将字典附加到循环Python中的列表中

Question

我是一个基本的python程序员,所以希望我的问题的答案很容易.我正在尝试拿一本字典并将其附加到列表中.然后字典更改值,然后在循环中再次附加.似乎每次执行此操作时,列表中的所有字典都会更改其值以匹配刚刚附加的字典.例如:

>>> dict = {}
>>> list = []
>>> for x in range(0,100):
...     dict[1] = x
...     list.append(dict)
... 
>>> print list

我会假设结果会是,[{1:1}, {1:2}, {1:3}... {1:98}, {1:99}]但我得到了:

[{1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}, {1: 99}]

任何帮助是极大的赞赏.

Answer 1

您需要附加一个副本,否则您只是一遍又一遍地添加对同一个字典的引用:

yourlist.append(yourdict.copy())

我使用yourdict而yourlist不是dict和list; 你不想掩盖内置类型.

Answer 2

当您adict在循环之外创建字典时，您会将相同的字典附加到您的alist列表中。这意味着所有副本都指向同一个字典，并且{1:99}每次都获得最后一个值。只需在循环中创建每个字典，现在您就有了 100 个不同的字典。

alist = []
for x in range(100):
    adict = {1:x}
    alist.append(adict)
print(alist)

Answer 3

只需放入dict = {}循环内即可。

>>> dict = {}
>>> list = []
>>> for x in range(0, 100):
       dict[1] = x
       list.append(dict)
       dict = {}

>>> print list

Answer 4

您还可以使用zip和列表理解来执行您需要的操作。

如果您希望 dict 值从一次使用开始 range(1,100)

l = [dict(zip([1],[x])) for x in range(1,100)]