) 无量纲化 > V1=v/sqrt(sum(v^2)/(length(v)-1)) > v2=scale(v,center=FALSE,scale=TRUE) 根据最大最小值进行归一化 > v3=(v-min(v))/(max(v)-min(v)) 进行标准正态化 > v1=(v-mean(v))/sd(v) > v2=scale(v,center = TRUE,scale=TRUE) Box-Cox变换 使用car包里的boxCox函数 > install.packages("car") > library(car) > boxCox(age~.,data=bank) 使用caret包，做Box-Cox变换 > install.packages("caret") > library(caret) > dat<-subset(bank,select="age") > trans<-preProcess(dat,method=C("BoxCox")) 数据预处理下 违反常识的异常值 基于数据分布的异常值(离群点)识别 bank.dirty=read.csv("bank-dirty.csv") summary(bank.dirty) age job marital education Min. : 17.00 admin. :10422 divorced: 4612 university.degree :12165 1st Qu.: 32.00 blue-collar: 9254 married :24928 high.school : 9515 Median : 38.00 technician : 6743 single :11568 basic.9y : 6043 Mean : 40.03 services : 3969 NA's : 80 professional.course: 5242 3rd Qu.: 47.00 management : 2924 basic.4y : 4175 Max. :123.00 (Other) : 7546 (Other) : 2310 NA's :2 NA's : 330 NA's : 1738 default housing loan contact month no :32588 no :18622 no :33950 cellular :26144 may :13769 yes : 3 yes :21576 yes : 6248 telephone:15044 jul : 7174 NA's: 8597 NA's: 990 NA's: 990 aug : 6178 jun : 5318 nov : 4101 apr : 2632 (Other): 2016 day_of_week duration campaign pdays previous fri:7827 Min. : 0.0 Min. : 1.000 Min. : 0.0 Min. :0.000 mon:8514 1st Qu.: 102.0 1st Qu.: 1.000 1st Qu.:999.0 1st Qu.:0.000 thu:8623 Median : 180.0 Median : 2.000 Median :999.0 Median :0.000 tue:8090 Mean : 258.3 Mean : 2.568 Mean :962.5 Mean :0.173 wed:8134 3rd Qu.: 319.0 3rd Qu.: 3.000 3rd Qu.:999.0 3rd Qu.:0.000 Max. :4918.0 Max. :56.000 Max. :999.0 Max. :7.000 poutcome emp.var.rate cons.price.idx cons.conf.idx failure : 4252 Min. :-3.40000 Min. :92.20 Min. :-50.8 nonexistent:35563 1st Qu.:-1.80000 1st Qu.:93.08 1st Qu.:-42.7 success : 1373 Median : 1.10000 Median :93.75 Median :-41.8 Mean : 0.08189 Mean :93.58 Mean :-40.5 3rd Qu.: 1.40000 3rd Qu.:93.99 3rd Qu.:-36.4 Max. : 1.40000 Max. :94.77 Max. :-26.9 euribor3m nr.employed y Min. :0.634 Min. :4964 no :36548 1st Qu.:1.344 1st Qu.:5099 yes: 4640 Median :4.857 Median :5191 Mean :3.621 Mean :5167 3rd Qu.:4.961 3rd Qu.:5228 Max. :5.045 Max. :5228 常识告诉我们，虽然123岁的老人存在，但概率也极低，也不太可能是银行的客户 找出在年龄这一列的上离群值和下离群值 > head(bank.dirty[order(bank.dirty$age,decreasing = TRUE),'age',drop=FALSE],n=5) age 39494 123 38453 98 38456 98 27827 95 38922 94 > tail(bank.dirty[order(bank.dirty$age,decreasing = TRUE),'age',drop=FALSE],n=5) age 37559 17 37580 17 38275 17 120 NA 156 NA 异常值的处理 当作缺失值处理 > bank.dirty$age[which(bank.dirty$age>98)]<-NA 删除或者插补 重编码 职业类型有12个分类，不利于后续分析，把除了unknown以外的分类进行重新编码，简化成4类 Month有12个分类，把它转化成季度 Education的分类，除了unknow之外有7类 进行重编码 levels(bank.dirty$job) <- c( "management","services","entrepreneur","entrepreneur", "management","unemployed", "entrepreneur","services", "unemployed","services","unemployed","unknown" ) > levels(bank.dirty$month) <- c("Q2","Q3","Q4","Q3","Q2", "Q1","Q2","Q4","Q4","Q3") > > levels(bank.dirty$education) <- c( "primary","primary","primary","secondary", "primary","tertiary","tertiary","unknown") 缺失值 分类较多，分类是unknown,不能给我们提供信息 有些模型不能处理缺失值，比如Logistic回归 缺失值插补的方法 1、 用中位数或众数插补 > library(imputeMissings) > bank.clean<-impute(bank.dirty,object = compute(bank.dirty,method = "median/mode")) 2、 最邻近(knn)插补 library(DMwR) bank.clean=knnImputation(bank.dirty,k=5) 3、 随机森林插补 library(missForest) Imp = missForest(bank.dirty) bank.clean = Imp$ximp 缺失值插补的R包 1、 imputeMissings包 2、 DMwR包 用Logistic回归建立客户响应模型 1、 广义线性模型 广义线性模型擅长于处理因变量不是连续变量的问题 1) Y是分类变量 2) Y是定序变量 3) Y是离散取值 2、 当Y取值是0-1二分类变量是，就是Logistic回归 Logistic回归在R中的实现 数据重编码 bank$y=ifelse(bank$y=='yes',1,0) 改成以Q1为参考因子 bank$month<-relevel(bank$month,ref="Q1") 构建Logistic回归模型 > model<-glm(y~.,data=bank,family = 'binomial') > summary(model) Call: glm(formula = y ~ ., family = "binomial", data = bank) Deviance Residuals: Min 1Q Median 3Q Max -5.9958 -0.3082 -0.1887 -0.1333 3.4283 Coefficients: (1 not defined because of singularities) Estimate Std. Error z value Pr(>|z|) (Intercept) -1.957e+02 1.935e+01 -10.116 < 2e-16 *** age 1.851e-03 2.415e-03 0.767 0.443289 jobblue-collar -2.659e-01 7.942e-02 -3.348 0.000814 *** jobentrepreneur -2.029e-01 1.248e-01 -1.626 0.103924 jobhousemaid -3.628e-02 1.475e-01 -0.246 0.805705 jobmanagement -8.054e-02 8.501e-02 -0.947 0.343423 jobretired 2.928e-01 1.067e-01 2.743 0.006092 ** jobself-employed -1.680e-01 1.176e-01 -1.428 0.153332 jobservices -1.497e-01 8.552e-02 -1.751 0.079969 . jobstudent 2.674e-01 1.106e-01 2.416 0.015680 * jobtechnician 3.462e-03 7.096e-02 0.049 0.961086 jobunemployed 8.514e-03 1.273e-01 0.067 0.946686 jobunknown -8.046e-02 2.390e-01 -0.337 0.736420 maritalmarried 1.567e-02 6.824e-02 0.230 0.818420 maritalsingle 6.620e-02 7.791e-02 0.850 0.395473 maritalunknown 6.303e-02 4.113e-01 0.153 0.878211 educationbasic.6y 9.647e-02 1.202e-01 0.803 0.422195 educationbasic.9y -2.154e-02 9.494e-02 -0.227 0.820557 educationhigh.school 3.381e-02 9.188e-02 0.368 0.712895 educationilliterate 1.132e+00 7.395e-01 1.531 0.125887 educationprofessional.course 1.136e-01 1.013e-01 1.121 0.262175 educationuniversity.degree 2.134e-01 9.188e-02 2.322 