从元组中删除字符

发布于 2024-12-09 12:13:54 字数 314 浏览 0 评论 0原文

我用来

Users = win32net.NetGroupGetUsers(IP,'none',0),

获取系统上的所有本地用户。输出是一个元组，

(([{'name': u'Administrator'}, {'name': u'Guest'}, {'name': u'Tom'}], 3, 0),)

我想清理它，以便它只打印出“管理员，来宾，汤姆”。我尝试使用剥离和替换，但不能在元组上使用它们。有没有办法将其转换为字符串，以便我可以操作它，或者是否有更简单的方法来处理它？

原文

Im using

Users = win32net.NetGroupGetUsers(IP,'none',0),

to get all the local users on a system. The output is a tuple,

(([{'name': u'Administrator'}, {'name': u'Guest'}, {'name': u'Tom'}], 3, 0),)

I want to clean this up so it just prints out "Administrator, Guest, Tom". I tried using strip and replace but you cant use those on tuples. Is there a way to convert this into a string so i can manipulate it or is there an even simpler way to go about it?

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み格子的夏天 2024-12-16 12:13:54

这不应该以逗号结尾：

Users = win32net.NetGroupGetUsers(IP,'none',0),

结尾的逗号将结果转换为包含结果的单个项目元组，该结果本身就是一个元组。

您想要的数据位于 Users[0] 中。

>>> print Users[0]
[{'name': u'Administrator'}, {'name': u'Guest'}, {'name': u'Tom'}]

为了解压这个字典列表，我们使用生成器表达式：

Users = win32net.NetGroupGetUsers(IP,'none',0)
print ', '.join(d['name'] for d in Users[0])

This should not end with a comma:

Users = win32net.NetGroupGetUsers(IP,'none',0),

The trailing comma turns the result into a single item tuple containing the result, which is itself a tuple.

The data you want is in Users[0].

>>> print Users[0]
[{'name': u'Administrator'}, {'name': u'Guest'}, {'name': u'Tom'}]

To unpack this list of dictionaries we use a generator expression:

Users = win32net.NetGroupGetUsers(IP,'none',0)
print ', '.join(d['name'] for d in Users[0])

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遗弃Ｍ 2024-12-16 12:13:54

', '.join(user['name'] for user in Users[0][0])

', '.join(user['name'] for user in Users[0][0])

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归途 2024-12-16 12:13:54

input = (([{'name': u'Administrator'}, {'name': u'Guest'}, {'name': u'Tom'}], 3, 0),)    
in_list = input[0][0]    
names = [x['name'] for x in in_list]    
print names

[u'Administrator', u'Guest', u'Tom']

input = (([{'name': u'Administrator'}, {'name': u'Guest'}, {'name': u'Tom'}], 3, 0),)    
in_list = input[0][0]    
names = [x['name'] for x in in_list]    
print names

[u'Administrator', u'Guest', u'Tom']

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~没有更多了~