python

set 集合  {}  无序

s = {1,2,3,4,5}

s = {}
print(type(s))   # 空{}就是字典

s = {1,2,3,4,5}
s = {1,2,3,'22','ss',False,(3,4)}
列表 和 字典 和 集合
print(s)

增
s.add(8)
print(s)

s.update('alex')   # 迭代添加
print(s)

删
s = {1,2,3,'22','ss',(3,4)}

s1 = s.pop()    #随机删除
print(s)    #返回值  返回的被删除的元素
print(s1)

s.clear()
print(s)      #清空 set()

s = {1,2,3,'22','ss',(3,4)}
del s           #删除整个集合
print(s)

s = {1,2,3,'22','ss',(3,4)}
s.remove()    #通过元素指定删除
print(s)

改:
s = {1,2,3,'22','ss',(3,4)}
s.remove(3)
s.add(7)
print(s)

查:

s = {1,2,3,'22','ss',(3,4)}
for i in s:
    print(i)

s = {1,2,3,4,33,21,4,1234,34,1234,5324,1234,12341,23,22,34112,2,21,23,34,123,4}
集合天然去重
print(s)

面试题:
lst = [1,2,3,42,3,45,67,534,2]
print(list(set(lst)))

s = set(lst)
print(list(s))

# 其他操作:

s = {'黑哥'}
s1 = {'虎哥','开哥','alex哥','黑哥','wu哥'}

print(s & s1)  #  交集  俩个集合都有的才输出

print(s | s1)    #   并集  两个集合 合并到一起

print(s1 - s)       # 差集   前面减后边的集合前边集合中有的输出出来   自己总结

print(s ^ s1)         # 反交集  自己总结

print(s < s1)     # 子集
print(s1 > s)     # 超(父)集

dic = {{1,2,3}:1}
a = frozenset((1,2,3))
dic ={a:1}
print(dic)
print(dic)

lst = [1,2,3,[4,5,6],7]
lst1 = lst    #赋值
lst2 = lst[:] # 切片
#
lst.append(7)
print(lst)
print(lst2)

import copy

lst4 = copy.deepcopy(lst)  # 深拷贝
lst[-2].append(7)
print(lst)   #[1, 2, 3, [4, 5, 6,7], 7]
print(lst4)  #[1, 2, 3, [4, 5, 6], 7]

浅拷贝:

    拷贝的是第一层元素

深拷贝:

    拷贝所有元素开辟新的空间存放,完全没有关系

lst  = [1,2,3]
lst1 = [1,2,3]

print(id(lst))
print(id(lst1))

面试题
a = [1, 2]
a[1] = a  # [1,[1,2]]
print(a[1])

答案:
[1, [...]]

lst = [1,3,[5,7],9]
lst1 = lst
lst2 = lst[:]
# print(id(lst))
# print(id(lst1))
# print(id(lst2))
lst.append(9)
print(lst)
print(lst1)
print(lst2)

是不是一个东西  长的一不一样

lst = [1,2,3,4,5]

for i in lst:

    lst.remove(i)

print(lst)
删除不行

lst = [1,2,3,4,5]
for i in lst:
    lst.pop()
print(lst)
不行

lst = [1,2,3,4,5]
for i in lst:
    lst.clear()
print(lst)
行,作弊

lst = [1,2,3,4,5]
for i in lst:
    del lst[0]
print(lst)
不行

lst = [1,2,3,4,5]
for i in range(len(lst)):
    lst.pop(0)

print(lst)
这个行

lst = [1,2,3,4,5]
for i in range(len(lst)):
    lst.pop()

print(lst)
这个行

lst = [1,2,3,4,5]
for i in lst:
    if type(i) == int:
        lst[i-1] = ' '
print(lst)
print(list(set(lst)))

lst = [1,2,3,4,5]
lst1 = []

for i in lst:
    lst1.append(i)
for k in lst1:
    lst.remove(k)
print(lst)
这个方式也可以

for i in dic:
    dic['c'] = 3

print(dic)

for i in dic:
    dic.pop(i)

print(dic)

lst = {}
dic ={'a':1,'b':2}
for i in dic:
    lst[i]= dic[i]

print(lst)

for i in lst:
    dic.pop(i)
print(dic)

################################深坑###################################

dic = dict.fromkeys(['alex','wusir','wenzhou','nvshen','kaige'],['太白','姗姗'])
fromkeys 第一是键  第二是值
print(dic)

dic['alex'].append('老村长')
print(dic)

ascii  # 不支持中文
unicode  万国码   中文4个字节  英文 2个
utf-8    最少使用1个字节   欧洲 2  亚洲 3
gbk     中文 2 英 1
windows  默认编码GBK

a = '你好黑'
a1 = 'alex'

a2 = a.encode('utf-8')   #编码
print(a)
print(a2)
a3 = a2.decode('utf-8')  #解码
print(a3)
bytes字节   用于传输和存储

a = '你好黑'
a1 = 'alex'
a2 = a.encode('gbk')  #编码
a3 = a.encode('utf-8')   # 最长用
print(a2)
print(a3)
a4 = a2.decode('gbk') #解码
a5 = a3.decode('utf-8')
print(a4,a5)


######坑########

lst = [1,2,3,4,5,[3,4,6,8],5]
lst1 = lst
lst1 = lst[:]   # 浅拷贝只拷贝第一层元素
print(id(lst))
print(id(lst1))
lst.append(6)
print(lst)
print(lst1)
lst[-2].append(9)
print(lst)
print(lst1)

dic = {'a':1,'b':1}
dic1 = dic
print(id(dic1))
print(id(dic))

import copy
lst = [1,2,3,4,5,[3,4,6,8],5]
lst1 = copy.deepcopy(lst)
print(lst)
print(lst1)
lst[-2].append(10)
print(lst)
print(lst1)

巴特西

python_6

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