用另一个列表中的键，值对更新python字典的列表 [英] Updating a list of python dictionaries with a key, value pair from another list

问题描述

假设我有以下python字典列表：

  dict1 = [{'domain'：'比率'} ，{'domain'：'Geometry'}] 
  

和一个列表，如：

  list1 = [3，6] 
  

我想更新 dict1 或创建另一个列表，如下所示：

  dict1 = [{'domain'：'比率'，'count'：3}，{'domain'：'几何'，'count'：6}] 
  

我该怎么做？

解决方案

 >>> l1 = [{'domain'：'Ratios'}，{'domain'：'Geometry'}] 
>>> l2 = [3，6] 
>>>对于d，zip中的num（l1，l2）：
 d ['count'] = num 
 
 
>>> l1 
 [{'count'：3，'domain'：'Ratios'}，{'count'：6，'domain'：'Geometry'}] 
  / pre> 
 
 
另一种做法，这次用列表理解不会使原来的变体：
 >>> d（n，d，n，n）在zip（l1，l2）] 
 [{'count'：3，'domain'：'比率'}，{'count'域'：'几何'}] 
  
 
Let's say I have the following list of python dictionary:
dict1 = [{'domain':'Ratios'},{'domain':'Geometry'}]
and a list like:
list1 = [3, 6]
I'd like to update dict1 or create another list as follows:
dict1 = [{'domain':'Ratios', 'count':3}, {'domain':'Geometry', 'count':6}]
How would I do this? 
解决方案
>>> l1 = [{'domain':'Ratios'},{'domain':'Geometry'}]
>>> l2 = [3, 6]
>>> for d,num in zip(l1,l2):
        d['count'] = num


>>> l1
[{'count': 3, 'domain': 'Ratios'}, {'count': 6, 'domain': 'Geometry'}]
Another way of doing it, this time with a list comprehension which does not mutate the original:
>>> [dict(d, count=n) for d, n in zip(l1, l2)]
[{'count': 3, 'domain': 'Ratios'}, {'count': 6, 'domain': 'Geometry'}]


                        
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