我有 2 个列表，我想合并它们，答案应该如下所示

Question

我有两个字典列表，我想合并它们。当两个列表中都存在字典时，我想在字典中添加一个“信心”键，以反映该字典存在于两个列表中。

列表 1

lst1 = [
    {'key': 'data_collected.service_data'},
    {'key': 'gdpr.gdpr_compliance'},
    {'key': 'disclosure_of_information.purpose_of_disclosure'},
    {'key': 'opt_out.choice_of_opt_out'}
]

列表 2

lst2 = [
    {'key': 'child_data_protection.parent_guardian_consent'},
    {'key': 'ccpa.ccpa_compliance'},
    {'key': 'disclosure_of_information.purpose_of_disclosure'},
    {'key': 'opt_out.choice_of_opt_out'}
]

当我在代码下面运行时，我没有得到正确的输出

res = []
for x in lst1:
    for y in lst2:
        if x["key"] == y["key"]:
            if x not in res and y not in res:
                res.append({"key": x["key"], "confidence": 1})
        else:
            if x not in res and y not in res:
                res.append(x)
                res.append(y)

print(res)

输出应该喜欢

[
    {'key': 'data_collected.service_data'},
    {'key': 'gdpr.gdpr_compliance'},
    {
        'key': 'disclosure_of_information.purpose_of_disclosure',
        'confidence': 1
    },
    {
        'key': 'opt_out.choice_of_opt_out',
        'confidence': 1
    },
    {'key': 'child_data_protection.parent_guardian_consent'},
    {'key': 'ccpa.ccpa_compliance'}
]

Answer 1

您可以使用set comprehension 收集列表中每个字典的“关键”元素。然后你可以遍历所有键并检查一个键是否在两个列表中。

keys_1 = {d["key"] for d in lst1}
keys_2 = {d["key"] for d in lst2}

output = []
for k in keys_1 | keys_2:
    d = {"key": k}
    if k in keys_1 and k in keys_2:
        d["confidence"] = 1
    output.append(d)

Answer 2

您可以使用set 上的intersection 和symmetric_difference 函数完全避免原始循环：

# Shortened key names for brevity
a = [{"key": "a"}, {"key": "b"}, {"key": "c"}]
b = [{"key": "a"}, {"key": "d"}, {"key": "e"}]

# Turn both lists into sets
a_keys = {entry["key"] for entry in a}
b_keys = {entry["key"] for entry in b}

# Add elements that are in both sets with confidence set to 1
result = [{"key": key, "confidence": 1} for key in a_keys.intersection(b_keys)]
# Add elements that are not in both sets
result += [{"key": key} for key in a_keys.symmetric_difference(b_keys)]

将导致：

[{'confidence': 1, 'key': 'a'},
 {'key': 'b'},
 {'key': 'd'},
 {'key': 'c'},
 {'key': 'e'}]

请注意，元素顺序会随着set 而改变。

Answer 3

lst1.extend(i for i in (i if i not in lst1 else lst1[lst1.index(i)].update({'confidence': 1}) for i in lst2) if i is not None)

lst1 将是你的结果

Answer 4

如果您不太担心性能。

intersection = [value for value in lst1 if value in lst2]
res = [val for val in lst1 if val not in intersection] + [val for val in lst2 if val not in intersection]
res += list(map(add_confidence, intersection))

Answer 5

另一种方法可以是：

lst1 = [
    {'key': 'data_collected.service_data'},
    {'key': 'gdpr.gdpr_compliance'},
    {'key': 'disclosure_of_information.purpose_of_disclosure'},
    {'key': 'opt_out.choice_of_opt_out'}
]
lst2 = [
    {'key': 'child_data_protection.parent_guardian_consent'},
    {'key': 'ccpa.ccpa_compliance'},
    {'key': 'disclosure_of_information.purpose_of_disclosure'},
    {'key': 'opt_out.choice_of_opt_out'}
]
for data in lst1:
    # If same data exists in lst2, add confidence key and remove it from lst2
    if data in lst2:
        lst2.remove(data)
        data['confidence']=1

# At the end of above for loop, lst2 contains unique data, now just add both the lists to get the final result            
lst1 = lst1+lst2        
print (lst1)

输出：

[{'key': 'data_collected.service_data'}, {'key': 'gdpr.gdpr_compliance'}, {'key': 'disclosure_of_information.purpose_of_disclosure', 'confidence': 1}, {'key': 'opt_out.choice_of_opt_out', 'confidence': 1}, {'key': 'child_data_protection.parent_guardian_consent'}, {'key': 'ccpa.ccpa_compliance'}]