忽略某些键的两个字典列表之间的区别

Question

我想通过仅比较字典的某些键来找到两个字典列表之间的区别。a = [    {"fruit": "apple", "colour": "green", "notes":"one a day"},    {"vegetable": "tomato", "colour": "red", "origin": "asia"}]b = [    {"fruit": "apple", "colour": "green", "origin": "central asia"},    {"fruit": "strawberry", "colour": "red", "notes":"vitamin c"}]在示例中，我想忽略比较逻辑notes，origin但仍将它们保留在输出中。预期结果：output = [    {"vegetable": "tomato", "colour": "red", "origin": "asia"},    {"fruit": "strawberry", "colour": "red", "notes":"vitamin c"}]我尝试使用in条件，但它比较了所有字典键：difference = [i for i in a if i not in b]我试图调整这个解决方案，但它比较单个字典条目而不是整个集合的问题：def equal_dicts(a, b):    ignore_keys = ("notes", "origin")    ka = set(a).difference(ignore_keys)    kb = set(b).difference(ignore_keys)    return ka == kb and all(a[k] == b[k] for k in ka) for item in a:        if not equal_dicts(a, b):            print('same dictionary')

Answer 1

你可以这样做：

def list_diff(a, b):

    return [a_item for a_item in a if not any(equal_dicts(a_item, b_item) for b_item in b)]

output = list_diff(a, b) # Everythin that's in a but not in b

output.extend(list_diff(b, a)) # Everythin that's in b but not in a 

这给了你：

output = [

  {'vegetable': 'tomato', 'colour': 'red', 'origin': 'asia'}, 

  {'fruit': 'strawberry', 'colour': 'red', 'notes': 'vitamin c'}

]

Answer 2

您可以先过滤掉要忽略的键，然后正常比较字典。

如果您仍然希望将忽略的键包含在比较结果中，则需要对该方法进行一些修改 - 例如将过滤后的结果与原始结果进行比较a并b添加缺失的字段。

ignored_keys = {'notes', 'origin'}

filtered_a = [{k:v for k,v in sub_dict.items() if k not in ignored_keys} for sub_dict in a]

filtered_b = [{k:v for k,v in sub_dict.items() if k not in ignored_keys} for sub_dict in b]

那么区别是：

in_a_not_b = [elem for elem in filtered_a if elem not in filtered_b]

in_b_not_a = [elem for elem in filtered_b if elem not in filtered_a]

full_diff = in_a_not_b + in_b_not_a

至于被忽略的键，我可能会在合并结果之前添加它们，因为我们知道结果来自哪里......

或者没有太多思考的懒惰版本（例如：可能有更有效/更智能的方法，但我喜欢 dict/list 理解并且讨厌“手动”循环）：从过滤的东西到原始 dict 的映射。

dicts 是不可散列的，但我们可以将过滤后的“dicts”变成成对的元组：

ignored_keys = {'notes', 'origin'}

filtered_a = {tuple((k,v) for k,v in sub_dict.items() if k not in ignored_keys):sub_dict for sub_dict in a}

filtered_b = {tuple((k,v) for k,v in sub_dict.items() if k not in ignored_keys):sub_dict for sub_dict in b}

#compare on keys but get the original as the result

in_a_not_b = [filtered_a[elem] for elem in filtered_a if elem not in filtered_b]

in_b_not_a = [filtered_b[elem] for elem in filtered_b if elem not in filtered_a]

full_diff = in_a_not_b + in_b_not_a

结果：

>>> full_diff

[{'vegetable': 'tomato', 'colour': 'red', 'origin': 'asia'}, {'fruit': 'strawberry', 'colour': 'red', 'notes': 'vitamin c'}]