1. 程式人生 > >python字典fromkeys()方法中的坑

python字典fromkeys()方法中的坑

form tip 本質 pre 聯動 字典 range end pen

  • 自定操作中的fromkeys()方法接收兩個參數,第一個參數為一個可叠代對象,作為返回字典的key,第二個參數為value,默認為None,具體用法如下:

    li = [1,2,3]
    dic1 = dict.fromkeys(li)
    dic2 = dict.fromkeys(li,[])
    print(dic1)    # {1: None, 2: None, 3: None}
    print(dic2)    # {1: [], 2: [], 3: []}
  • 此時我為dic2中key為1的列表增加一個元素‘test’,如下:

    dic2[1].append(‘test‘)
    print(dic2)      # {1: [‘test‘], 2: [‘test‘], 3: [‘test‘]}
  • 竟然把三個列表的值都給改了,這是為啥呢?先打印下他們的內存地址

    print("dic2[1]:{}\ndic2[2]:{}\ndic2[3]:{}".format(id(dic2[1]),id(dic2[2]),id(dic2[3])))
    # dic2[1]:1714986428808
    # dic2[2]:1714986428808
    # dic2[3]:1714986428808
  • 原來它的所有鍵都指向了同一個內存地址,這也就不難怪修改其中一個而引發聯動了,因為本質上只有一個列表。因此,在字典中定義不同的列表不要用fromkeys方法,還是老老實實定義吧

    dic2 =  {1: [], 2: [], 3: []}
    print(id(dic2[1]))  # 1657985662344
    print(id(dic2[2]))  # 1657986500680
    print(id(dic2[3]))  # 1657986501960
    dic2[1].append(‘test‘)
    print(dic2)
    # {1: [‘test‘], 2: [], 3: []}
  • 使用循環來產生多key的字典:

    dic2 = {}
    for k in range(10):
    dic2[k] = []
    print(dic2)   # {0: [], 1: [], 2: [], 3: [], 4: [], 5: [], 6: [], 7: [], 8: [], 9: []}
  • tips: fromkeys方法會返回一個新的字典,對原字典無影響
    dic1 = {1:2}
    dic2 = dic1.fromkeys([1,2,3],‘test‘)
    print(dic1)    #  {1: 2}
    print(dic2)   # {1: ‘test‘, 2: ‘test‘, 3: ‘test‘}
  • python字典fromkeys()方法中的坑