0.020211 * educationunknown 1.361e-01 1.196e-01 1.138 0.255314 defaultunknown -3.055e-01 6.712e-02 -4.552 5.32e-06 *** defaultyes -7.150e+00 1.135e+02 -0.063 0.949784 housingunknown -7.385e-02 1.390e-01 -0.531 0.595260 housingyes -3.740e-03 4.121e-02 -0.091 0.927695 loanunknown NA NA NA NA loanyes -6.362e-02 5.725e-02 -1.111 0.266454 contacttelephone -6.068e-01 7.124e-02 -8.518 < 2e-16 *** monthQ2 -2.192e+00 1.125e-01 -19.479 < 2e-16 *** monthQ3 -1.463e+00 1.148e-01 -12.747 < 2e-16 *** monthQ4 -1.995e+00 1.240e-01 -16.088 < 2e-16 *** day_of_weekmon -1.216e-01 6.588e-02 -1.846 0.064887 . day_of_weekthu 6.375e-02 6.382e-02 0.999 0.317842 day_of_weektue 6.867e-02 6.545e-02 1.049 0.294118 day_of_weekwed 1.436e-01 6.530e-02 2.199 0.027911 * duration 4.667e-03 7.397e-05 63.092 < 2e-16 *** campaign -4.543e-02 1.158e-02 -3.922 8.77e-05 *** pdays -9.627e-04 2.162e-04 -4.452 8.50e-06 *** previous -5.806e-02 5.879e-02 -0.988 0.323369 poutcomenonexistent 4.507e-01 9.372e-02 4.809 1.51e-06 *** poutcomesuccess 9.371e-01 2.106e-01 4.451 8.56e-06 *** emp.var.rate -1.389e+00 7.693e-02 -18.057 < 2e-16 *** cons.price.idx 1.815e+00 1.193e-01 15.218 < 2e-16 *** cons.conf.idx 3.353e-02 6.664e-03 5.033 4.84e-07 *** euribor3m 6.054e-02 1.126e-01 0.537 0.590987 nr.employed 4.937e-03 1.873e-03 2.635 0.008413 ** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 28999 on 41187 degrees of freedom Residual deviance: 17199 on 41141 degrees of freedom AIC: 17293 Number of Fisher Scoring iterations: 10 > exp(coef(model)) (Intercept) age jobblue-collar 9.856544e-86 1.001853e+00 7.665077e-01 jobentrepreneur jobhousemaid jobmanagement 8.163314e-01 9.643733e-01 9.226187e-01 jobretired jobself-employed jobservices 1.340142e+00 8.453874e-01 8.609387e-01 jobstudent jobtechnician jobunemployed 1.306514e+00 1.003468e+00 1.008550e+00 jobunknown maritalmarried maritalsingle 9.226922e-01 1.015789e+00 1.068445e+00 maritalunknown educationbasic.6y educationbasic.9y 1.065061e+00 1.101276e+00 9.786948e-01 educationhigh.school educationilliterate educationprofessional.course 1.034388e+00 3.101297e+00 1.120248e+00 educationuniversity.degree educationunknown defaultunknown 1.237856e+00 1.145744e+00 7.367445e-01 defaultyes housingunknown housingyes 7.851906e-04 9.288126e-01 9.962671e-01 loanunknown loanyes contacttelephone NA 9.383587e-01 5.450980e-01 monthQ2 monthQ3 monthQ4 1.116739e-01 2.314802e-01 1.360620e-01 day_of_weekmon day_of_weekthu day_of_weektue 8.854888e-01 1.065828e+00 1.071082e+00 day_of_weekwed duration campaign 1.154380e+00 1.004678e+00 9.555850e-01 pdays previous poutcomenonexistent 9.990378e-01 9.435960e-01 1.569466e+00 poutcomesuccess emp.var.rate cons.price.idx 2.552531e+00 2.493091e-01 6.140533e+00 cons.conf.idx euribor3m nr.employed 1.034103e+00 1.062408e+00 1.004949e+00 Job变量的基准水平是management,从上面的结果看，服务业和自主劳动者购买银行产品的几率(odds)是管理岗从业人员的0.88倍，未就业人员购买银行产品的几率是管理岗人员的1.25倍 > summary(model.step) 向前逐步回归 > model.step=step(model,direction = "backward") 向后逐步回归 > model.step = step(model, direction = "forward") 双向逐步回归 > model.step = step(model, direction = "both") > summary(model.step) Call: glm(formula = y ~ job + education + default + contact + month + day_of_week + duration + campaign + pdays + poutcome + emp.var.rate + cons.price.idx + cons.conf.idx + nr.employed, family = "binomial", data = bank) Deviance Residuals: Min 1Q Median 3Q Max -5.9884 -0.3088 -0.1887 -0.1332 3.4026 Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -2.031e+02 1.426e+01 -14.246 < 2e-16 *** jobblue-collar -2.700e-01 7.917e-02 -3.411 0.000648 *** jobentrepreneur -2.043e-01 1.242e-01 -1.645 0.100003 jobhousemaid -2.832e-02 1.464e-01 -0.193 0.846590 jobmanagement -8.368e-02 8.409e-02 -0.995 0.319670 jobretired 3.234e-01 9.130e-02 3.542 0.000397 *** jobself-employed -1.670e-01 1.176e-01 -1.421 0.155435 jobservices -1.528e-01 8.545e-02 -1.789 0.073666 . jobstudent 2.682e-01 1.046e-01 2.565 0.010316 * jobtechnician 4.389e-03 7.093e-02 0.062 0.950665 jobunemployed 8.975e-03 1.271e-01 0.071 0.943715 jobunknown -6.363e-02 2.378e-01 -0.268 0.789057 educationbasic.6y 8.993e-02 1.196e-01 0.752 0.452024 educationbasic.9y -2.716e-02 9.416e-02 -0.288 0.772992 educationhigh.school 2.890e-02 9.053e-02 0.319 0.749573 educationilliterate 1.118e+00 7.398e-01 1.511 0.130744 educationprofessional.course 1.084e-01 1.004e-01 1.079 0.280686 educationuniversity.degree 2.103e-01 9.017e-02 2.332 0.019678 * educationunknown 1.363e-01 1.195e-01 1.140 0.254110 defaultunknown -3.017e-01 6.666e-02 -4.526 6.02e-06 *** defaultyes -7.141e+00 1.135e+02 -0.063 0.949831 contacttelephone -6.011e-01 7.069e-02 -8.504 < 2e-16 *** monthQ2 -2.210e+00 1.108e-01 -19.939 < 2e-16 *** monthQ3 -1.475e+00 1.146e-01 -12.869 < 2e-16 *** monthQ4 -1.982e+00 1.183e-01 -16.755 < 2e-16 *** day_of_weekmon -1.210e-01 6.584e-02 -1.837 0.066174 . day_of_weekthu 6.208e-02 6.374e-02 0.974 0.330066 day_of_weektue 6.851e-02 6.538e-02 1.048 0.294651 day_of_weekwed 1.420e-01 6.525e-02 2.176 0.029592 * duration 4.667e-03 7.396e-05 63.099 < 2e-16 *** campaign -4.587e-02 1.158e-02 -3.960 7.49e-05 *** pdays -8.822e-04 2.024e-04 -4.358 1.31e-05 *** poutcomenonexistent 5.219e-01 6.356e-02 8.211 < 2e-16 *** poutcomesuccess 9.996e-01 2.028e-01 4.928 8.31e-07 *** emp.var.rate -1.376e+00 6.885e-02 -19.980 < 2e-16 *** cons.price.idx 1.845e+00 1.041e-01 17.725 < 2e-16 *** cons.conf.idx 3.622e-02 4.853e-03 7.464 8.42e-14 *** nr.employed 5.883e-03 9.765e-04 6.024 1.70e-09 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 28999 on 41187 degrees of freedom Residual deviance: 17203 on 41150 degrees of freedom AIC: 17279 Number of Fisher Scoring iterations: 10 模型预测 用predict函数，参数type=’response’ Newdata参数是要预测的数据集 > prob<-predict(model.step,type = 'response') > head(prob) 1 2 3 4 5 6 0.015029328 0.006044212 0.011640349 0.010173952 0.016897254 0.007174804 假设以0.5为临界值 > pre<-ifelse(prob>0.5,1,0) > table(pre,bank$y) pre 0 1 0 35596 2667 1 952 1973 > 预测的准确率 > (35592+1964)/(35592+2676+956+1964) [1] 0.911819 实际有响应的客户被识别出了多少 > 1964/(1964+2676) [1] 0.4232759 模型评估 > confusionMatrix(bank$y,pre,pos='1') Confusion Matrix and Statistics Reference Prediction 0 1 0 35596 952 1 2667 1973 Accuracy : 0.9121 95% CI : (0.9094, 0.9149) No Information Rate : 0.929 P-Value [Acc > NIR] : 1 Kappa : 0.476 Mcnemar's Test P-Value : <2e-16 Sensitivity : 0.67453 Specificity : 0.93030 Pos Pred Value : 0.42522 Neg Pred Value : 0.97395 Prevalence : 0.07102 Detection Rate : 0.04790 Detection Prevalence : 0.11265 Balanced Accuracy : 0.80241 'Positive' Class : 1 Kappa 统计量(kappa statistic) 用于评判分类器的分类结果与随机分类的差异度 用Kappa统计量评价： 较差：小于0.20 一般：0.20至0.40 稳健：0.40至0.60 好的：0.60至0.80 很好的：0.80至1.00 ROC曲线 pred<-prediction(prob,bank$y) perf<-performance(pred,measure = "tpr",x="fpr") plot(perf) RandomForest 加载数据列 > data=read.table("input.txt",header = TRUE) > str(data) 'data.frame': 222 obs. of 23 variables: $ Acti_Profile : num 0 0 0 0 0 0 0 0 0 0 ... $ Activity : num 1.25 0 0.938 6.562 0 ... $ Diastolic_PTT : num 256 240 253 0 241 ... $ Diastolic : num 73.2 78.6 74 0 78.4 ... $ Heart_Rate_Curve : num 81.2 69.7 77.6 95 83.6 ... $ Heart_Rate_Variability_HF: num 131 250 135 144 141 ... $ Heart_Rate_Variability_LF: num 311 218 203 301 244 ... $ MAP : num 86 93.5 86.9 0 91.7 ... $ Position : num 0 0 0 1 0 0 0 0 0 0 ... $ PTT_Raw : num 308 288 308 0 295 ... $ RR_Interval : num 734 878 773 632 714 ... $ Sleep_Wake : num 1 1 1 1 1 0 1 1 0 0 ... $ SpO2 : num 0 0 99 0 98.4 ... $ Sympatho_Vagal_Balance : num 23 8.17 14.5 20.4 16.88 ... $ Systolic_PTT : num 308 288 307 0 295 ... $ Systolic : num 113 124 113 0 119 ... $ Autonomic_arousals : num 0 0 0 0 0 0 0 0 0 0 ... $ Cardio_complex : num 0 0 0 1 0 0 0 0 0 0 ... $ Cardio_rhythm : num 0 0 2 0 0 0 0 0 0 0 ... $ Classification_Arousal : num 0 0 0 0 0 0 0 0 0 0 ... $ PTT_Events : num 1 0 2 0 0 0 0 0 0 0 ... $ Systolic_Events : num 1 0 1 0 0 0 0 0 0 0 ... $ y : num 1 0 1 0 0 0 0 0 0 0 ... 加载随机森林包 > library(randomForest) 进行训练 以y作为因变量，其余数据作为自变量 > rf <- randomForest(y ~ ., data=data, ntree=100, proximity=TRUE,importance=TRUE) > plot(rf) 重要性检测 衡量把一个变量的取值变为随机数，随机森林预测准确性的降低程度 > importance(rf,type=1) %IncMSE Acti_Profile 0.00000000 Activity 0.99353251 Diastolic_PTT 0.32193611 Diastolic 1.99891809 Heart_Rate_Curve 0.92001352 Heart_Rate_Variability_HF 2.07870722 Heart_Rate_Variability_LF -0.24957163 MAP 0.48142975 Position 1.86876751 PTT_Raw 1.94648914 RR_Interval 0.60557964 Sleep_Wake 1.00503782 SpO2 0.25396165 Sympatho_Vagal_Balance 1.42906765 Systolic_PTT 1.27965813 Systolic 0.77382673 Autonomic_arousals 0.00000000 Cardio_complex 1.00503782 Cardio_rhythm 1.14283152 Classification_Arousal -0.04383997 PTT_Events 4.63980680 Systolic_Events 33.29461169 输出随机森林的模型 > print(rf) Call: randomForest(formula = y ~ ., data = data, ntree = 100, proximity = TRUE, importance = TRUE) Type of random forest: regression Number of trees: 100 No. of variables tried at each split: 7 Mean of squared residuals: 0.003226897 残差平方和SSE % Var explained: 98.7 > 总平方和（SST）：（样本数据-样本均值）的平方和 回归平方和（SSR）：（预测数据-样本均值）的平方和 残差平方和（SSE）：（样本数据-预测数据均值）的平方和 SST = SSR + SSE 基尼指数： > importance(rf,type=2) IncNodePurity Acti_Profile 0.000000000 Activity 0.445181480 Diastolic_PTT 0.452221870 Diastolic 0.449372186 Heart_Rate_Curve 0.473113852 Heart_Rate_Variability_HF 0.226815300 Heart_Rate_Variability_LF 0.205457353 MAP 0.536977574 Position 0.307333210 PTT_Raw 0.656726800 RR_Interval 0.452738011 Sleep_Wake 0.014423077 SpO2 1.793361279 Sympatho_Vagal_Balance 0.352759689 Systolic_PTT 0.851951505 Systolic 0.823955781 Autonomic_arousals 0.000000000 Cardio_complex 0.008047619 Cardio_rhythm 0.141907084 Classification_Arousal 0.085739429 PTT_Events 7.468690820 Systolic_Events 39.000163018 > 进行预测 prediction <- predict(rf, data[,],type="response") 输出预测结果 table(observed =data$y,predicted=prediction) plot(prediction) 支持向量机 library(e1071) svmfit<-svm(y~.,data=data,kernel="linear",cost=10,scale=FALSE) > print(svmfit) Call: svm(formula = y ~ ., data = data, kernel = "linear", cost = 10, scale = FALSE) Parameters: SVM-Type: eps-regression SVM-Kernel: linear cost: 10 gamma: 0.04545455 epsilon: 0.1 Number of Support Vectors: 20 > plot(svmfit,data) 神经网络 > concrete<-read_excel("Concrete_Data.xls") > str(concrete) Classes ‘tbl_df’, ‘tbl’ and 'data.frame': 1030 obs. of 9 variables: $ Cement : num 540 540 332 332 199 ... $ Slag : num 0 0 142 142 132 ... $ Ash : num 0 0 0 0 0 0 0 0 0 0 ... $ water : num 162 162 228 228 192 228 228 228 228 228 ... $ superplastic: num 2.5 2.5 0 0 0 0 0 0 0 0 ... $ coarseagg : num 1040 1055 932 932 978 ... $ fineagg : num 676 676 594 594 826 ... $ age : num 28 28 270 365 360 90 365 28 28 28 ... $ strength : num 80 61.9 40.3 41.1 44.3 ... > normalize <- function(x){ return ((x-min(x))/(max(x)-min(x)))} > concrete_norm <- as.data.frame(lapply(concrete,normalize)) > concrete_train <- concrete_norm[1:773,] > concrete_test <- concrete_norm[774:1030,] > library(neuralnet) > concrete_model <- neuralnet(strength ~ Cement+Slag+Ash+water+superplastic+coarseagg+fineagg+age,data=concrete_train) > plot(concrete_model) model_results <- compute(concrete_model,concrete_test[1:8]) predicted_strength <- model_results$net.result > cor(predicted_strength,concrete_test$strength) [,1] [1,] 0.7205120076 > concrete_model2 <- neuralnet(strength ~ Cement+Slag+Ash+water+superplastic+coarseagg+fineagg+age,data=concrete_train,hidden=5) > plot(concrete_model2) 计算误差 > model_results2 <- compute(concrete_model2,concrete_test[1:8]) > predicted_strength2 <- model_results2$net.result > cor(predicted_strength2,concrete_test$strength) [,1] [1,] 0.6727155609 > 主成分分析 身高、体重、胸围、坐高 > test<-data.frame( + X1=c(148, 139, 160, 149, 159, 142, 153, 150, 151, 139, + 140, 161, 158, 140, 137, 152, 149, 145, 160, 156, + 151, 147, 157, 147, 157, 151, 144, 141, 139, 148), + X2=c(41, 34, 49, 36, 45, 31, 43, 43, 42, 31, + 29, 47, 49, 33, 31, 35, 47, 35, 47, 44, + 42, 38, 39, 30, 48, 36, 36, 30, 32, 38), + X3=c(72, 71, 77, 67, 80, 66, 76, 77, 77, 68, + 64, 78, 78, 67, 66, 73, 82, 70, 74, 78, + 73, 73, 68, 65, 80, 74, 68, 67, 68, 70), + X4=c(78, 76, 86, 79, 86, 76, 83, 79, 80, 74, + 74, 84, 83, 77, 73, 79, 79, 77, 87, 85, + 82, 78, 80, 75, 88, 80, 76, 76, 73, 78) + ) > test.pr<-princomp(test,cor=TRUE) > summary(test.pr,loadings=TRUE) Importance of components: Comp.1 Comp.2 Comp.3 Comp.4 Standard deviation 1.8817805390 0.55980635717 0.28179594325 0.25711843909 Proportion of Variance 0.8852744993 0.07834578938 0.01985223841 0.01652747293 Cumulative Proportion 0.8852744993 0.96362028866 0.98347252707 1.00000000000 Loadings: Comp.1 Comp.2 Comp.3 Comp.4 X1 0.497 0.543 -0.450 0.506 X2 0.515 -0.210 -0.462 -0.691 X3 0.481 -0.725 0.175 0.461 X4 0.507 0.368 0.744 -0.232 前两个主成分的累计贡献率已经达到96% 可以舍去另外两个主成分 达到降维的目的 因此可以得到函数表达式 Z1=-0.497X'1-0.515X'2-0.481X'3-0.507X'4 Z2= 0.543X'1-0.210X'2-0.725X'3-0.368X'4 4.画主成分的碎石图并预测 > screeplot(test.pr,type="lines") > p<-predict(test.pr) > p Comp.1 Comp.2 Comp.3 Comp.4 [1,] -0.06990949737 -0.23813701272 -0.35509247634 -0.266120139417 [2,] -1.59526339772 -0.71847399061 0.32813232022 -0.118056645885 [3,] 2.84793151061 0.38956678680 -0.09731731272 -0.279482487139 [4,] -0.75996988424 0.80604334819 -0.04945721875 -0.162949297761 [5,] 2.73966776853 0.01718087263 0.36012614873 0.358653043787 [6,] -2.10583167924 0.32284393414 0.18600422367 -0.036456083707 [7,] 1.42105591247 -0.06053164925 0.21093320662 -0.044223092351 [8,] 0.82583976981 -0.78102575640 -0.27557797533 0.057288571933 [9,] 0.93464401954 -0.58469241699 -0.08814135786 0.181037745585 [10,] -2.36463819933 -0.36532199291 0.08840476284 0.045520127461 [11,] -2.83741916086 0.34875841111 0.03310422938 -0.031146930047 [12,] 2.60851223537 0.21278727930 -0.33398036623 0.210157574387 [13,] 2.44253342081 -0.16769495893 -0.46918095412 -0.162987829937 [14,] -1.86630668724 0.05021383642 0.37720280364 -0.358821916178 [15,] -2.81347420580 -0.31790107093 -0.03291329149 -0.222035112399 [16,] -0.06392982655 0.20718447599 0.04334339948 0.703533623798 [17,] 1.55561022242 -1.70439673831 -0.33126406220 0.007551878960 [18,] -1.07392250663 -0.06763418320 0.02283648409 0.048606680158 [19,] 2.52174211878 0.97274300950 0.12164633439 -0.390667990681 [20,] 2.14072377494 0.02217881219 0.37410972458 0.129548959692 [21,] 0.79624421805 0.16307887263 0.12781269571 -0.294140762463 [22,] -0.28708320594 -0.35744666106 -0.03962115883 0.080991988802 [23,] 0.25151075072 1.25555187663 -0.55617324819 0.109068938725 [24,] -2.05706031616 0.78894493512 -0.26552109297 0.388088642937 [25,] 3.08596854773 -0.05775318018 0.62110421208 -0.218939612456 [26,] 0.16367554630 0.04317931667 0.24481850312 0.560248997030 [27,] -1.37265052598 0.02220972121 -0.23378320040 -0.257399715466 [28,] -2.16097778154 0.13733232981 0.35589738735 0.093123683044 [29,] -2.40434826507 -0.48613137190 -0.16154440788 -0.007914021222 [30,] -0.50287467640 0.14734316507 -0.20590831261 -0.122078819188 >
加载数据
> w<-read.table("test.prn",header = T)
> w
X.. X...1
1 A 2
2 B 3
3 C 5
4 D 5
> library(readxl)
> dat<-read_excel("test.xlsx")
> dat
# A tibble: 4 x 2
`商品` `价格`
<chr> <dbl>
1 A 2
2 B 3
3 C 5
4 D 5
> bank=read.table("bank-full.csv",header = TRUE,sep=",")
查看数据结构
> str(bank)
'data.frame': 41188 obs. of 21 variables:
$ age : int 56 57 37 40 56 45 59 41 24 25 ...
