在python中列出重复的数据串联 [英] List duplicate data concatenation in python

问题描述

我是python的新手，在获取此输出时遇到问题

i am a newbie in python and facing issue in getting this output

 a = [('textVerify', 'AH', 'SELECT SERVICES'), ('textVerify', 'F7', 'test1>'),('audioVerify', '091;0'), ('imageVerify', 'duck.gif'),('imageVerify', 'egg.gif')]

我想创建一个新列表，其中应包含所有第0个唯一元素，例如

i want to create a new list which should hold all the 0th unique element like

  audioVerify,imageVerify,textVerify

所以预期的结果是

 ['textVerify',(('AH', 'SELECT SERVICES'), ('F7', 'test1>'))  'audioVerify', ('091;0'),  ('imageVerify', ('duck.gif','egg.gif')]

推荐答案

您最好使用 defaultdict 为此:

You'd better use a defaultdict for this:

>>> from collections import defaultdict
>>> d = defaultdict(list)
>>> for item in a:
...     d[item[0]].append(item[1:])
...
>>> d
defaultdict(<class 'list'>, {'textVerify': [('AH', 'SELECT SERVICES'), 
('F7', 'est1>')], 'imageVerify': [('duck.gif',), ('egg.gif',)], 
'audioVerify': [('091;0',)]})

现在您可以按名称/索引访问其元素:

Now you can access its elements by name/index:

>>> d['textVerify']
[('AH', 'SELECT SERVICES'), ('F7', 'test1>')]
>>> d['textVerify'][0][0]
'AH'

如果您需要保留字典键的顺序，则可以使用

If you need to preserve the order of the dictionary keys, you can use an OrderedDict, together with the .setdefault() method, as described by Ashwini Chaudhary:

>>> d = OrderedDict()
>>> for x in a:
...     d.setdefault(x[0],[]).append(x[1:])
...
>>> d
OrderedDict([('textVerify', [('AH', 'SELECT SERVICES'), ('F7', 'test1>')]), 
('audioVerify', [('091;0',)]), ('imageVerify', [('duck.gif',), ('egg.gif',)])])

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