速战速决 Python - python 第三方库（pandas）: DataFrame连接，去重，空值处理
速战速决 Python https://github.com/webabcd/PythonSample
作者 webabcd
速战速决 Python - python 第三方库（pandas）: DataFrame连接，去重，空值处理
示例如下:
thirdLib/pandas/sample3.py
# pandas
# DataFrame 连接：append(), concat(), merge(), join()
# DataFrame 去重
# DataFrame 空值处理
import pandas as pd
data1 = {
"name": ["zhao", "qian"],
"age": [40, 25],
"city": ["beijing ", "beijing"],
"gender": ["M", "F"]
}
a = pd.DataFrame(data=data1)
data2 = {
"name": ["qian", "sun"],
"age": [25, 22],
"city": ["beijing", "shanghai"],
"salary": [100, 150]
}
b = pd.DataFrame(data=data2)
print(a)
'''
name age city gender
0 zhao 40 beijing M
1 qian 25 beijing F
'''
print(b)
'''
name age city salary
0 qian 25 beijing 100
1 sun 22 shanghai 150
'''
# append() 在 DataFrame 数据的末尾追加指定的 DataFrame 数据
print(a.append(b))
'''
name age city gender salary
0 zhao 40 beijing M NaN
1 qian 25 beijing F NaN
0 qian 25 beijing NaN 100.0
1 sun 22 shanghai NaN 150.0
'''
# drop_duplicates() 去重
# subset - 根据指定的字段去重
# keep - first保留第一条重复数据，last保留最后一条重复数据，False删除全部重复数据
# inplace - 是否直接修改原对象
# False 默认值，原对象不变，返回数据修改后的副本
# True 直接修改原对象，返回值为 None
print(a.append(b).drop_duplicates(subset=['age','city'], keep='first', inplace=False))
'''
name age city gender salary
0 zhao 40 beijing M NaN
1 qian 25 beijing F NaN
1 sun 22 shanghai NaN 150.0
'''
# concat() 拼接多个 DataFrame 数据
print(pd.concat([a, b]))
'''
name age city gender salary
0 zhao 40 beijing M NaN
1 qian 25 beijing F NaN
0 qian 25 beijing NaN 100.0
1 sun 22 shanghai NaN 150.0
'''
# concat() 拼接多个 DataFrame 数据
# ignore_index=True 重建索引
print(pd.concat([a, b], ignore_index=True))
'''
name age city gender salary
0 zhao 40 beijing M NaN
1 qian 25 beijing F NaN
2 qian 25 beijing NaN 100.0
3 sun 22 shanghai NaN 150.0
'''
# concat() 拼接多个 DataFrame 数据
# join='outer' 默认值，各方列名不相同的也集成到一起
# join='inner' 各方列名不相同的就舍弃，只保留列名相同的数据
print(pd.concat([a, b], ignore_index=True, join='inner'))
'''
name age city
0 zhao 40 beijing
1 qian 25 beijing
2 qian 25 beijing
3 sun 22 shanghai
'''
# concat() 拼接多个 DataFrame 数据
# axis=1 横向拼接
print(pd.concat([a, b], axis=1))
'''
name age city gender name age city salary
0 zhao 40 beijing M qian 25 beijing 100
1 qian 25 beijing F sun 22 shanghai 150
'''
# merge() 整合 2 个 DataFrame 数据
# on='name' 以 name 列为关联关键字整 2 个 DataFrame 数据
# 如果 2 个 DataFrame 数据的关联关键字不相同，则类似这么设置 left_on="name1", right_on="name2"
print(pd.merge(a, b, on='name'))
'''
name age_x city_x gender age_y city_y salary
0 qian 25 beijing F 25 beijing 100
'''
# merge() 整合 2 个 DataFrame 数据
# how='inner' 默认值，只整合两边都存在的数据
# how='outer' 整合两边的全部数据
# how='left' 只整合左边存在的数据
# how='right' 只整合右边存在的数据
print(pd.merge(a, b, on='name', how='outer'))
'''
name age_x city_x gender age_y city_y salary
0 zhao 40.0 beijing M NaN NaN NaN
1 qian 25.0 beijing F 25.0 beijing 100.0
2 sun NaN NaN NaN 22.0 shanghai 150.0
'''
# merge() 整合 2 个 DataFrame 数据
# suffixes - 相同列名整合时，为列名加上指定的后缀，以便区分列是来自哪个 DataFrame 的
print(pd.merge(a, b, on='name', how='outer', suffixes=("_left", "_right")))
'''
name age_left city_left gender age_right city_right salary
0 zhao 40.0 beijing M NaN NaN NaN
1 qian 25.0 beijing F 25.0 beijing 100.0
2 sun NaN NaN NaN 22.0 shanghai 150.0
'''
# join() 整合 2 个 DataFrame 数据
# 相当于 how='left' 的 merge()
print(a.join(b.set_index("name"), on="name", lsuffix="_left", rsuffix='_right'))
'''
name age_left city_left gender age_right city_right salary
0 zhao 40 beijing M NaN NaN NaN
1 qian 25 beijing F 25.0 beijing 100.0
'''
# 以下用于说明空值如何处理
index3 = ['zhao', 'qian', 'sun', 'li', 'zhou']
data3 = {
"age": [None, None, 22, 28, 28],
"city": ['beijing', 'beijing', None, 'beijing', 'shanghai']
}
c = pd.DataFrame(data=data3, index=index3)
print(c)
'''
NaN 数字类型的空值（来自 numpy 的 nan），None 其他类型的空值
age city
zhao NaN beijing
qian NaN beijing
sun 22.0 None
li 28.0 beijing
zhou 28.0 shanghai
'''
# isnull() 判断数据是否有空值
# notnull() 判断数据是否没有空值
print(c.isnull())
'''
age city
zhao True False
qian True False
sun False True
li False False
zhou False False
'''
# 获取指定列没有空值的数据
print(c[c.age.notnull()])
'''
age city
sun 22.0 None
li 28.0 beijing
zhou 28.0 shanghai
'''
# dropna() 删除空值数据
# how="any" 有一个字段空值，则整行删除
# how="all" 所有字段都空值，才整行删除
# subset=["column1", "column2"] 只从指定的列中查询
print(c.dropna(how="any"))
'''
age city
li 28.0 beijing
zhou 28.0 shanghai
'''
print(c.dropna(how="any", subset=["age"]))
'''
age city
sun 22.0 None
li 28.0 beijing
zhou 28.0 shanghai
'''
# 为指定的字段中的空值填充一个指定的值
# inplace=False 默认值，原对象不变，返回数据修改后的副本
# inplace=True 直接修改原对象，返回值为 None
print(c.fillna({'age':0,'city':'unknown'}, inplace=True)) # None
print(c)
'''
age city
zhao 0.0 beijing
qian 0.0 beijing
sun 22.0 unknown
li 28.0 beijing
zhou 28.0 shanghai
'''
速战速决 Python https://github.com/webabcd/PythonSample
作者 webabcd
