我如何通过从相同索引的其他列表中获取值并根据条件形成新列表来获得 append 值

Question

（我的输入）：我有如下 4 个列表，每个列表包含 8 个值。

eventIdList=['112','114','115','117','198','125','138','107'] ## No duplicate value inside, like primary key. 

eventTypeList=[fuel,driver,driver,trax,driver,fuel,lux,driver] ## duplicate values exist inside.

partnerIdList= ['aux','box','box','disc','box','cot','top','box' ] ## duplicate values exist inside.

customerIdList= ['dell','apple','apple','amazon','apple','microsoft','dell','apple'] ## duplicate values exist inside.

我为索引 0 获取的第一个有效负载如下所示。 类似地，for 循环内的其他索引。

"message":{

"eventId": '112',

"eventType": 'fuel',

"partnerId": 'aux'

"customerId": 'dell'

}

观察 Input 中的 4 个列表，除了 eventId，所有有效负载（eventType、partnerId 和 customerId 值）的第 1、第 2 和第 7 个索引都是相同的 {'driver','box','apple'}。 所以我需要在列表中显示这些事件 ID，通过迭代附加 eventId 值，这里的 eventId 应该是 [114,115,107]。 我怎样才能实现它？

OUTPUT 期望在 for 循环下

当索引为 1 “消息”时：{

eventId: [114],

eventType: "driver",

partnerId: "box"

customerId: "apple"

}

当索引为 2 “消息”时：{

eventId: [114,115],

eventType: "driver",

partnerId: "box"

customerId: "apple"

}

当索引为 7 “消息”时：{

eventId: [114,115,107],

eventType: "driver",

partnerId: "box"

customerId: "apple"

}

---

我的代码是这样的。 没完成。

def myAppendFunction(eventIdList, eventTypeList, customerIdList, partnerIdList):

    formedEventIdList=[]

    for inc , eventIdelem in  enumerate(eventIdList):

      

       print('inc**',inc) 

       print('eventid**',eventIdelem) 

       print('eventType**', eventTypeList[inc])

       print('customerId**', customerIdList[inc])

       print('partnerId**', partnerIdList[inc])

           

            #LOGIC WANT

            # if (eventTypeList[inc]==eventTypeList[inc+1]) and (customerIdList[inc]==customerIdList[inc+1]) and (partnerIdList[inc]==partnerIdList[inc+1]) :           

 

       #formedEventIdList.append(eventIdelem)

 

            message = {"Type": "Notification",

            "Message": {"eventType": eventType,"customerId": customerId,"partnerId": partnerId, "eventIds": formedEventIdList}

            }

#for end

再次提供我的程序， OUTPUT和预期的 OUTPUT下面。 请给我可以在这里为我工作的代码。

我的 Python 程序如下：

eventIdList=['112','114','115','117','198','125','138','107'] 
eventTypeList=['fuel','driver','driver','trax','driver','fuel','lux','driver'] ## duplicate values (like driver, fuel etc) is exist in eventTypeList, 
partnerIdList= ['aux','box','box','disc','box','cot','top','box' ] ## duplicate values exist in partnerIdList like box
customerIdList= ['dell','apple','apple','amazon','apple','microsoft','dell','apple'] ## duplicate values is exist like apple, dell etc in customerIdList
for inc , eventIdelem in  enumerate(eventIdList):
   formTheEventIdList=[]
   message = {
    "Message": {"eventType": eventTypeList[inc],"customerId": customerIdList[inc],"eventIds": eventIdelem,"partnerId": partnerIdList[inc]}       
    }
   print('inc=',inc,'OUTPUT*********:: ', message)

下面这个程序的OUTPUT：

inc= 0 OUTPUT*********:: {'Message': {'eventType': 'fuel', 'customerId': 'dell', 'eventIds': '112', 'partnerId': '辅助'}}

inc= 1 OUTPUT*********:: {'Message': {'eventType': 'driver', 'customerId': 'apple', 'eventIds': '114', 'partnerId': '盒子'}}

inc= 2 OUTPUT*********:: {'Message': {'eventType': 'driver', 'customerId': 'apple', 'eventIds': '115', 'partnerId': '盒子'}}

inc= 3 OUTPUT*********:: {'Message': {'eventType': 'trax', 'customerId': 'amazon', 'eventIds': '117', 'partnerId': '光盘'}}

inc= 4 OUTPUT*********:: {'Message': {'eventType': 'driver', 'customerId': 'apple', 'eventIds': '198', 'partnerId': '盒子'}}

inc= 5 OUTPUT*********:: {'Message': {'eventType': 'fuel', 'customerId': 'microsoft', 'eventIds': '125', 'partnerId': '婴儿床'}}

inc= 6 OUTPUT*********:: {'Message': {'eventType': 'lux', 'customerId': 'dell', 'eventIds': '138', 'partnerId': '最佳'}}

inc= 7 OUTPUT*********:: {'Message': {'eventType': 'driver', 'customerId': 'apple', 'eventIds': '107', 'partnerId': '盒子'}}

但我想要 OUTPUT 如下：

inc= 0 OUTPUT*********:: {'Message': {'eventType': 'fuel', 'customerId': 'dell', 'eventIds': ['112'], 'partnerId' : '辅助'}}

inc= 1 OUTPUT*********:: {'Message': {'eventType': 'driver', 'customerId': 'apple', 'eventIds': ['114'] , 'partnerId' ： '盒子'}}

inc= 2 OUTPUT*********:: {'Message': {'eventType': 'driver', 'customerId': 'apple', 'eventIds': ['115',114] , ' partnerId': 'box'}}

inc= 3 OUTPUT*********:: {'Message': {'eventType': 'trax', 'customerId': 'amazon', 'eventIds': ['117'], 'partnerId' : '光盘'}}

inc= 4 OUTPUT*********:: {'Message': {'eventType': 'driver', 'customerId': 'apple', 'eventIds': ['198'], 'partnerId' ： '盒子'}}

inc= 5 OUTPUT*********:: {'Message': {'eventType': 'fuel', 'customerId': 'microsoft', 'eventIds': ['125'], 'partnerId' : '婴儿床'}}

inc= 6 OUTPUT*********:: {'Message': {'eventType': 'lux', 'customerId': 'dell', 'eventIds': ['138'], 'partnerId' ： '顶部'}}

inc= 7 OUTPUT*********:: {'Message': {'eventType': 'driver', 'customerId': 'apple', 'eventIds': ['107',115,114] , ' partnerId': 'box'}}

========================

注意：只有那些具有相同“eventType”、“customerId”和“partnerId”的记录。 存储这些 eventIds 并显示在列表中。 [当'eventType','customerId'和'partnerId'值相同时，与其他索引匹配，则eventID为append。 这里当index=1,2 and 7时。所以在这种情况下需要将append'eventIds'放在一个列表中并显示]

[当 index=0,3,4,5,6] 'eventType'、'customerId' 和 'partnerId' 值 3 都与其他记录/索引不匹配。 因此在这种情况下不需要 append。

Answer 1

使用zip可以方便地同时迭代四个列表。

可以通过列表理解来过滤满足特定条件的元素。

eventIdList=['112','114','115','117','198','125','138','107'] ## No duplicate value inside, like primary key. 

eventTypeList=['fuel','driver','driver','trax','driver','fuel','lux','driver'] ## duplicate values exist inside.

partnerIdList= ['aux','box','box','disc','box','cot','top','box' ] ## duplicate values exist inside.

customerIdList= ['dell','apple','apple','amazon','apple','microsoft','dell','apple'] ## duplicate values exist inside.

def filterPartnerId(goodPartnerId, eventIdList, eventTypeList, partnerIdList, customerIdList):
    return [(evId, evType, pId, cId) for (evId, evType, pId, cId) in zip(eventIdList, eventTypeList, partnerIdList, customerIdList) if pId == goodPartnerId]

print(filterPartnerId("box", eventIdList, eventTypeList, partnerIdList, customerIdList))
# [('114', 'driver', 'box', 'apple'),
#  ('115', 'driver', 'box', 'apple'),
#  ('198', 'driver', 'box', 'apple'),
#  ('107', 'driver', 'box', 'apple')]

您还可以将四个列表转换为一个字典列表：

def fourlists_to_onelist(eventIdList, eventTypeList, partnerIdList, customerIdList):
    return [{"eventId": evId, "eventType": evType, "partnerId": pId, "customerId": cId} for (evId, evType, pId, cId) in zip(eventIdList, eventTypeList, partnerIdList, customerIdList)]

def filterPartnerId(goodPartnerId, onelist):
    return [x for x in onelist if x['partnerId'] == goodPartnerId]

onelist = fourlists_to_onelist(eventIdList, eventTypeList, partnerIdList, customerIdList)
print(filterPartnerId("box", onelist))
# [{'eventId': '114', 'eventType': 'driver', 'partnerId': 'box', 'customerId': 'apple'},
#  {'eventId': '115', 'eventType': 'driver', 'partnerId': 'box', 'customerId': 'apple'},
#  {'eventId': '198', 'eventType': 'driver', 'partnerId': 'box', 'customerId': 'apple'},
#  {'eventId': '107', 'eventType': 'driver', 'partnerId': 'box', 'customerId': 'apple'}]

您还可以使用statistics.mode查找列表中最常见的元素：

from statistics import mode

goodPartnerId = mode(partnerIdList)
print(goodPartnerId)
# box