Question

例如我有一本字典：

dictA={"nest1":{"01feb":[1,2,3,4,5],"02feb":[1,7,8,9,10]},
       "nest2":{"01feb":[1,2,3,4,5],"02feb":[6,4,8,10,10]}}

里面的列表具有相同的长度。我需要将 nest1 和 nest2 合并为一个字典，结果应如下所示：

dictA={"nest":{"01feb":[2,4,6,8,10],"02feb":[7,11,16,19,20]}}

Answer 1

您可以像这个示例一样使用dict comprehension，map()和zip()（适用于Python 2和Python 3）。

$result='<select id="type" name="type" class="form-control">';
foreach ($types as $type) {
        $result.= '<option value="' . $type . '">' . $type . '</option>';
    }
$result.='</select>';

echo $result;

输出：

dictA = {'nest1': {'01feb': [1, 2, 3, 4, 5], '02feb': [1, 7, 8, 9, 10]},
 'nest2': {'01feb': [1, 2, 3, 4, 5], '02feb': [6, 4, 8, 10, 10]}}

a = (v.items() for _, v in map(list, dictA.items()))
# You can also use another map():
# final = {'nest': {k: list(map(sum, zip(v,j))) for (k, v), (_, j) in zip(*a)}}
final = {'nest': {k: [m+n for m, n in zip(v, j)] for (k, v), (_, j) in zip(*a)}}

print(final)

Answer 2

请为您的查询找到以下代码。

dictA={"nest1":{"01feb":[1,2,3,4,5],"02feb":[1,7,8,9,10]},
       "nest2":{"01feb":[1,2,3,4,5],"02feb":[6,4,8,10,10]}}
result ={}
final_op = {}
for k,v in dictA.iteritems():
    for nk,nv in v.iteritems():
        if result.has_key(nk):
            i=0
            while i < len(result[nk]):
                result[nk][i] += nv[i]
                i += 1
        else:
            result[nk] = nv
final_op['nest'] = result
print final_op

输出：

{'nest': {'02feb': [7, 11, 16, 19, 20], '01feb': [2, 4, 6, 8, 10]}}

Answer 3

您必须遍历dict并更新迭代中的值。

dictA={"nest1":{"01feb":[1,2,3,4,5],"02feb":[1,7,8,9,10]},
       "nest2":{"01feb":[1,2,3,4,5],"02feb":[6,4,8,10,10]}}

def merge(dictA):
    merge_dict = {}
    for key in dictA:
        for sub_key in dictA[key]:
            if sub_key in merge_dict:
                # update the nested value
                merge_dict[sub_key] = [sum(x) for x in zip(*[merge_dict[sub_key], dictA[key][sub_key]])]
            else:
                merge_dict[sub_key] = dictA[key][sub_key]
    return merge_dict

merge_dict = merge(dictA)
dictA.clear()
dictA["nest"] = merge_dict

print(dictA)

Answer 4

这是一个执行您要求的功能：

def merge_nested(dictionary):
    result = dict()
    for nested in dictionary.values():
        for key in nested.keys():
            if key in result:
                # We've already found it, add our values to it's
                for i in range(len(result[key])):
                    result[key][i] += nested[key][i]
            else:
                result[key] = nested[key]
    return {"nest":result}

使用此功能，您将获得以下输出：

>>> print(merge_nested({"nest1":{"01feb":[1,2,3,4,5],"02feb":[1,7,8,9,10]},"nest2":{"01feb":[1,2,3,4,5],"02feb":[6,4,8,10,10]}}))
{'nest': {'01feb': [2, 4, 6, 8, 10], '02feb': [7, 11, 16, 19, 20]}}

这是@Arockia的答案here

的修改版本

Answer 5

groupby itertools的解决方案：

from itertools import chain, groupby
import operator

dictA={"nest1":{"01feb":[1,2,3,4,5],"02feb":[1,7,8,9,10]},
       "nest2":{"01feb":[1,2,3,4,5],"02feb":[6,4,8,10,10]}}

A = {
    "nest": dict(
        (key, list(map(operator.add, *(v for _, v in group))))
        for key, group in groupby(
            sorted(
                chain(*(v.iteritems() for v in dictA.values()))  # Python 2
                # chain(*(v.items() for v in dictA.values()))    # Python 3
            ),
            lambda x: x[0],
        )
    )
}

print(A)

结果：

{'nest': {'02feb': [7, 11, 16, 19, 20], '01feb': [2, 4, 6, 8, 10]}}

如何在同一个字典下合并两个嵌套字典

5 个答案: