基于Python中的另一列更新列的值

Question

我有一个作为 df 的数据框：

现在基于第一列 ['Message'] 上的字符串处理函数，test(l1,l2) 我得到一个像 rect_d{} 这样的字典：

{'ISIN': ':',
 'ISIN :': 'KE5000008986',
 'SETTLEMENT DATE': '-',
 'SETTLEMENT DATE -': '06/01/2020',
 'TRADE DATE': '-'}

基于dict的值，我打算更新不同列上同一行的相似信息，比如基于结算日期的最后一个值，我想更新同一行的结算日期。

我已经相应地编写了我的代码：

for idx, row in df.iterrows():
    split_t=df['Message'][idx].split()
    ret_d= test(items, split_t)
    for key in ret_d:
        print(key)
        if key=='SETTLEMENT DATE' or 'SETTLEMENT DATE:' or 'Settlement date :' or 'SETTLEMENT DATE -':
            df.loc[idx,'SETTLEMENT DATE']=ret_d[key]
        elif key=='ISIN:' or 'ISIN' or 'ISIN :':
            df.loc[idx,'ISIN']=ret_d[key]
        elif key=='CASH ACCOUNT':
            df.loc[idx,'CASH ACCOUNT']=ret_d[key]
        else: print('done')
    break

理想情况下应该用这些值填充我的 df。但它不仅填充了第一个值。

我做不到，因为价值观不是那么直截了当：

df['key']=df['rect_d'].apply(lambda x:rect_d[x])

我在更新 df 时有什么遗漏吗？任何见解都受到高度赞赏。

df.to_dict() 格式供参考：

{'CASH ACCOUNT': {0: None, 1: None, 2: None, 3: None},
 'CLEARING BIC': {0: None, 1: None, 2: None, 3: None},
 'CLIENT NAME': {0: None, 1: None, 2: None, 3: None},
 'ISIN': {0: None, 1: None, 2: None, 3: None},
 'MARKET': {0: None, 1: None, 2: None, 3: None},
 'Message': {0: '{1:F01SCBLMUMUBSSU0422020871}{2:O5990735200109UBSWUS33BSCI12525443762001091635N}{3:{108:AMDBHDVHU7RUGRHA}}{4:\n:20:WICTIN010Q55PK19\n:79:ACCOUNT/2EVERECFNG\nSENDER/WICTIN01/Ravali Nandi\nRECEIVER/SCB Mauritius\nHI TEAM,\n.\nCAN YOU PLEASE PROCESS BELOW\nTRADE AS RECEIVE FREE OF PAYMENT\nAS PER BELOW DETAILS\n.\nTRADE DETAILS\nSAFE ACC NO : 045891400139\nQUANTITY : 2,840,100\nISIN : KE5000008986\nPSET - XNAIKEN1\nDEAG - SCBLKENX\n70E - SHARE RECEIVED FROM RWANDA TO KENYA\nTRADE DATE - 06/01/2020\nSETTLEMENT DATE - 06/01/2020\nRegards\nRavali\n-}{5:{CHK:97709F6E1473}}',
  1: '{1:F01SCBLMUMUXSSU0092740678}{2:O5991259200113SCBLKENXXSSU00927406782001131359N}{4:\n:20:CANC\n:79:KE0022282/MK1\n.\nKINDLY CANCEL\nMT543\nREF KE0022282/MK1\n.\nKIND\nREGARDS,\nSCB SECURITIES SERVICES\nKENYA\n-}',
  2: '{1:F01SCBLMUMUXXXX0092826091}{2:O5992027200114SCBLHKHHXXXX00928260912001141627N}{3:{108:UIGMU18026A00799}}{4:\n:20:LEKO14012020RVP\n:21:UIGMU18026A00799\n:79:Safekeep Account : 3CEXOCUSCL                    \nPlease accept the below trade instruction TYPE RVP\nPORTFOLIO ACCOUNT - 045887500023 ISIN KYG5462G107\n3 SECURITY NAME - LEKOIL LTD TRADE DATE 14 JANUARY\n2020 SETTLEMENT DATE 16 JANUARY 2020 QUANTITY - 1\nSETTLEMENT AMOUNT - 7.73 GBP COUNTERPARTY BIC PAR\nBGB2L CREST CODE - FFQAQ SELLER BIC - CVGXUS33 STA\nMP DUTY - GBOX PSET - CRSTGB22\n-}',
  3: '{1:F01SCBLMUMUXXXX0088474512}{2:O5992028200114SCBLHKHHXXXX00884745122001141628N}{3:{108:UIGMU18026A00800}}{4:\n:20:LEKO14012020DVP\n:21:UIGMU18026A00800\n:79:Safekeep Account : 3CEXOCUSCL                    \nPlease accept the below trade instruction TYPE DVP\nPORTFOLIO ACCOUNT - 045887500023 ISIN KYG5462G107\n3 SECURITY NAME - LEKOIL LTD TRADE DATE 14 JANUARY\n2020 SETTLEMENT DATE 16 JANUARY 2020 QUANTITY - 1\nSETTLEMENT AMOUNT - 7.66 GBP COUNTERPARTY BIC PAR\nBGB2L CREST CODE - FFQAQ SELLER BIC - CVGXUS33 STA\nMP DUTY - GBOX PSET - CRSTGB22\n-}'},
 'New Context': {0: 'SSTM', 1: 'SSTM', 2: 'SSTM', 3: 'SSTM'},
 'PRICE CFA': {0: None, 1: None, 2: None, 3: None},
 'SAFEKEEP ACCOUNT': {0: None, 1: None, 2: None, 3: None},
 'SCA': {0: None, 1: None, 2: None, 3: None},
 'SECURITY NAME': {0: None, 1: None, 2: None, 3: None},
 'SETTLEMENT DATE': {0: '-', 1: None, 2: None, 3: None},
 'TRADE DATE': {0: None, 1: None, 2: None, 3: None},
 'TRADE TYPE': {0: None, 1: None, 2: None, 3: None}}

项目：

items=['SETTLEMENT DATE', 'SETTLEMENT DATE:','Settlement date :','SETTLEMENT DATE -', 'CASH ACCOUNT', 'CASH ACCOUNT:',
       'ISIN:',  'ISIN', 'ISIN :',
       'TRADE DATE','TRADE DATE:']

Answer 1

DataFrame.update

我们可以通过迭代消息列中的字符串并使用test 函数将每个字符串映射到dictionary 来创建记录，然后我们可以从这些记录中创建一个新的数据帧并更新原始数据帧中的值这个新创建的数据框

df.update(pd.DataFrame([test(items, msg.split()) for msg in df['Message']], index=df.index))