按功能对列表进行排序会导致错误的结果 [英] Sorting a list by function leads to wrong result

问题描述

在函数中将列表作为参数传递时，为什么以下列表没有更改?

When passing a list as an argument in function, why is the following list not changed?

def foo(*x):
    y=sorted(x)
    print(y)

a=[3,2,1]

该函数返回[[3, 2, 1]]，而不是[[1,2,3]].为什么会这样呢?这与按值调用有关吗?

The function is returning [[3, 2, 1]], not [[1,2,3]]. Why is this happening? Is this something to do with call by value?

推荐答案

因为函数的参数指定为* a，这就像说您的参数是未定义维的元组

because the argument of your function is specified as *a, which is like saying your argument is a tuple of undefined dimension

当您尝试对带有嵌套列表的元组进行排序时，该值不会更改

when you try to sort a tuple with a nested list, the value will not change

事实上，您获得了列表列表(您获得了[[3，2，1]]而不是[3，2，1])

infact as result you got a list of list (you got [[3, 2, 1]] not [3, 2, 1])

如果您尝试这样做，它将起作用

if you try this, it will work

def foo(*x): 
    y=sorted(x[0]) 
    print(y)

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