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 boutput.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 resultin_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'}]