Question

我具有嵌套函数，这些函数旨在对列表列表的索引进行分区，这些列表对应于相同的元素（列表）。

我的第一次尝试不是从内部函数返回所需的列表。我认为这是因为仅在函数内部创建了输出列表。所以我修改了脚本。现在，我从外部定义我的输出列表，如titissuesets。它传递给内部函数并由内部函数修改。但是，正确修改的列表并未从ind_list_renew函数传递给find_like_tissues_set函数！

#!/usr/bin/env python

import numpy as np

l1 = [1,2,3]

l2=[2,3]

l3=[1,2,3]

l4=[2,3,4]

l5=[2,3]

mylist = [l1,l2,l3,l4,l5]
liketissuesets = []

def listidentity(v,b,f):
    if len(v) != len(b):
        return []
    else:
        for j in range(len(b)):
            if v[j]!=b[j]:
                return []
            else:
                return [f]


def ind_list_renew(changinglist, liketissuesets):
    a=changinglist[0]
    b=mylist[a]
    common = []

    for f,v in enumerate(mylist):
        common = common + listidentity(v,b,f)

    print(common)
    liketissuesets = liketissuesets + [common]
    print(liketissuesets)
    changinglist = changinglist.tolist()
    indtodelete = [j for j,k in enumerate(changinglist) if k in common]
    changinglist = np.delete(changinglist, indtodelete)

    if len(changinglist) != 0:
        ind_list_renew(changinglist, liketissuesets)
    else:
        print('yay', liketissuesets)
        return liketissuesets


def find_like_tissues_set(mylist, liketissuesets):

    indoriginal = np.arange(len(mylist))
    c=ind_list_renew(indoriginal, liketissuesets)
    print(c)
    return c

b=find_like_tissues_set(mylist, liketissuesets)

print(b)

Answer 1

确实，您不需要全局liketissuesets。您可以像这样重组代码：

def ind_list_renew(changinglist, liketissuesets):
    # stuff omitted

    if len(changinglist) != 0:
        return ind_list_renew(changinglist, liketissuesets)
    else:
        print('yay', liketissuesets)
        return liketissuesets


def find_like_tissues_set(mylist):
    indoriginal = np.arange(len(mylist))
    c = ind_list_renew(indoriginal, [])
    print(c)
    return c

b = find_like_tissues_set(mylist)

print(b)

函数不返回外部定义的变量

1 个答案: