集合

s = {1, 2, 3, 4, 5, 6}
s.add(7)
# s.add(3)#集合中只能有一个3
print(s)

s = {1, 2, 3, 4, 5, 6}
s.clear()
print(s)

s = {1, 2, 3, 4, 5, 6}
s1 = s.copy()
print(s1)
# 删除  pop    随机删除
s = {1, 2, 3, 4, 5, 6}
s.pop()
print(s)

# 删除 remove    指定删除
# 删除元素不存在会报错
s = {1, 2, 3, 4, 5, 6}
s.remove(6)
print(s)

# 删除  discard
# 删除元素不存在不会报错
s = {1, 2, 3, 4, 5, 6}
s.discard(7)
print(s)
a = []
python_l = ["gb", "cmq", "zxt"]
linux_l = ["gb", "cmq"]
for i in python_l:
    for x in linux_l:
        if i == x:
            a.append(i)
print(a)

python_l = ["gb", "cmq", "zxt"]
linux_l = ["gb", "cmq"]
# 集合可以用来去重
i = set(python_l)
y = set(linux_l)

print(i, y)
print(i.intersection(y))  # intersection 是求交集

python_l = ["gb", "cmq", "zxt"]
linux_l = ["gb", "cmq", "cy"]
i = set(python_l)
y = set(linux_l)
print(i.union(y))
print(i | y)  # union是求并集

# 差集
python_l = ["gb", "cmq", "zxt"]
linux_l = ["gb", "cmq", "cy"]
i = set(python_l)
y = set(linux_l)
print("差集", i - y)
print(i.difference(y))
print("差集", y - i)
print(y.difference(i))

# 交叉补集
python_l = ["gb", "cmq", "zxt"]
linux_l = ["gb", "cmq", "cy"]
i = set(python_l)
y = set(linux_l)
print("交叉补集", i.symmetric_difference(y))
print(i ^ y)

sl = ["gb", "cmq", "zxt"]
s2 = ["gb", "cmq"]
print(s1.issubset(s2))  # s1是s2的子集
print(s2.issubset(s1))  # False
print(s2.issuperset(s1))  # s1是s2的父集

sl = ["gb", "cmq", "zxt"]
s2 = ["gb", "cmq"]
s1.update(s2)  # 更新多个值
# s1.add(1,2,3)只能更新一个值
# s1.union(s2) 不更新
print(s1)
s1.update((3, 4))
print(s1)

 

posted @ 2019-03-06 09:35  点燃  阅读(111)  评论(0)    收藏  举报