$ job : Factor w/ 12 levels "admin.","blue-collar",..: 4 8 8 1 8 8 1 2 10 8 ...
$ marital : Factor w/ 4 levels "divorced","married",..: 2 2 2 2 2 2 2 2 3 3 ...
$ education : Factor w/ 8 levels "basic.4y","basic.6y",..: 1 4 4 2 4 3 6 8 6 4 ...
$ default : Factor w/ 3 levels "no","unknown",..: 1 2 1 1 1 2 1 2 1 1 ...
$ housing : Factor w/ 3 levels "no","unknown",..: 1 1 3 1 1 1 1 1 3 3 ...
$ loan : Factor w/ 3 levels "no","unknown",..: 1 1 1 1 3 1 1 1 1 1 ...
$ contact : Factor w/ 2 levels "cellular","telephone": 2 2 2 2 2 2 2 2 2 2 ...
$ month : Factor w/ 10 levels "apr","aug","dec",..: 7 7 7 7 7 7 7 7 7 7 ...
$ day_of_week : Factor w/ 5 levels "fri","mon","thu",..: 2 2 2 2 2 2 2 2 2 2 ...
$ duration : int 261 149 226 151 307 198 139 217 380 50 ...
$ campaign : int 1 1 1 1 1 1 1 1 1 1 ...
$ pdays : int 999 999 999 999 999 999 999 999 999 999 ...
$ previous : int 0 0 0 0 0 0 0 0 0 0 ...
$ poutcome : Factor w/ 3 levels "failure","nonexistent",..: 2 2 2 2 2 2 2 2 2 2 ...
$ emp.var.rate : num 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 ...
$ cons.price.idx: num 94 94 94 94 94 ...
$ cons.conf.idx : num -36.4 -36.4 -36.4 -36.4 -36.4 -36.4 -36.4 -36.4 -36.4 -36.4 ...
$ euribor3m : num 4.86 4.86 4.86 4.86 4.86 ...
$ nr.employed : num 5191 5191 5191 5191 5191 ...
$ y : Factor w/ 2 levels "no","yes": 1 1 1 1 1 1 1 1 1 1 ...
查看数据的最小值，最大值，中位数，平均数，分位数
> summary(bank)
age job marital
Min. :17.00 admin. :10422 divorced: 4612
1st Qu.:32.00 blue-collar: 9254 married :24928
Median :38.00 technician : 6743 single :11568
Mean :40.02 services : 3969 unknown : 80
3rd Qu.:47.00 management : 2924
Max. :98.00 retired : 1720
(Other) : 6156
education default housing
university.degree :12168 no :32588 no :18622
high.school : 9515 unknown: 8597 unknown: 990
basic.9y : 6045 yes : 3 yes :21576
professional.course: 5243
basic.4y : 4176
basic.6y : 2292
(Other) : 1749
loan contact month day_of_week
no :33950 cellular :26144 may :13769 fri:7827
unknown: 990 telephone:15044 jul : 7174 mon:8514
yes : 6248 aug : 6178 thu:8623
jun : 5318 tue:8090
nov : 4101 wed:8134
apr : 2632
(Other): 2016
duration campaign pdays
Min. : 0.0 Min. : 1.000 Min. : 0.0
1st Qu.: 102.0 1st Qu.: 1.000 1st Qu.:999.0
Median : 180.0 Median : 2.000 Median :999.0
Mean : 258.3 Mean : 2.568 Mean :962.5
3rd Qu.: 319.0 3rd Qu.: 3.000 3rd Qu.:999.0
Max. :4918.0 Max. :56.000 Max. :999.0
previous poutcome emp.var.rate
Min. :0.000 failure : 4252 Min. :-3.40000
1st Qu.:0.000 nonexistent:35563 1st Qu.:-1.80000
Median :0.000 success : 1373 Median : 1.10000
Mean :0.173 Mean : 0.08189
3rd Qu.:0.000 3rd Qu.: 1.40000
Max. :7.000 Max. : 1.40000
cons.price.idx cons.conf.idx euribor3m
Min. :92.20 Min. :-50.8 Min. :0.634
1st Qu.:93.08 1st Qu.:-42.7 1st Qu.:1.344
Median :93.75 Median :-41.8 Median :4.857
Mean :93.58 Mean :-40.5 Mean :3.621
3rd Qu.:93.99 3rd Qu.:-36.4 3rd Qu.:4.961
Max. :94.77 Max. :-26.9 Max. :5.045
nr.employed y
Min. :4964 no :36548
1st Qu.:5099 yes: 4640
Median :5191
Mean :5167
3rd Qu.:5228
Max. :5228
> psych::describe(bank)
方差 个数 平均值 标准差 均值 去掉最大 中位数 最小值 最大值 极差 偏差 峰度
绝对偏差
最小值
之后
的平均数
vars n mean sd median trimmed mad min max range skew kurtosis
age 1 41188 40.02 10.42 38.00 39.30 10.38 17.00 98.00 81.00 0.78 0.79
job* 2 41188 4.72 3.59 3.00 4.48 2.97 1.00 12.00 11.00 0.45 -1.39
marital* 3 41188 2.17 0.61 2.00 2.21 0.00 1.00 4.00 3.00 -0.06 -0.34
education* 4 41188 4.75 2.14 4.00 4.88 2.97 1.00 8.00 7.00 -0.24 -1.21
default* 5 41188 1.21 0.41 1.00 1.14 0.00 1.00 3.00 2.00 1.44 0.07
housing* 6 41188 2.07 0.99 3.00 2.09 0.00 1.00 3.00 2.00 -0.14 -1.95
loan* 7 41188 1.33 0.72 1.00 1.16 0.00 1.00 3.00 2.00 1.82 1.38
contact* 8 41188 1.37 0.48 1.00 1.33 0.00 1.00 2.00 1.00 0.56 -1.69
month* 9 41188 5.23 2.32 5.00 5.31 2.97 1.00 10.00 9.00 -0.31 -1.03
day_of_week* 10 41188 3.00 1.40 3.00 3.01 1.48 1.00 5.00 4.00 0.01 -1.27
duration 11 41188 258.29 259.28 180.00 210.61 139.36 0.00 4918.00 4918.00 3.26 20.24
campaign 12 41188 2.57 2.77 2.00 1.99 1.48 1.00 56.00 55.00 4.76 36.97
pdays 13 41188 962.48 186.91 999.00 999.00 0.00 0.00 999.00 999.00 -4.92 22.23
previous 14 41188 0.17 0.49 0.00 0.05 0.00 0.00 7.00 7.00 3.83 20.11
poutcome* 15 41188 1.93 0.36 2.00 2.00 0.00 1.00 3.00 2.00 -0.88 3.98
emp.var.rate 16 41188 0.08 1.57 1.10 0.27 0.44 -3.40 1.40 4.80 -0.72 -1.06
cons.price.idx 17 41188 93.58 0.58 93.75 93.58 0.56 92.20 94.77 2.57 -0.23 -0.83
cons.conf.idx 18 41188 -40.50 4.63 -41.80 -40.60 6.52 -50.80 -26.90 23.90 0.30 -0.36
euribor3m 19 41188 3.62 1.73 4.86 3.81 0.16 0.63 5.04 4.41 -0.71 -1.41
nr.employed 20 41188 5167.04 72.25 5191.00 5178.43 55.00 4963.60 5228.10 264.50 -1.04 0.00
y* 21 41188 1.11 0.32 1.00 1.02 0.00 1.00 2.00 1.00 2.45 4.00
se
age 0.05
job* 0.02
marital* 0.00
education* 0.01
default* 0.00
housing* 0.00
loan* 0.00
contact* 0.00
month* 0.01
day_of_week* 0.01
duration 1.28
campaign 0.01
pdays 0.92
previous 0.00
poutcome* 0.00
emp.var.rate 0.01
cons.price.idx 0.00
cons.conf.idx 0.02
euribor3m 0.01
