如何将字典列表与内部列表连接起来

Question

我有一个带有内部列表和字典的字典列表，我需要将它们连接到一个大字典中。名单是这样的：[{'total': {'last_day': '7'}},  {'total': {'progress': '07/04'}},  {'total': {'piecies': '3008'}},  {'total': {'week': ['16.0']}},  {'total': {'week': ['17.0']}},  {'total': {'week': ['15.0']}},  {'total': {'week': ['17.0']}},  {'total': {'week': ['16.0']}},  {'total': {'week': ['13.0']}},  {'total': {'week': ['6.0']}},  {'tkts': [{'tktvalue': '13.5'}]},  {'tkts': [{'month': {'consuntivato_pezzi': '2346'}}]},  {'tkts': [{'month': {'consuntivato_euro': '31671.00'}}]},  {'tkts': [{'month': {'preventivato_pezzi': '9897'}}]}]我尝试了一些 for 循环和递归函数，但没有很好的结果for vars in temporary_var:    for title in vars:        try:            try:                table_var[title].update(vars[title])            except KeyError:                table_var[title] = vars[title]        except AttributeError:            table_var[title].append(vars[title][0])但我只能得到这个：{'total': {'last_day': '7', 'progress': '07/04', 'volumes': '3008', 'week': ['6.0']}, 'tkts': [{'service': 'SOSPC'}, {'tktvalue': '13.5'}, {'month': {'volumes1': '2346'}}, {'month': {'volumes2': '31671.00'}}, {'month': {'volumes3': '9897'}}]}但我需要这个：{'total': {'last_day': '7', 'progress': '07/04', 'volumes': '3008', 'week': ['16.0', '17.0', '15.0', '17.0', '16.0', '13.0','6.0']}, 'tkts': [{'service': 'SOSPC', 'tktvalue': '13.5', 'month': {'volumes1': '2346', 'volumes2': '31671.00', 'volumes3': '9897'}}]}

Answer 1

试试这个，它会得到你想要的输出：

lst=[{'total': {'last_day': '7'}},

 {'total': {'progress': '07/04'}},

 {'total': {'piecies': '3008'}},

 {'total': {'week': ['16.0']}},

 {'total': {'week': ['17.0']}},

 {'total': {'week': ['15.0']}},

 {'total': {'week': ['17.0']}},

 {'total': {'week': ['16.0']}},

 {'total': {'week': ['13.0']}},

 {'total': {'week': ['6.0']}},

 {'tkts': [{'tktvalue': '13.5'}]},

 {'tkts': [{'month': {'consuntivato_pezzi': '2346'}}]},

 {'tkts': [{'month': {'consuntivato_euro': '31671.00'}}]},

 {'tkts': [{'month': {'preventivato_pezzi': '9897'}}]}

]

d={} # the dictionary that will hold the result

for dd in lst: # for each dictionary in the list of dictionaries

    for key,value in dd.items():

        if key not in d: # key does not exist in dictionary d

            d[key]=value

        else: # key exists in dictionary d

            if isinstance(value,dict): # check if the value is a dictionary or a list

                for key1,value2 in value.items():

                    if key1 not in d[key]:

                        d[key]={**d[key],**value} # combine the dictionaries

                    else:

                        d[key][key1].append(value2[0])

            elif isinstance(value,list): # check if the value is a  list

                if isinstance(value[0],dict): # check if the value is a dictionary or a list

                    for key1,value2 in value[0].items():

                        if key1 not in d[key][0]:

                            d[key][0]={**d[key][0],**value[0]}

                        else:

                            d[key][0][key1]={**d[key][0][key1],**value2}

print(d)

输出：{'total': {'last_day': '7', 'progress': '07 / 04 ',' piecies ':' 3008 ',' week ': ['16 .0', '17 .0' , '15 .0', '17 .0', '16 .0', '13 .0',' 6.0 ']},' tkts ': [{' tktvalue ': '13 .5', '月' ：{'consuntivato_pizzi'：'2346'，'final_euro'：'31671.00'，'quoted_pieces'：'9897'}}]}

Answer 2

尝试在此处使用默认列表。您可以了解有关它们的更多信息collections.defaultdict

from collections import defaultdict

result = defaultdict(list)

for sequence in (yourdict):

    for keys,dictionary in sequence.items():

          result[keys].append(dictionary)

print(dict(result))

输出 ：

{'total': [{'last_day': '7'}, {'progress': '07/04'}, {'piecies': '3008'}, {'week': ['16.0']}, {'week': ['17.0']}, {'week': ['15.0']}, {'week': ['17.0']}, {'week': ['16.0']}, {'week': ['13.0']}, {'week': ['6.0']}], 'tkts': [[{'tktvalue': '13.5'}], [{'month': {'consuntivato_pezzi': '2346'}}], [{'month': {'consuntivato_euro': '31671.00'}}], [{'month': {'preventivato_pezzi': '9897'}}]]}