使用 += 但未附加列表时出现 UnboundLocalError

问题描述

我不太明白下面两个类似代码的区别:

I do not quite understand the difference between the following two similar codes:

def y(x):
    temp=[]
    def z(j):
        temp.append(j)
    z(1)
    return temp

调用 y(2) 返回 [1]

def y(x):
    temp=[]
    def z(j):
        temp+=[j]
    z(1)
    return temp

调用 y(2) 返回 UnboundLocalError: local variable 'temp' referenced before assignment.为什么 + 运算符会产生错误?谢谢

calling y(2) returns UnboundLocalError: local variable 'temp' referenced before assignment. Why + operator generates the error? Thanks

推荐答案

回答标题，+和"append"的区别是:

Answer to the heading, the difference between + and "append" is:

[11, 22] + [33, 44,] 

会给你:

[11, 22, 33, 44]

和.

b = [11, 22, 33]
b.append([44, 55, 66]) 

会给你

[11, 22, 33 [44, 55, 66]] 

错误答案

这是因为当您对作用域中的变量进行赋值时，该变量将成为该作用域的本地变量，并隐藏外部作用域中任何类似命名的变量

This is because when you make an assignment to a variable in a scope, that variable becomes local to that scope and shadows any similarly named variable in the outer scope

这里的问题是 temp+=[j] 等于 temp = temp +[j].临时变量在分配之前在此处读取.这就是它给出这个问题的原因.这实际上包含在 python 常见问题解答中.

The problem here is temp+=[j] is equal to temp = temp +[j]. The temp variable is read here before its assigned. This is why it's giving this problem. This is actually covered in python FAQ's.

如需进一步阅读，请单击此处.:)

For further readings, click here. :)

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