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  • set,三元,拷贝

    # l1 = [11,22,33,44,11]
    # a = set(l1)
    # a.add(55)
    # a.clear()
    # print(a)
    #
    # l2 = {22,77,88}
    # l3 = {11,22,33,44}
    # b = l3.difference(l2)
    # print(b)#a存在b不存在,并赋值给c
    # l3.difference_update(l2)#更新a,返回none
    # print(l3)
    # l3.discard(11)#不报错
    # l3.remove(44)
    # print(l3)
    # print(l2)
    # l4 = l3.intersection(l2)#取交集,并赋值,updata是更新自己
    # z = l3.isdisjoint(l2)#判断是否有交集,没交集返回True
    # x = l3.issubset(l2)#判断是否是子集
    # c = l3.issuperset(l2)#父集合
    # print(z,x,c)
    # ret = l2.pop()#移除最后一个,不报错,remove报错
    # print(ret,l2)
    # q = {11,22,33,44}
    # w = {33,44}
    # e = q.symmetric_difference(w)#对称交集,a存在b不存在,b存在a不存在的,赋值给e,updata是更新a
    # print(e)
    # r = q.union(w)#并
    # print(r)
    # w.update([55,66])#更新,后接可迭代
    # print(w)
    # def set_1():
    # old_dict = {
    # "#1": 11,
    # "#2": 22,
    # "#3": 100,
    # }
    # new_dict = {
    # "#1": 33,
    # "#4": 22,
    # "#7": 100,
    # }
    # old_set = set(old_dict.keys())
    # new_set = set(new_dict.keys())
    # del_key = old_set.difference(new_set)
    # add_key = new_set.difference(old_set)
    # up_key = old_set.intersection(new_set)
    # print(del_key,add_key,up_key)
    # for up_keys in up_key:
    # old_dict[up_keys] = new_dict[up_keys]
    # print(old_dict)
    # for add_keys in add_key:
    # old_dict[add_keys] = new_dict[add_keys]
    # print(old_dict)
    # for del_keys in del_key:
    # del old_dict[del_keys]
    # print(old_dict)
    # name ="ciahsd" if 1==2 else "asd"#三元运算
    # print(name)
    # str一次性创建,不能被修改,如果修改只能再创建
    # list链表,下一个元素的位置,上一个元素的位置
    import copy
    # n1 = 123
    # print(id(n1))
    # n2 = n1
    #由于python内部对数字和字符串的优化,无论如何拷贝赋值,字符串和数字的地址在内存中都不变
    #对于列表,元组,字典,集合,其他,浅拷贝,copy.cpoy()只拷贝第一层,内存地址改变
    #深拷贝除最后一层数字字符串,其他层都拷贝,内存地址都要改变

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  • 原文地址:https://www.cnblogs.com/currynashinians000/p/8590511.html
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