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：输入（linux系统上的常规/ 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

我正在寻找文件或返回到屏幕的输出：

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

我将非常感谢这个例子如何有效地完成或应用任何排序算法。谢谢！

Answer 1

你可以打开文件，把它扔进一个列表然后把所有用户都扔进某个哈希表

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)))

Answer 2

这应该遍历您的/etc/passwd并按组对用户进行排序。你不必做任何想要解决这个问题的事情。

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))

在Python中对文件的内容进行排序

2 个答案: