我有一个如下数据框，我想删除方括号和单引号（'）和逗号。
id currentTitle1
1 ['@@@0000070642@@@']
2 ['@@@0000082569@@@']
3 ['@@@0000082569@@@']
4 ['@@@0000082569@@@']
5 ['@@@0000060910@@@', '@@@0000039198@@@']
6 ['@@@0000060910@@@']
7 ['@@@0000129849@@@']
8 ['@@@0000082569@@@']
9 ['@@@0000082569@@@', '@@@0000060905@@@', '@@@0000086889@@@']
10 ['@@@0000082569@@@']
我想要输出如下
id currentTitle1
1 @@@0000070642@@@
2 @@@0000082569@@@
3 @@@0000082569@@@
4 @@@0000082569@@@
5 @@@0000060910@@@ @@@0000039198@@@
6 @@@0000060910@@@
7 @@@0000129849@@@
8 @@@0000082569@@@
9 @@@0000082569@@@ @@@0000060905@@@ @@@0000086889@@@
10 @@@0000082569@@@
我从正则表达式清理操作中获取数据为df['currentTitle']=df['currentTitle'].str.findall(r'@{3}\d+@{3}')
编辑：发布不干净的数据。请记住，还有空行也没有包含
id currentTitle currentTitle_unclean
1 @@@0000070642@@@ accompanying functions of @@@0000070642@@@ and business risk assessment - director
2 @@@0000082569@@@ account @@@0000082569@@@ - sales agent /representative at pronovias fashion group
3 @@@0000082569@@@ account manager/product @@@0000082569@@@ - handbags and accessories
4 @@@0000082569@@@ account @@@0000082569@@@ for entrepreneurs and small size companies
5 @@@0000060910@@@ @@@0000039198@@@ academic @@@0000060910@@@ , administrative, and @@@0000039198@@@ liaison coordinator
6 @@@0000060910@@@ account executive at bluefin insurance @@@0000060910@@@ limited
7 @@@0000129849@@@ account executive for interior @@@0000129849@@@ magazine inex
8 @@@0000082569@@@ account @@@0000082569@@@ high potential secondment programme
9 @@@0000082569@@@ @@@0000060905@@@ @@@0000086889@@@ account @@@0000082569@@@ @@@0000060905@@@ -energy and commodities @@@0000086889@@@ candidate
10 @@@0000082569@@@ account @@@0000082569@@@ paints, coatings, adhesives - ser, slo, cro
答案 0 :(得分：4)
您可以apply使用join：
df['currentTitle1'] = df['currentTitle1'].apply(' '.join)
print (df)
id currentTitle currentTitle_unclean \
0 1 @@@0000070642@@@ accompanying functions of @@@0000070642@@@ and...
1 2 @@@0000082569@@@ account @@@0000082569@@@ - sales agent /repres...
2 3 @@@0000082569@@@ account manager/product @@@0000082569@@@ - han...
3 4 @@@0000082569@@@ account @@@0000082569@@@ for entrepreneurs and...
4 5 @@@0000060910@@@ @@@0000039198@@@ academic @@@0000060910@@@ ,...
5 6 @@@0000060910@@@ account executive at bluefin insurance @@@0000...
6 7 @@@0000129849@@@ account executive for interior @@@0000129849@@...
7 8 @@@0000082569@@@ account @@@0000082569@@@ high potential second...
8 9 @@@0000082569@@@ @@@0000060905@@@ @@@0000086889@@@ account @@@...
9 10 @@@0000082569@@@ account @@@0000082569@@@ paints, coatings, adh...
currentTitle1
0 @@@0000070642@@@
1 @@@0000082569@@@
2 @@@0000082569@@@
3 @@@0000082569@@@
4 @@@0000039198@@@ @@@0000060910@@@ @@@000003919...
5 @@@0000060910@@@
6 @@@0000129849@@@
7 @@@0000082569@@@
8 @@@0000060905@@@ @@@0000086889@@@ @@@000008256...
9 @@@0000082569@@@
或者如上所述not_a_robot：
df['currentTitle1'].map(lambda x: ' '.join(x))
如果错误：
TypeError：只能加入可迭代的
然后可以添加条件，如果没有列表让原始值：
df['currentTitle1'] = df['currentTitle1'].apply(lambda x: ' '.join(x) if type(x) == list
else x)
或创建空字符串：
df['currentTitle1'] = df['currentTitle1'].apply(lambda x: ' '.join(x) if type(x) == list
else '')
答案 1 :(得分：1)
这适用于我的机器，同时也创建了dataframe：
import pandas as pd
import re
data = ['accompanying functions of @@@0000070642@@@ and business risk assessment - director',
'account @@@0000082569@@@ - sales agent /representative at pronovias fashion group',
'account manager/product @@@0000082569@@@ - handbags and accessories',
'account @@@0000082569@@@ for entrepreneurs and small size companies',
'academic @@@0000060910@@@ , administrative, and @@@0000039198@@@ liaison coordinator',
'account executive at bluefin insurance @@@0000060910@@@ limited',
'account executive for interior @@@0000129849@@@ magazine inex',
'account @@@0000082569@@@ high potential secondment programme',
'account @@@0000082569@@@ @@@0000060905@@@ -energy and commodities @@@0000086889@@@ candidate',
'account @@@0000082569@@@ paints, coatings, adhesives - ser, slo, cro']
df = pd.DataFrame({'currentTitle_unclean': data})
df['currentTitle'] = df['currentTitle_unclean'].apply(lambda x: ' '.join(re.findall(r'@{3}\d+@{3}', x)))