python列表的字典根据价值找到重复的 [英] python list of dictionaries find duplicates based on value

问题描述

我在python 2.7中有一个列表。

  a = [{'id'：1，'desc'：'smth'}，
 {'id '：2，'desc'：'smthelse'}，
 {'id'：1，'desc'：'smthelse2'}，
 {'id'：1，'desc'：'smthelse3 '}，....] 
  

我想通过列表，找到那些具有相同的值 - id（例如id = 1）并创建一个新的dict

  b = [{'id'：1 ，'desc'：[smth，smthelse2，smthelse3]}，
 {'id'：2，'desc'：'smthelse'}] 
  

我希望我很清楚

非常感谢你的建议

解决方案

即使它们只包含一个元素，最好将desc值保持为列表。这样你可以在b中为d中的

  
 print d ['id'] 
在d ['desc']中描述：
 print desc 
  

这也适用于字符串只需返回个人字符，这不是你想要的。

现在，该解决方案为您提供列表列表列表：

  a = [{'id'：1，'desc'：'smth'}，{'id'：2，'desc'：'smthelse'}， {'id'：1，'desc'：'smthelse2'}，{'id'：1，'desc'：'smthelse3'}] 
 
c = {} 
 for d a：
 c.setdefault（d ['id']，[]）。append（d ['desc']）
b = [{'id'：k，'desc'：v} k，v in c.iteritems（）] 
  

b 现在：

  [{'desc'：['smth'，'smthelse2'，'smthelse3']， 'id'：1}，
 {'desc'：['smthelse']，'id'：2}] 
  

I have a list of dicts in python 2.7.

a =[{'id': 1,'desc': 'smth'},
    {'id': 2,'desc': 'smthelse'},
    {'id': 1,'desc': 'smthelse2'},
    {'id': 1,'desc': 'smthelse3'},....]

I would like to go trough the list and find those dicts that have the same value - id (e.g. id = 1) and create a new dict

b = [{'id':1, 'desc' : [smth, smthelse2,smthelse3]}, 
     {'id': 2, 'desc': 'smthelse'}]

I hope I was clear enough

thank you very much for your suggestions

解决方案

It is better to keep the "desc" values as lists everywhere even if they contain a single element only. This way you can do

for d in b:
    print d['id']
    for desc in d['desc']:
        print desc

This would work for strings too, just returning individual characters, which is not what you want.

And now the solution giving you a list of dicts of lists:

a =[{'id': 1,'desc': 'smth'},{'id': 2,'desc': 'smthelse'},{'id': 1,'desc': 'smthelse2'},{'id': 1,'desc': 'smthelse3'}]

c = {}
for d in a:
    c.setdefault(d['id'], []).append(d['desc'])
b = [{'id': k, 'desc': v} for k,v in c.iteritems()]

b is now:

[{'desc': ['smth', 'smthelse2', 'smthelse3'], 'id': 1},
 {'desc': ['smthelse'], 'id': 2}]

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