python - 为什么这段代码会失败？在for循环中将变量分配给自身列表

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'，但应为 ['hello']。但是，这有效:

a = 'hello'
a = [a]
a

返回 ['hello']。

Answer 1

要实现这一点，首先您必须了解您有两个不同的引用，a 和 x(针对每个元素)，以及列表 [a,b,c] 的引用，仅在 for 循环中使用，并且再也不会了。

为了实现您的目标，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

['hello']

但正如我所说，如果您执行 print(a) 它将再次显示:

'hello' #here i printed the value of the variable a, wich has no modification

因为你从来没有对它做过任何事情。

看看这个问题，了解更多关于引用文献的信息 How do I pass a variable by reference?