比较字典的多个值并汇总结果

如何解决比较字典的多个值并汇总结果

我有一个用例，在比较这些值时，我需要忽略sku，targetFlag和qualifierFlag，并且列表中的任何字典是否具有与这三个值相同的值，并且它应该与所有这三个键和值一起汇总在一个字典中。

有人可以告诉我我需要做些什么修改，或者有其他更好的解决方案来获得正确的结果。

到目前为止，与上述选项进行比较时，我尝试过的解决方案只能取一个值：

from collections import OrderedDict
import pandas as pd

d = OrderedDict()

d = (pd.DataFrame(lst)
            .groupby(['endDate','storeCode','startDate','promoName','targetFlag','qualifierFlag'])
            .sku
            .agg(set)
            .reset_index()
            .to_dict('r'))

lt = [{
   "sku":"40428200","storeCode":"316","endDate":"05445100","promoName":"2201:316 SUPER 37100548200","startDate":"40505000","targetFlag":"0","qualifierFlag":"0",},{
   "sku":"4567","targetFlag":"1","qualifierFlag":"1",{
   "sku":"12345","storeCode":"123","promoName":"2201:316 GRAND 0548200",{
   "sku":"40428200","endDate":"18840100","promoName":"2201:316 AVI 37125790200",}
]

预期结果将是这样

[
 {'endDate': '05445100','promoName': '2201:316 GRAND 37100548200','startDate': '40505000','storeCode': '123','res': [
      {
        'qualifierFlag': '0','sku': '12345','targetFlag': '1'
      }
  ]
  },{'endDate': '05445100','promoName': '2201:316 SUPER 37100548200','storeCode': '316','sku': '40428200','targetFlag': '0'
      },{
        'qualifierFlag': '1','sku': '4567','targetFlag': '1'
      },]
  },{'endDate': '18840100','promoName': '2201:316 AVI 37125790200','targetFlag': '1'
      }
  ]
]

解决方法

以下代码能解决问题吗？

d = (pd.DataFrame(lst)
       .groupby(['endDate','storeCode','startDate','promoName'])['sku','targetFlag','qualifierFlag']
       .agg(set)
       .reset_index()
       .to_dict('r'))

,

使用itertools.groupby：

from itertools import groupby

def key_func(x):
    return (x["storeCode"],x["promoName"],x["startDate"],x["endDate"])

lst = sorted(lst,key=key_func)

result = [
    {
        "storeCode": k[0],"promoName": k[1],"startDate": k[2],"endDate": k[3],"res": [
            {
                "sku": x["sku"],"qualifierFlag": x["qualifierFlag"],"targetFlag": x["targetFlag"],}
            for x in list(v)
        ],}
    for k,v in groupby(lst,key=key_func)
]

结果：

[
    {
        "storeCode": "123","promoName": "2201:316 GRAND 0548200","startDate": "40505000","endDate": "05445100","res": [{"sku": "12345","qualifierFlag": "0","targetFlag": "1"}],},{
        "storeCode": "316","promoName": "2201:316 AVI 37125790200","endDate": "18840100","res": [{"sku": "40428200","promoName": "2201:316 SUPER 37100548200","res": [
            {"sku": "40428200","targetFlag": "0"},{"sku": "4567","qualifierFlag": "1","targetFlag": "1"},],]

Jean-Marc Billod的方法使用set作为聚合函数，因此仅返回唯一值。