为什么'remove'在本地调用时会改变列表？

Question

可能重复:
了解Python的传递函数参数的逐个调用样式

我最近遇到过这个:

x = [1,2,3]

def change_1(x):
    x = x.remove(x[0])
    return x

结果如下:

>>> change_1(x)
>>> x
[2, 3]

我发现这种行为令人惊讶,因为我认为函数内部的任何内容都不会对外部变量产生影响.此外,我构建了一个示例,基本上做同样的事情,但没有使用remove:

x = [1,2,3]

def change_2(x):
    x = x[1:]
    return x

结果如下:

>>> change_2(x)
[2, 3] # Also the output prints out here not sure why this is
>>> x
[1, 2, 3]

我得到了我期望的结果,该函数不会改变x.

所以它必须是remove具有特定效果的东西.这里发生了什么？

Answer 1

其中一个令人困惑的事情是,你已经调用了很多不同的东西'x'.例如:

def change_1(x):       # the x parameter is a reference to the global 'x' list
    x = x.remove(x[0]) # on the left, though, is a new 'x' that is local to the function
    return x           # the local x is returned

>>> x = [1, 2, 3]
>>> y = change_1(x)    # assign the return value to 'y'
>>> print y
None                   # this is None because x.remove() assigned None to the local 'x' inside the function
>>> print x
[2, 3]                 # but x.remove() modified the global x inside the function

def change_2(x):
    x = x[1:]          # again, x on left is local, it gets a copy of the slice, but the 'x' parameter is not changed
    return x           # return the slice (copy)

>>> x = [1, 2, 3] 
>>> y = change_2(x)
>>> print x
[1, 2, 3]             # the global 'x' is not changed!
>>> print y
[2, 3]                # but the slice created in the function is assigned to 'y'