对数据列表/字典进行排序和展平

如何解决对数据列表/字典进行排序和展平

我觉得这必须相当简单，但为了理智，我发现自己嵌套了太多循环。

给出这样的字典列表

[{'id': 101,'value1': 'yes','value2': '','value3': '','value4': 'no' },{'id': 102,'value1': '','value2': 'yes','value4': 'no'},{'id': 103,'value1': 'no','value3': 'yes','value4': ''   },{'id': 104,'value3': 'no','value4': 'yes'},{'id': 105,{'id': 106,'value4': ''   }]

我想创建值匹配或为空的字典列表。

所需的结果（列表列表，但可以是列表/字典/任何东西，注意有重复）

[[{'id': 101,'value4': ''   }],[{'id': 102,[{'id': 104,'value4': ''   }]]

我认为必须有一些方法可以使用 itertools groupby 来做到这一点，但我无法弄清楚。这个答案 Sort and group a list of dictionaries 非常相似，但不完全是我需要的。如果它只是一个值，但倍数让我适合，那将非常简单。有什么想法吗？

编辑：所以这个可怕的建筑是有效的。问题是我至少有 13 个字段 (value1...value13) 来执行此操作，因此需要使其更加灵活。

list = [ {'id': 101,'value4': ''   }]

final_list = []
matched = False
for dict1 in list:
    sub_list = []
    for dict2 in list:
        if dict1 == dict2:
            continue
        print(dict1)
        print(dict2)
        print('---')
        if ((dict1['value1'] == dict2['value1'] or dict1['value1'] == '' or dict2['value1'] == '') and
            (dict1['value2'] == dict2['value2'] or dict1['value2'] == '' or dict2['value2'] == '') and
            (dict1['value3'] == dict2['value3'] or dict1['value3'] == '' or dict2['value3'] == '') and
            (dict1['value4'] == dict2['value4'] or dict1['value4'] == '' or dict2['value4'] == '')):
            # so these two match
            # Now make sure it doesn't conflict with the other entries already there
            if sub_list:
                subsublist = sub_list
                sub_conflict = False
                for dict3 in subsublist:
                    if dict2 == dict3:
                        continue
                    print("    ",dict2)
                    print("    ",dict3)
                    if ((dict2['value1'] == dict3['value1'] or dict2['value1'] == '' or dict3['value1'] == '') and
                        (dict2['value2'] == dict3['value2'] or dict2['value2'] == '' or dict3['value2'] == '') and
                        (dict2['value3'] == dict3['value3'] or dict2['value3'] == '' or dict3['value3'] == '') and
                        (dict2['value4'] == dict3['value4'] or dict2['value4'] == '' or dict3['value4'] == '')):
                        print('no conflict for this one')
                    else:
                        sub_conflict = True
                if not sub_conflict:
                    sub_list.append(dict2)
            else:
                sub_list.append(dict1)
                sub_list.append(dict2)
                print('appending both to list')
    #if not matched and [dict1 not in list3 for list3 in final_list]:
    # if not matched:
    #     print('not matched')
    #     sub_list=[dict1]
    sorted_list = sorted(sub_list,key=lambda k: k['id'])
    print('-------------------------------')
    final_list.append(tuple(sorted_list))

#print(*final_list)

final_final_list = []
for list in final_list:
    if list not in final_final_list:
        final_final_list.append(list)

for list in final_final_list:
    # print(list)
    for list2 in list:
        print(list2)
    print('')

解决方法

假设您的字典列表名为 dict_list，是否这样做

from itertools import product

values = ['yes','no']
num_value_fields = 4
keys = [f'value{i}' for i in range(1,num_value_fields + 1)]
results = []
for combo in product(values,repeat=num_value_fields):
    result = [d for d in dict_list if all(d[key] in {value,''}
                                          for key,value in zip(keys,combo))]
    if len(result) > 1:
        results.append(result)

产生预期的输出

[[{'id': 105,'value1': '','value2': 'yes','value3': 'yes','value4': ''},{'id': 106,'value1': 'yes','value2': '','value3': '','value4': ''}],[{'id': 101,'value4': 'no'},{'id': 102,{'id': 105,[{'id': 104,'value3': 'no','value4': 'yes'},[{'id': 103,'value1': 'no',[{'id': 102,{'id': 103,'value4': ''}]]

？