解决Pandas中apply与lambda引发的TypeError: string indices must be integers错误

我有如下Pandas DataFrame，需求为：当CompanyAd列包含"OTC+"或"OTC +"时，将Name列对应值修改为"ASD"。但运行以下代码时出现错误TypeError: string indices must be integers，请问该如何解决？

DataFrame定义代码 #

import pandas as pd
from pandas import Timestamp

dictA = {'Event_ID': {0: 'G-00001',   1: 'G-00002 ',   2: 'G-00003',   3: 'G-00004',   4: 'G-00005',   5: 'G-00006',   6: 'G-00007',   7: 'G-00008',   8: 'G-00009',   9: 'G-00010',   10: 'G-00011',   11: 'G-00012',   12: 'G-00013',   13: 'G-00014',   14: 'G-00015',   15: 'G-00016',   16: 'G-00017'},
 'Name': {0: 'ABC',   1: 'CSA',   2: 'CSA',   3: 'VSX',   4: 'ABC',   5: 'ABC',   6: 'CSA',   7: 'ABC',   8: 'VSX',   9: 'CSA',   10: 'VSX',   11: 'ABC',   12: 'VSX',   13: 'VSX',   14: 'ABC',   15: 'ABC',   16: 'CSA'},
 'CompanyAd      ': {0: '51Job, Inc.Cayman Islands NMS              ',   1: "724 Solution's Inc. Canada NMS             ",   2: 'A B SKF Sweden OTC+              ',   3: 'A/S Steamship Company Torm Denmark"s NMS OTC+       ',   4: 'ABB Ltd. Switzerland OTC+             ',   5: 'Aber Diamond Ltd. Canada CAP MKT                    ',   6: 'Abitibi Consolidated Inc. Canada OTC +              ',   7: 'ABN Amro Bank N.V. Netherlands AMEX - Preferred OTC+',   8: 'ABN Amro Holding N.V. Netherlands NYSE              ',   9: 'Acambis plc United Kingdom OTC +              ',   10: "Ace Aviation Holdings'aed Inc. Canada OTC           ",   11: 'Acetex Corp. Canada OTC - Debt+              ',   12: 'Acrex Ventures, Ltd. Canada OTC+             ',   13: 'ACS-Tech 80 Ltd. Israel CAP MKT              ',   14: 'Actions Semiconductor Co. Ltd. Cayman Islands NMS   ',   15: 'Adastra Minerals Inc. Canada OTC*              ',   16: 'ADB Systems International Inc. Canada OTC           '},
 'ticket': {0: 671,   1: 5,   2: 5,   3: 23,   4: 4,   5: 60,   6: 60,   7: 89,   8: 0,   9: 6,   10: 3,   11: 2,   12: 4,   13: 32,   14: 3,   15: 1,   16: 23},
 'Revenue': {0: 6720,   1: 56,   2: 78,   3: 34,   4: 89,   5: 73,   6: 345,   7: 890,   8: 0,   9: 45,   10: 39,   11: 34,   12: 89,   13: 127,   14: 84,   15: 100,   16: 525},
 'Expences': {0: 150.0,   1: 18.0,   2: 38.0,   3: 23.0,   4: 150.0,   5: 55.0,   6: 110.0,   7: 150.0,   8: 0.0,   9: 16.0,   10: 23.0,   11: 150.0,   12: 48.0,   13: 35.0,   14: 55.0,   15: 150.0,   16: float('nan')},
 'expect': {0: 50.0,   1: 100.0,   2: 100.0,   3: float('nan'),   4: 40.0,   5: 60.0,   6: 60.0,   7: float('nan'),   8: 50.0,   9: 60.0,   10: 30.0,   11: 20.0,   12: 40.0,   13: 10.0,   14: 120.0,   15: 140.0,   16: 90.0},
 'Signed_Date': {0: Timestamp('2021-06-01 00:00:00'),   1: Timestamp('2021-06-05 00:00:00'),   2: Timestamp('2021-06-03 00:00:00'),   3: Timestamp('2021-06-03 00:00:00'),   4: Timestamp('2021-06-02 00:00:00'),   5: Timestamp('2021-04-15 00:00:00'),   6: Timestamp('2021-06-12 00:00:00'),   7: Timestamp('2021-06-02 00:00:00'),   8: Timestamp('2021-04-30 00:00:00'),   9: Timestamp('2021-06-22 00:00:00'),   10: Timestamp('2021-06-10 00:00:00'),   11: Timestamp('2021-06-03 00:00:00'),   12: Timestamp('2021-06-12 00:00:00'),   13: Timestamp('2021-04-24 00:00:00'),   14: Timestamp('2021-04-21 00:00:00'),   15: Timestamp('2021-06-07 00:00:00'),   16: Timestamp('2021-04-02 00:00:00')}}
df = pd.DataFrame.from_dict(dictA)

报错的运行代码 #

df['account_name_e']=df['CompanyAd      '].apply(str).str.replace(" ","")
df['account_name_e'] = df['account_name_e'].apply(str).str.replace("(?i)[^0-9a-z+]",'')
df['NameE'] = df['Name']
df['NameE'] = df['NameE'].apply(lambda row: str('ASD') if 'OTC+' in str(row['account_name_e']) else row)

df

错误原因 #

df['NameE'].apply()中的lambda函数接收的是**NameE列的单个字符串元素**，而非整行数据。此时尝试用row['account_name_e']去索引字符串，自然会触发string indices must be integers错误——字符串只能用整数下标访问，不能用字典键。

解决方案 #

方法一：按行使用df.apply() #

如果需要同时访问多列数据，可以用df.apply()并指定axis=1按行处理，此时lambda接收的是整行数据：

# 预处理CompanyAd列，生成account_name_e
df['account_name_e'] = df['CompanyAd      '].str.replace(" ", "").str.replace("(?i)[^0-9a-z+]", '')
# 按行判断并赋值
df['NameE'] = df.apply(lambda row: 'ASD' if 'OTC+' in row['account_name_e'] else row['Name'], axis=1)

方法二：布尔索引直接修改（推荐，更高效） #

Pandas的布尔索引比apply()性能更优，适合大数据集。直接用正则匹配原列生成掩码，再批量修改：

# 生成匹配掩码：匹配"OTC+"或"OTC +"（\s*匹配任意数量的空格）
mask = df['CompanyAd      '].str.contains(r'OTC\s*\+', regex=True, na=False)
# 初始化NameE列为Name的副本
df['NameE'] = df['Name'].copy()
# 对符合条件的行赋值为"ASD"
df.loc[mask, 'NameE'] = 'ASD'

这种方法无需生成中间列account_name_e，直接匹配原列即可完成需求，代码更简洁高效。

内容的提问来源于stack exchange，提问作者rra