Why can't Python reference passing be sorted

Question

import os
import tqdm
list_name = []
path_dir = ['sofile_ads_og_pt']
path_so = "/Users/PycharmProjects/"

def listdir(path,list_name):  # 传入存储的list    
    for file in os.listdir(path):
        filecheck = file.split('.')
        if filecheck[0] == 'so' and filecheck[len(filecheck)-1] == 'log':
            file_path = os.path.join(path, file)
            # if os.path.isdir(file_path):
            #     listdir(file_path, list_name)
            # else:
            #     list_name.append(file_path)
        list_name.append(file_path)
    print(list_name)

    list_name = sorted(list_name, key=lambda x: (int(x.split('.')[1])))
    print(list_name)
    print("载入文件完成：",len(list_name))

if __name__ == '__main__':

    for dirname in (path_dir):
        list_name.clear()
        path_so_all = path_so + dirname
        listdir(path_so_all,list_name)
        print("main部分：")
        print(list_name)
        list_name = sorted(list_name, key=lambda x: (int(x.split('.')[1])))
        print(list_name)
        print(path_so_all)


    import os
    import tqdm
    list_name = []
    path_dir = ['sofile_ads_og_pt']
    path_so = "/Users/PycharmProjects/"
    def listdir(path,list_name):  # 传入存储的list
    
        for file in os.listdir(path):
            filecheck = file.split('.')
            if filecheck[0] == 'so' and filecheck[len(filecheck)-1] == 'log':
                file_path = os.path.join(path, file)
                # if os.path.isdir(file_path):
                #     listdir(file_path, list_name)
                # else:
                #     list_name.append(file_path)
            list_name.append(file_path)
        print(list_name)
    
        list_name = sorted(list_name, key=lambda x: (int(x.split('.')[1])))
        print(list_name)
        print("载入文件完成：",len(list_name))
    
    if __name__ == '__main__':
    
        for dirname in (path_dir):
            list_name.clear()
            path_so_all = path_so + dirname
            listdir(path_so_all,list_name)
            print("main 部分：")
            print(list_name)
            list_name = sorted(list_name, key=lambda x: (int(x.split('.')[1])))
            print(list_name)
            print(path_so_all)

output:

listdir funtion:

list_name = [not sorted]

list_name = [sorted]

main funtion:

list_name = [not sorted]

list_name = [sorted]

why the main part of list_name is not sorted ???

And if the list_name is local variable, why global list_name have items ???

Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. — Community, Jun 15 '22 at 12:33
Does this answer your question? [How do I clone a list so that it doesn't change unexpectedly after assignment?](https://stackoverflow.com/questions/2612802/how-do-i-clone-a-list-so-that-it-doesnt-change-unexpectedly-after-assignment) — Iguananaut, Jun 15 '22 at 12:36
To answer the question in the title: because Python does not perform reference passing. Values are always passed by name. — Konrad Rudolph, Jun 15 '22 at 13:04

score 2 · Answer 1 · answered Jun 15 '22 at 12:49

2

By using sorted you are creating a brand new list. This new list is assigned to to the variable list_name, but that name is local to the function; it doesn't change the list_name back out in main.

If you use sort instead, it will modify the list in place and that modification will be seen in both places.

answered Jun 15 '22 at 12:49

Mark Ransom

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But if list_name is local variable, why in main part list_name have some items to output not empty list ? I'm fuzzy about this. And i add print(id(list_name)) in function listdir and main all output : 140649046215744, which means is pass-by-reference – Iaoceot Jun 15 '22 at 13:07
1

@Iaoceot because `append` modifies the list object without having to reassign it. And try printing the `id` after you sort. – Mark Ransom Jun 15 '22 at 13:10
oh which means sorted function change the list_name in Memory place? before sorted:id(list_name) = 140677502236544 after sorted:id(list_name) = 140676958919552 – Iaoceot Jun 15 '22 at 13:14

Why can't Python reference passing be sorted

1 Answers1