day 5 集合

# -*- coding: utf_8 _*_
# Author:Vi
#集合是无序的

list_1 = [1,2,3,2,3,5,7]
list_1 = set(list_1)#将列表转变成集合
list_2 = set([2,6,3,888,8])
print(list_1,type(list_1))#type为打印类型
print(list_1.intersection(list_2))#交集,两个都有的值取出来
print(list_1.union(list_2))#并集
print(list_1.difference(list_2))#1有2没有的,差集
list_3 = set([2,6])
print(list_3.issubset(list_2))#判断是否为子集
print(list_2.issuperset(list_3))#判断前面是否为后面父集
print(list_1.symmetric_difference(list_2))#对称差集,两个里面都没有的取出来去掉都有的
list_4 = set([3,5])
print(list_3.isdisjoint(list_4))#判断两个是否没有交集
#交集
print(list_1&list_2)

#并集
print(list_1|list_2)

#差集
print(list_1-list_2)

#对称差集
print(list_1^list_2 )

#添加
list_1.add(999)
print(list_1)

#添加多项
list_1.update([34,36,38])
print(list_1)

#删除
list_1.remove(34)
print(list_1)


print(list_1.discard("dddd"))#不报错
#随机删除
print(list_1.pop())

#查找
print(38 in list_1)
#-------------文件操作

pro = open("hellowworld",encoding='utf-8').read()#打开文件后面要加encoding...因为默认是gbk
print(pro)
f = open("DIO",'w',encoding='utf-8')#文件句柄

f.write("不老不死,\n")
f.write("替身之力!")#这就是写,上面加w

f.close()


'''
f = open("hellowworld2","w",encoding="utf-8")
f.write("明明是两倍的快乐,\n")
f.write("it would be twice happiness")#写内容
'''
#f = open("hellowworld","a",encoding="utf-8")#append可以追加,不能读;
#f.write("\n明明是两倍的快乐,\n")
#f.write("it would be twice happiness")
f = open("hellowworld2","r",encoding='utf-8')
#for i in range(3):
# print(f.readline())#读一行
#for line in f.readlines():
# print(line.strip())#去空格换行,打印每一个readlines里面的每行
for index,line in enumerate(f.readlines()):
if index == 2:
print('-------------')
continue
print(line)
posted @ 2019-04-28 21:02  arnold2029  阅读(121)  评论(0)    收藏  举报