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
解决方案
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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