假设字典像您的示例输入中那样排列,您可以使用该zip()
函数获取相关的成对字典的列表,然后可以使用它any()
来检查是否存在差异:
>>> 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
或获得不同的对:
>>> [(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'})]
您甚至可以获得每对都不匹配的密钥:
>>> [[k for k in x if x[k] != y[k]] for x, y in pairs if x != y]
[['key2']]
可能连同相关值:
>>> [[(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')]]
如果您尚未对输入列表进行排序,则也可以轻松地做到这一点:
>>> from operator import itemgetter
>>> list_1, list_2 = [sorted(l, key=itemgetter('unique_id'))
for l in (list_1, list_2)]