nr.employed 0.36
y* 0.00
查看数据是否有缺失值
> sapply(bank,anyNA)
age job marital education
FALSE FALSE FALSE FALSE
default housing loan contact
FALSE FALSE FALSE FALSE
month day_of_week duration campaign
FALSE FALSE FALSE FALSE
pdays previous poutcome emp.var.rate
FALSE FALSE FALSE FALSE
cons.price.idx cons.conf.idx euribor3m nr.employed
FALSE FALSE FALSE FALSE
y
FALSE
成功与不成功的个数
> table(bank$y)
no yes
36548 4640
在是否结婚这个属性的取值与
是否成功的数量比较
> table(bank$y,bank$marital)
divorced married single unknown
no 4136 22396 9948 68
yes 476 2532 1620 12
> xtabs(~y+marital,data=bank)
marital
y divorced married single unknown
no 4136 22396 9948 68
yes 476 2532 1620 12
> tab=table(bank$y,bank$marital)
> tab
divorced married single unknown
no 4136 22396 9948 68
yes 476 2532 1620 12
在是否结婚这个属性上的取值
> margin.table(tab,2)
divorced married single unknown
4612 24928 11568 80
> margin.table(tab,1)
no yes
36548 4640
在是否结婚这个属性上横向看概率
> prop.table(tab,1)
divorced married single unknown
no 0.113166247 0.612783189 0.272189997 0.001860567
yes 0.102586207 0.545689655 0.349137931 0.002586207
在是否结婚这个属性上纵向看概率
> prop.table(tab,2)
divorced married single unknown
no 0.8967910 0.8984275 0.8599585 0.8500000
yes 0.1032090 0.1015725 0.1400415 0.1500000
平的列联表
以第一列和第二列，展开分类group by 1,2
以col.vars 的取值进行次数统计
> ftable(bank[,c(3,4,21)],row.vars = 1:2,col.vars = "y")
y no yes
marital education
divorced basic.4y 406 83
basic.6y 169 13
basic.9y 534 31
high.school 1086 107
illiterate 1 1
professional.course 596 61
university.degree 1177 160
unknown 167 20
married basic.4y 2915 313
basic.6y 1628 139
basic.9y 3858 298
high.school 4683 475
illiterate 12 3
professional.course 2799 357
university.degree 5573 821
unknown 928 126
single basic.4y 422 31
basic.6y 301 36
basic.9y 1174 142
high.school 2702 448
illiterate 1 0
professional.course 1247 177
university.degree 3723 683
unknown 378 103
unknown basic.4y 5 1
basic.6y 6 0
basic.9y 6 2
high.school 13 1
illiterate 0 0
professional.course 6 0
university.degree 25 6
unknown 7 2
卡方检验，在p值小于2.2e-16时，拒绝原假设，认为数据不服从卡方分布
> chisq.test(tab)
Pearson's Chi-squared test
data: tab
X-squared = 122.66, df = 3, p-value < 2.2e-16
画直方图
> hist(bank$age)
> library(lattice)
画连续变量的分布，就是把直方图的中位数连接起来
以年龄为横轴，y为纵轴，数据是bank,画图，auto.key是否有图例
> densityplot(~age,groups = y,data=bank,plot.point=FALSE,auto.key = TRUE)
画Box图
> boxplot(age~y,data=bank)
双样本t分布检验，p值小于0.05时拒绝原假设
这里的原假设是两个样本没有相关性
得到的结果是p值为1.805e-06,拒绝两个样本没有相关性的假设
这里认为两个样本有相关性
> t.test(age~y,data=bank,alternative="two.sided",var.equal=FALSE)
Welch Two Sample t-test
data: age by y
t = -4.7795, df = 5258.5, p-value = 1.805e-06
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-1.4129336 -0.5909889
sample estimates:
mean in group no mean in group yes
39.91119 40.91315
数据可视化
画饼图
> tab=table(bank$marital)
> pie(tab)
画直方图
> tab=table(bank$marital)
> barplot(tab)
画下面这个图
> tab=table(bank$marital,bank$y)
> plot(tab)
画层叠直方图
> tab=table(bank$marital,bank$y)
> lattice::barchart(tab,auto.key=TRUE)
加载这个包，准备画图
> library(dplyr)
> data=group_by(bank,marital,y)
> data=tally(data)
！！！！！！！！！！！！！
> ggplot2::ggplot(data=data,mapping=aes(marital,n))+geom_bar(mapping=aes(fill=y),position="dodge",stat="identity")
数据预处理
分组之后再画图
> labels=c('青年','中年','老年')
> bank$age_group=cut(bank$age,breaks = c(0,35,55,100),right = FALSE,labels = labels)
> library(ggplot2)
> ggplot(data=bank,mapping = aes(age_group))+geom_bar(mapping = aes(fill=y),position="dodge",stat="count")
衍生变量
直接使用$符向原数据框添加新的变量
> bank$log.cons.price.idx=log(bank$cons.price.idx)
使用transform函数向原数据框添加变量
> bank<-transform(bank,log.cons.price.idx=log(cons.price.idx),log.nr.employed=log(nr.employed))
使用dplyr包里的mutate函数增加变量
> bank<-dplyr::mutate(bank,log.cons.price.idx=log(cons.price.idx))
使用dplyr包里的transmute函数只保留新生成的变量
> bank2<-dplyr::transmute(bank,log.cons.price.idx=log(cons.price.idx),log.nr.employed=log(nr.employed))
中心化
> v=1:10
> v1=v-mean(v)
> v2=scale(v,center=TRUE,scale = FALSE)
无量纲化
> V1=v/sqrt(sum(v^2)/(length(v)-1))
> v2=scale(v,center=FALSE,scale=TRUE)
根据最大最小值进行归一化
> v3=(v-min(v))/(max(v)-min(v))
进行标准正态化
> v1=(v-mean(v))/sd(v)
> v2=scale(v,center = TRUE,scale=TRUE)
Box-Cox变换
使用car包里的boxCox函数
> install.packages("car")
> library(car)
> boxCox(age~.,data=bank)
使用caret包，做Box-Cox变换
> install.packages("caret")
> library(caret)
> dat<-subset(bank,select="age")
> trans<-preProcess(dat,method=C("BoxCox"))
数据预处理下
违反常识的异常值
基于数据分布的异常值(离群点)识别
bank.dirty=read.csv("bank-dirty.csv")
summary(bank.dirty)
age job marital education
Min. : 17.00 admin. :10422 divorced: 4612 university.degree :12165
1st Qu.: 32.00 blue-collar: 9254 married :24928 high.school : 9515
Median : 38.00 technician : 6743 single :11568 basic.9y : 6043
Mean : 40.03 services : 3969 NA's : 80 professional.course: 5242
3rd Qu.: 47.00 management : 2924 basic.4y : 4175
Max. :123.00 (Other) : 7546 (Other) : 2310
NA's :2 NA's : 330 NA's : 1738
default housing loan contact month
no :32588 no :18622 no :33950 cellular :26144 may :13769
