Python 中的文件内容排序

Question

大家好，我是 python 的新手，我正在尝试根据以下标准使用 PYTHON 3.4 对 /etc/passwd 文件进行排序：在 linux 系统上输入（常规 /etc/passwd 文件：

raj:x:501:512::/home/raj:/bin/ksh
ash:x:502:502::/home/ash:/bin/zsh
jadmin:x:503:503::/home/jadmin:/bin/sh
jwww:x:504:504::/htdocs/html:/sbin/nologin
wwwcorp:x:505:511::/htdocs/corp:/sbin/nologin
wwwint:x:506:507::/htdocs/intranet:/bin/bash
scpftp:x:507:507::/htdocs/ftpjail:/bin/bash
rsynftp:x:508:512::/htdocs/projets:/bin/bash
mirror:x:509:512::/htdocs:/bin/bash
jony:x:510:511::/home/jony:/bin/ksh
amyk:x:511:511::/home/amyk:/bin/ksh

我要查找的输出到文件或返回到屏幕：

Group 511 : jony, amyk, wwwcorp
Group 512 : mirror, rsynftp, raj
Group 507 : wwwint, scpftp 
and so on

这是我的计划：

1) Open and read the whole file or do it line by line 
2) Loop through the file using python regex 
3) Write it into temp file or create a dictionary 
4) Print the dictionary keys and values

我真的很感激这个例子如何有效地完成或应用任何排序算法。谢谢！

Answer 1

这应该遍历您的 /etc/passwd 并按组对用户进行排序。你不需要做任何花哨的事情来解决这个问题。

with open('/etc/passwd', 'r') as f:
    res = {}

    for line in f:
        parts = line.split(':')

        try:
            name, gid = parts[0], int(parts[3])
        except IndexError:
            print("Invalid line.")
            continue

        try:
            res[gid].append(name)
        except KeyError:
            res[gid] = [name]

for key, value in res.items():
    print(str(key) + ': ' + ', '.join(value))

Answer 2

您可以打开文件，将其放入一个列表中，然后将所有用户放入某种散列中 table

with open("/etc/passwd") as f:
    lines = f.readlines()

group_dict = {}
for line in lines:
    split_line = line.split(":")
    user = split_line[0]
    gid = split_line[3]
    # If the group id is not in the dict then put it in there with a list of users 
    # that are under that group
    if gid not in group_dict:
        group_dict[gid] = [user]
    # If the group id does exist then add the new user to the list of users in 
    # the group
    else:
        group_dict[gid].append(user)

# Iterate over the groups and users we found. Keys (group) will be the first item in the tuple, 
# and the list of users will be the second item. Print out the group and users as we go
for group, users in group_dict.iteritems():
    print("Group {}, users: {}".format(group, ",".join(users)))

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