在循环中将字典添加到列表的末尾

Question

我正在尝试将一个字典添加到列表中。然后在循环中更改字典的值，再次将其附加到列表中。似乎每次这样做时，列表中的所有字典都会更改它们的值以匹配刚刚附加的那个字典。 例如：

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

我原本认为结果会是 [{1:1}, {1:2}, {1:3}... {1:98}, {1:99}]，但实际上得到的却是：

[{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}]

- Acoop

---

你正在用相同的引用填充“列表”（不应该命名为“列表”）和“字典”（不应该称为“字典”）。 - jonrsharpe

Answer 1

你需要追加一个副本，否则你只是不断地添加对同一字典的引用：

yourlist.append(yourdict.copy())

我使用了yourdict和yourlist而不是dict和list；你不想掩盖内置类型。

- Martijn Pieters

Answer 2

当你在循环外创建adict字典时，你将相同的字典附加到alist列表中。这意味着所有的副本都指向同一个字典，每次都会得到最后一个值{1:99}。只需在循环内部创建每个字典，现在你有了100个不同的字典。

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

- chapelo

Answer 3

只需要把 dict = {} 放在循环里面。

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

>>> print list

- docjag

Answer 4

您可以使用 zip 和列表推导式来完成您需要的操作。

如果您希望字典值从1开始，请使用 range(1,100)。

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

- Padraic Cunningham

Answer 5

假设d是您的字典。在这里，如果您执行d.copy()。它会返回浅层副本，当您将嵌套字典放入d字典中时，它不起作用。为了克服这个问题，我们必须使用deepcopy。

from copy import deepcopy
list.append(deepcopy(d))

它完美地工作了！！！

- iNikkz