yes : 3 yes :21576 yes : 6248 telephone:15044 jul : 7174
NA's: 8597 NA's: 990 NA's: 990 aug : 6178
jun : 5318
nov : 4101
apr : 2632
(Other): 2016
day_of_week duration campaign pdays previous
fri:7827 Min. : 0.0 Min. : 1.000 Min. : 0.0 Min. :0.000
mon:8514 1st Qu.: 102.0 1st Qu.: 1.000 1st Qu.:999.0 1st Qu.:0.000
thu:8623 Median : 180.0 Median : 2.000 Median :999.0 Median :0.000
tue:8090 Mean : 258.3 Mean : 2.568 Mean :962.5 Mean :0.173
wed:8134 3rd Qu.: 319.0 3rd Qu.: 3.000 3rd Qu.:999.0 3rd Qu.:0.000
Max. :4918.0 Max. :56.000 Max. :999.0 Max. :7.000
poutcome emp.var.rate cons.price.idx cons.conf.idx
failure : 4252 Min. :-3.40000 Min. :92.20 Min. :-50.8
nonexistent:35563 1st Qu.:-1.80000 1st Qu.:93.08 1st Qu.:-42.7
success : 1373 Median : 1.10000 Median :93.75 Median :-41.8
Mean : 0.08189 Mean :93.58 Mean :-40.5
3rd Qu.: 1.40000 3rd Qu.:93.99 3rd Qu.:-36.4
Max. : 1.40000 Max. :94.77 Max. :-26.9
euribor3m nr.employed y
Min. :0.634 Min. :4964 no :36548
1st Qu.:1.344 1st Qu.:5099 yes: 4640
Median :4.857 Median :5191
Mean :3.621 Mean :5167
3rd Qu.:4.961 3rd Qu.:5228
Max. :5.045 Max. :5228
常识告诉我们，虽然123岁的老人存在，但概率也极低，也不太可能是银行的客户
找出在年龄这一列的上离群值和下离群值
|
> head(bank.dirty[order(bank.dirty$age,decreasing = TRUE),'age',drop=FALSE],n=5) age 39494 123 38453 98 38456 98 27827 95 38922 94 > tail(bank.dirty[order(bank.dirty$age,decreasing = TRUE),'age',drop=FALSE],n=5) age 37559 17 37580 17 38275 17 120 NA 156 NA |
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|
模型评估
> confusionMatrix(bank$y,pre,pos='1')
Confusion Matrix and Statistics
Reference
Prediction 0 1
0 35596 952
1 2667 1973
Accuracy : 0.9121
95% CI : (0.9094, 0.9149)
No Information Rate : 0.929
P-Value [Acc > NIR] : 1
Kappa : 0.476
Mcnemar's Test P-Value : <2e-16
Sensitivity : 0.67453
Specificity : 0.93030
Pos Pred Value : 0.42522
Neg Pred Value : 0.97395
Prevalence : 0.07102
Detection Rate : 0.04790
Detection Prevalence : 0.11265
Balanced Accuracy : 0.80241
'Positive' Class : 1
Kappa 统计量(kappa statistic)
用于评判分类器的分类结果与随机分类的差异度
用Kappa统计量评价：
较差：小于0.20
一般：0.20至0.40
稳健：0.40至0.60
好的：0.60至0.80
很好的：0.80至1.00
ROC曲线
pred<-prediction(prob,bank$y)
perf<-performance(pred,measure = "tpr",x="fpr")
plot(perf)
![]()
RandomForest
加载数据列
> data=read.table("input.txt",header = TRUE)
> str(data)
'data.frame': 222 obs. of 23 variables:
$ Acti_Profile : num 0 0 0 0 0 0 0 0 0 0 ...
$ Activity : num 1.25 0 0.938 6.562 0 ...
$ Diastolic_PTT : num 256 240 253 0 241 ...
$ Diastolic : num 73.2 78.6 74 0 78.4 ...
$ Heart_Rate_Curve : num 81.2 69.7 77.6 95 83.6 ...
$ Heart_Rate_Variability_HF: num 131 250 135 144 141 ...
$ Heart_Rate_Variability_LF: num 311 218 203 301 244 ...
$ MAP : num 86 93.5 86.9 0 91.7 ...
$ Position : num 0 0 0 1 0 0 0 0 0 0 ...
$ PTT_Raw : num 308 288 308 0 295 ...
$ RR_Interval : num 734 878 773 632 714 ...
$ Sleep_Wake : num 1 1 1 1 1 0 1 1 0 0 ...
$ SpO2 : num 0 0 99 0 98.4 ...
$ Sympatho_Vagal_Balance : num 23 8.17 14.5 20.4 16.88 ...
$ Systolic_PTT : num 308 288 307 0 295 ...
$ Systolic : num 113 124 113 0 119 ...
$ Autonomic_arousals : num 0 0 0 0 0 0 0 0 0 0 ...
$ Cardio_complex : num 0 0 0 1 0 0 0 0 0 0 ...
$ Cardio_rhythm : num 0 0 2 0 0 0 0 0 0 0 ...
$ Classification_Arousal : num 0 0 0 0 0 0 0 0 0 0 ...
$ PTT_Events : num 1 0 2 0 0 0 0 0 0 0 ...
$ Systolic_Events : num 1 0 1 0 0 0 0 0 0 0 ...
$ y : num 1 0 1 0 0 0 0 0 0 0 ...
加载随机森林包
> library(randomForest)
进行训练 以y作为因变量，其余数据作为自变量
> rf <- randomForest(y ~ ., data=data, ntree=100, proximity=TRUE,importance=TRUE)
> plot(rf)
重要性检测
衡量把一个变量的取值变为随机数，随机森林预测准确性的降低程度
> importance(rf,type=1)
%IncMSE
Acti_Profile 0.00000000
Activity 0.99353251
Diastolic_PTT 0.32193611
Diastolic 1.99891809
Heart_Rate_Curve 0.92001352
Heart_Rate_Variability_HF 2.07870722
Heart_Rate_Variability_LF -0.24957163
MAP 0.48142975
Position 1.86876751
PTT_Raw 1.94648914
RR_Interval 0.60557964
Sleep_Wake 1.00503782
SpO2 0.25396165
Sympatho_Vagal_Balance 1.42906765
Systolic_PTT 1.27965813
Systolic 0.77382673
Autonomic_arousals 0.00000000
Cardio_complex 1.00503782
Cardio_rhythm 1.14283152
Classification_Arousal -0.04383997
PTT_Events 4.63980680
Systolic_Events 33.29461169
输出随机森林的模型
|
> print(rf)
Call: randomForest(formula = y ~ ., data = data, ntree = 100, proximity = TRUE, importance = TRUE) Type of random forest: regression Number of trees: 100 No. of variables tried at each split: 7
Mean of squared residuals: 0.003226897 残差平方和SSE % Var explained: 98.7 |
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支持向量机
library(e1071)
svmfit<-svm(y~.,data=data,kernel="linear",cost=10,scale=FALSE)
> print(svmfit)
Call:
svm(formula = y ~ ., data = data, kernel = "linear", cost = 10, scale = FALSE)
Parameters:
SVM-Type: eps-regression
SVM-Kernel: linear
cost: 10
gamma: 0.04545455
epsilon: 0.1
Number of Support Vectors: 20
> plot(svmfit,data)
神经网络
> concrete<-read_excel("Concrete_Data.xls")
> str(concrete)
Classes ‘tbl_df’, ‘tbl’ and 'data.frame': 1030 obs. of 9 variables:
$ Cement : num 540 540 332 332 199 ...
