Question

a = 'hello'
b = None
c = None
for x in [a, b, c]:
    if isinstance(x, (basestring, dict, int, float)) or x is None:
        x = [x]
a

返回'hello'，但预期['hello']。但是，这有效：

a = 'hello'
a = [a]
a

返回['hello']

Answer 1

要实现这一点，首先你必须明白你有两个不同的结果，a和x（对于每个元素），以及列表[a，b，c]的引用，仅在for循环中使用，以及永远不会更多。

要实现目标，您可以这样做：

a = 'hello'
b = None
c = None
lst = [a, b, c] #here I create a reference for a list containing the three references above
for i,x in enumerate(lst):
    if isinstance(x, (str,dict, int, float)) or x is None: #here I used str instead basestring
        lst[i] = [x]

print(lst[0]) #here I print the reference inside the list, that's why the value is now showed modified

[ '你好']

但正如我所说，如果你打印（a）它会再次显示：

'你好'＃我打印了变量a的值，没有修改

因为你从未做过任何事情。

请查看此问题，了解有关参考How do I pass a variable by reference?

的更多信息

Answer 2

让我们一次完成这一行;

a = 'hello'  # assigns 'hello' to a.
b = None  # etc...
c = None
for x in [a, b, c]:  # Creates list from the values in 'a' 'b' and 'c', then iterates over them.
    # This if statement is always True. x is always either a string, or None.
    if isinstance(x, (basestring, dict, int, float)) or x is None:
        x = [x]  # On the first iteration, sets variable to the expected ['hello']
        # After that though - it replaces it with [None]
a  # = 'hello' - you have not assigned anything to the variable except on the first line.

唯一设置为['hello']的变量是x，它会被None快速覆盖。如果您更改了if检查以排除or x is None并分配到a而不是x，那么您将获得所需的结果。

还值得注意的是，启动for循环时会创建列表[a, b, c]。在for循环期间更改a b或c将不起作用 - 列表已经创建。

Answer 3

其他答案很棒。我认为，概括来说，list元素是不可变的/可散列的，因此for循环返回一个副本，而不是对原始对象的引用。我应该发现这一点，但教我编码太长时间没有睡觉！

我最终使用了这个：

def lst(x):
    if isinstance(x, (basestring, dict, int, float)) or x is None:
        x = [x]
    return x
[a, b, c] = map(lst, [a, b, c])

Python：为什么这段代码失败了？将变量分配给for循环中的自身列表

3 个答案: