将列表中的列表更改为字符串-python

我使用列表理解创建了一个列表，我想知道如何将列表中的列表更改为字符串

这是我的代码：

limit = 1000
searchPage = 0
tagId = 493
table = 'ContactGroupAssign'
selectedFields = ['Contact.Id', 'DateCreated', 'Contact.OwnerID', 'Contact.Groups']
queryDataUnknown = {'GroupId': tagId, 'DateCreated': "~>=~" + str(twoyearsago)}

tagsDic = {'Unknown': ',10738', 
'Cant Afford': ',10712', 
'Doesnt Want To Pay': ',10714', 
'No 3rd Party': ',10716', 
'No Longer Qualifies': ',10718', 
'Scam': ',10720', 
'No Contact': ',10734', 
'Collections - Non Payment': ',7878', 
'Billing Cancellation': ',10748', 
'NBDK': ',8744', 
'Within 5 Days: Cant Afford CTF': ',10744', 
'Within 5 Days: Changed Mind/Persuaded': ',10746'}

allcancelledmembers = []
while True:
        queryResults = infusionsoft.DataService('query', table, limit, searchPage, queryDataUnknown, selectedFields)
        allcancelledmembers += queryResults
        searchPage += 1
        if len(queryResults) < 1000:
            break
lc_allcancelmembers = [[record.get("Contact.Id"), convert_is_datetime(str(record.get("DateCreated"))), record.get("Contact.OwnerID"), record.get("Contact.Groups"), [k for k, v in tagsDic.items() if v in (record.get("Contact.Groups"))]] for record in allcancelledmembers]

这是我想要的结果：

[68505, datetime.datetime(2019, 1, 9, 9, 10, 52), 384114, '130,134,136,138,152,206,208,286,290,312,436,493,495,728,760,772,774,788,952,968,1088,1226,1254,2362,2584,2598,2972,2974,3102,3560,4552,4554,5430,6566,6950,7226,8156,8160,8412,8418,8672,8744,9179,9273,10738', 'Unknown','NBDK']

新结果中唯一的区别是只将每条记录的最后一项从['Unknown'，'NBDK']更改为'Unknown'，'NBDK'

有没有办法通过修改我代码中的这一行来做到这一点：

lc_allcancelmembers = [[record.get("Contact.Id"), convert_is_datetime(str(record.get("DateCreated"))), record.get("Contact.OwnerID"), record.get("Contact.Groups"), [k for k, v in tagsDic.items() if v in (record.get("Contact.Groups"))]] for record in allcancelledmembers]

如果没有，实现这一目标的最佳方式是什么

---

谢谢

由于我不认识您正在使用的模块，因此这有点不规范。但仍然有效

import datetime

def cleanContacts(A):
    return A[:-1] + A[-1]

A = [68505, datetime.datetime(2019, 1, 9, 9, 10, 52), 384114, '130,134,136,138,152,206,208,286,290,312,436,493,495,728,760,772,774,788,952,968,1088,1226,1254,2362,2584,2598,2972,2974,3102,3560,4552,4554,5430,6566,6950,7226,8156,8160,8412,8418,8672,8744,9179,9273,10738', ['Unknown', 'NBDK']]
C = [68505, datetime.datetime(2019, 1, 9, 9, 10, 52), 384114, '130,134,136,138,152,206,208,286,290,312,436,493,495,728,760,772,774,788,952,968,1088,1226,1254,2362,2584,2598,2972,2974,3102,3560,4552,4554,5430,6566,6950,7226,8156,8160,8412,8418,8672,8744,9179,9273,10738', 'Unknown','NBDK']
assert cleanContacts(A) == C

另外，如果我正确地猜测要弹出打开的元素来自最后一个列表理解，那么这颗星将起作用

lc_allcancelmembers = [[record.get("Contact.Id"), convert_is_datetime(str(record.get("DateCreated"))), record.get("Contact.OwnerID"), record.get("Contact.Groups"), *[k for k, v in tagsDic.items() if v in (record.get("Contact.Groups"))]] for record in allcancelledmembers]

您可以创建一个通用（递归）函数来展平嵌套列表，如下图所示，并使用它展平您得到的列表

def flatten_list(lst):
    for member in lst:
        if isinstance(member, list):
            for mem in flatten_list(member):
                yield mem
        else:
            yield member

if __name__ == '__main__':

    import datetime
    lst = [68505, datetime.datetime(2019, 1, 9, 9, 10, 52), 384114, '130,134,136,138,152,206,208,286,290,312,436,493,495,728,760,772,774,788,952,968,1088,1226,1254,2362,2584,2598,2972,2974,3102,3560,4552,4554,5430,6566,6950,7226,8156,8160,8412,8418,8672,8744,9179,9273,10738', ['Unknown', 'NBDK']]

    print(list(flatten_list(lst)))

给出：

[68505, datetime.datetime(2019, 1, 9, 9, 10, 52), 384114, '130,134,136,138,152,206,208,286,290,312,436,493,495,728,760,772,774,788,952,968,1088,1226,1254,2362,2584,2598,2972,2974,3102,3560,4552,4554,5430,6566,6950,7226,8156,8160,8412,8418,8672,8744,9179,9273,10738', 'Unknown', 'NBDK']

---

谢谢你@JimmyCarlos！并非每个记录都有两个值，如['Unknown'，'NBDK']。如果最后一项有两个值，有没有办法只获取最后一项？这样：:[k代表k，v在tagsDic.items（）中，如果v在（record.get（“Contact.Groups”）][-1]]中，这给了我一个列表索引超出范围的错误，所以我只是想知道这是否可能。如果您只想要最后一个值，您可以像[k代表k，v在tagsDic.items（）中，如果v在（record.get（“Contact.Groups”）][-1]中一样，不需要*作为[-1]已断开列表以获取最后一个元素。这将导致列表索引超出范围错误。我尝试了[k代表k，v在tagsDic.items（）中，如果v在（record.get（“Contact.Groups”）][-1]]中代表所有取消成员中的record]。去掉最后一个方括号，我会看到未解决的引用错误，最好开始把它分成两行。将主要内容放在第1行，然后在第二行做列表+=[k代表k，v….]，将两个列表连接在一起。

def flatten_list(lst):
    for member in lst:
        if isinstance(member, list):
            for mem in flatten_list(member):
                yield mem
        else:
            yield member

if __name__ == '__main__':

    import datetime
    lst = [68505, datetime.datetime(2019, 1, 9, 9, 10, 52), 384114, '130,134,136,138,152,206,208,286,290,312,436,493,495,728,760,772,774,788,952,968,1088,1226,1254,2362,2584,2598,2972,2974,3102,3560,4552,4554,5430,6566,6950,7226,8156,8160,8412,8418,8672,8744,9179,9273,10738', ['Unknown', 'NBDK']]

    print(list(flatten_list(lst)))

[68505, datetime.datetime(2019, 1, 9, 9, 10, 52), 384114, '130,134,136,138,152,206,208,286,290,312,436,493,495,728,760,772,774,788,952,968,1088,1226,1254,2362,2584,2598,2972,2974,3102,3560,4552,4554,5430,6566,6950,7226,8156,8160,8412,8418,8672,8744,9179,9273,10738', 'Unknown', 'NBDK']