$ Slag : num 0 0 142 142 132 ...
$ Ash : num 0 0 0 0 0 0 0 0 0 0 ...
$ water : num 162 162 228 228 192 228 228 228 228 228 ...
$ superplastic: num 2.5 2.5 0 0 0 0 0 0 0 0 ...
$ coarseagg : num 1040 1055 932 932 978 ...
$ fineagg : num 676 676 594 594 826 ...
$ age : num 28 28 270 365 360 90 365 28 28 28 ...
$ strength : num 80 61.9 40.3 41.1 44.3 ...
> normalize <- function(x){ return ((x-min(x))/(max(x)-min(x)))}
> concrete_norm <- as.data.frame(lapply(concrete,normalize))
> concrete_train <- concrete_norm[1:773,]
> concrete_test <- concrete_norm[774:1030,]
> library(neuralnet)
> concrete_model <- neuralnet(strength ~ Cement+Slag+Ash+water+superplastic+coarseagg+fineagg+age,data=concrete_train)
> plot(concrete_model)
model_results <- compute(concrete_model,concrete_test[1:8])
predicted_strength <- model_results$net.result
> cor(predicted_strength,concrete_test$strength)
[,1]
[1,] 0.7205120076
> concrete_model2 <- neuralnet(strength ~ Cement+Slag+Ash+water+superplastic+coarseagg+fineagg+age,data=concrete_train,hidden=5)
> plot(concrete_model2)
计算误差
|
> model_results2 <- compute(concrete_model2,concrete_test[1:8]) > predicted_strength2 <- model_results2$net.result > cor(predicted_strength2,concrete_test$strength) [,1] [1,] 0.6727155609 |
|
|
主成分分析
身高、体重、胸围、坐高
> test<-data.frame(
+ X1=c(148, 139, 160, 149, 159, 142, 153, 150, 151, 139,
+ 140, 161, 158, 140, 137, 152, 149, 145, 160, 156,
+ 151, 147, 157, 147, 157, 151, 144, 141, 139, 148),
+ X2=c(41, 34, 49, 36, 45, 31, 43, 43, 42, 31,
+ 29, 47, 49, 33, 31, 35, 47, 35, 47, 44,
+ 42, 38, 39, 30, 48, 36, 36, 30, 32, 38),
+ X3=c(72, 71, 77, 67, 80, 66, 76, 77, 77, 68,
+ 64, 78, 78, 67, 66, 73, 82, 70, 74, 78,
+ 73, 73, 68, 65, 80, 74, 68, 67, 68, 70),
+ X4=c(78, 76, 86, 79, 86, 76, 83, 79, 80, 74,
+ 74, 84, 83, 77, 73, 79, 79, 77, 87, 85,
+ 82, 78, 80, 75, 88, 80, 76, 76, 73, 78)
+ )
> test.pr<-princomp(test,cor=TRUE)
> summary(test.pr,loadings=TRUE)
Importance of components:
Comp.1 Comp.2 Comp.3 Comp.4
Standard deviation 1.8817805390 0.55980635717 0.28179594325 0.25711843909
Proportion of Variance 0.8852744993 0.07834578938 0.01985223841 0.01652747293
Cumulative Proportion 0.8852744993 0.96362028866 0.98347252707 1.00000000000
Loadings:
Comp.1 Comp.2 Comp.3 Comp.4
X1 0.497 0.543 -0.450 0.506
X2 0.515 -0.210 -0.462 -0.691
X3 0.481 -0.725 0.175 0.461
X4 0.507 0.368 0.744 -0.232
前两个主成分的累计贡献率已经达到96% 可以舍去另外两个主成分达到降维的目的
因此可以得到函数表达式 Z1=-0.497X'1-0.515X'2-0.481X'3-0.507X'4
Z2= 0.543X'1-0.210X'2-0.725X'3-0.368X'4
4.画主成分的碎石图并预测
> screeplot(test.pr,type="lines")
> p<-predict(test.pr)
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> p Comp.1 Comp.2 Comp.3 Comp.4 [1,] -0.06990949737 -0.23813701272 -0.35509247634 -0.266120139417 [2,] -1.59526339772 -0.71847399061 0.32813232022 -0.118056645885 [3,] 2.84793151061 0.38956678680 -0.09731731272 -0.279482487139 [4,] -0.75996988424 0.80604334819 -0.04945721875 -0.162949297761 [5,] 2.73966776853 0.01718087263 0.36012614873 0.358653043787 [6,] -2.10583167924 0.32284393414 0.18600422367 -0.036456083707 [7,] 1.42105591247 -0.06053164925 0.21093320662 -0.044223092351 [8,] 0.82583976981 -0.78102575640 -0.27557797533 0.057288571933 [9,] 0.93464401954 -0.58469241699 -0.08814135786 0.181037745585 [10,] -2.36463819933 -0.36532199291 0.08840476284 0.045520127461 [11,] -2.83741916086 0.34875841111 0.03310422938 -0.031146930047 [12,] 2.60851223537 0.21278727930 -0.33398036623 0.210157574387 [13,] 2.44253342081 -0.16769495893 -0.46918095412 -0.162987829937 [14,] -1.86630668724 0.05021383642 0.37720280364 -0.358821916178 [15,] -2.81347420580 -0.31790107093 -0.03291329149 -0.222035112399 [16,] -0.06392982655 0.20718447599 0.04334339948 0.703533623798 [17,] 1.55561022242 -1.70439673831 -0.33126406220 0.007551878960 [18,] -1.07392250663 -0.06763418320 0.02283648409 0.048606680158 [19,] 2.52174211878 0.97274300950 0.12164633439 -0.390667990681 [20,] 2.14072377494 0.02217881219 0.37410972458 0.129548959692 [21,] 0.79624421805 0.16307887263 0.12781269571 -0.294140762463 [22,] -0.28708320594 -0.35744666106 -0.03962115883 0.080991988802 [23,] 0.25151075072 1.25555187663 -0.55617324819 0.109068938725 [24,] -2.05706031616 0.78894493512 -0.26552109297 0.388088642937 [25,] 3.08596854773 -0.05775318018 0.62110421208 -0.218939612456 [26,] 0.16367554630 0.04317931667 0.24481850312 0.560248997030 [27,] -1.37265052598 0.02220972121 -0.23378320040 -0.257399715466 [28,] -2.16097778154 0.13733232981 0.35589738735 0.093123683044 [29,] -2.40434826507 -0.48613137190 -0.16154440788 -0.007914021222 [30,] -0.50287467640 0.14734316507 -0.20590831261 -0.122078819188 |
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