比较由python中的字典和唯一键组成的2个列表 [英] Comparing 2 lists consisting of dictionaries with unique keys in python

问题描述

我有2个列表，两个列表包含相同数量的字典。每个字典都有一个唯一的键。对于第二列表中的第一列表的每个字典存在匹配，即具有唯一键的字典存在于另一列表中。但是这2个字典的其他元素可能不同。例如：

I have 2 lists, both of which contain same number of dictionaries. Each dictionary has a unique key. There is a match for each dictionary of the first list in the second list, that is a dictionary with a unique key exists in the other list. But the other elements of such 2 dictionaries may vary. For example:

list_1 = [
            {
                'unique_id': '001',
                'key1': 'AAA',
                'key2': 'BBB',
                'key3': 'EEE'
             },
             {
                'unique_id': '002',
                'key1': 'AAA',
                'key2': 'CCC',
                'key3': 'FFF'
             }
         ]

 list_2 = [
             {
                'unique_id': '001',
                'key1': 'AAA',
                'key2': 'DDD',
                'key3': 'EEE'
             },
             {
                'unique_id': '002',
                'key1': 'AAA',
                'key2': 'CCC',
                'key3': 'FFF'
             }
         ]

我想比较2个匹配字典的所有元素。如果任何元素不相等，我想打印不等于的元素。

I want to compare all elements of 2 matching dictionaries. If any of the elements are not equal, I want to print the none-equal elements.

请帮助，

谢谢
最好的问候

Thanks Best Regards

推荐答案

假设dicts像你的示例输入一样排列，使用 zip（）函数获取关联的对的列表，然后可以使用 any（）检查是否有区别：

Assuming that the dicts line up like in your example input, you can use the zip() function to get a list of associated pairs of dicts, then you can use any() to check if there is a difference:

>>> list_1 = [{'unique_id':'001', 'key1':'AAA', 'key2':'BBB', 'key3':'EEE'}, 
              {'unique_id':'002', 'key1':'AAA', 'key2':'CCC', 'key3':'FFF'}]
>>> list_2 = [{'unique_id':'001', 'key1':'AAA', 'key2':'DDD', 'key3':'EEE'},
              {'unique_id':'002', 'key1':'AAA', 'key2':'CCC', 'key3':'FFF'}]
>>> pairs = zip(list_1, list_2)
>>> any(x != y for x, y in pairs)
True

不同的对：

>>> [(x, y) for x, y in pairs if x != y]
[({'key3': 'EEE', 'key2': 'BBB', 'key1': 'AAA', 'unique_id': '001'}, {'key3': 'EEE', 'key2': 'DDD', 'key1': 'AAA', 'unique_id': '001'})]

您甚至可以得到不匹配每个键的键：

You can even get the keys which don't match for each pair:

>>> [[k for k in x if x[k] != y[k]] for x, y in pairs if x != y]
[['key2']]

可能与相关值一起：

>>> [[(k, x[k], y[k]) for k in x if x[k] != y[k]] for x, y in pairs if x != y]
[[('key2', 'BBB', 'DDD')]]

strong>如果你的输入列表还没有排序，你也可以这样做：

NOTE: In case you're input lists are not sorted yet, you can do that easily as well:

>>> from operator import itemgetter
>>> list_1, list_2 = [sorted(l, key=itemgetter('unique_id')) 
                      for l in (list_1, list_2)]

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