python - 如何将第一个字符串与列匹配并打印匹配？

Question

我的数据框是

data = {
        'company_name' : ['auckland suppliers', 'Octagone', 'SodaBottel','Shimla Mirch'],
        'year' : [2000, 2001, 2003, 2004],
        'desc' : [' auckland has some good reviews','Octagone','we shall update you','we have varities of shimla mirch'],
}

df = pd.DataFrame(data)

我试过这段代码

df['CompanyMatch'] = df ['company_name'] == df ['desc']

如果 company_name 列的第一个单词与 desc 列匹配，我想打印“匹配”。我很困惑将索引 [0] 放在哪里以便它以这种方式打印:

> company_name         desc                                 CompanyMatch
> auckland suppliers   auckland has some good reviews       Match
> Octagone             Octagone                             Match
> SodaBottel           we shall update you                  NA
> Shimla Mirch         we have varities of shimla mirch     Match

Answer 1

您可以使用 numpy.where与 apply对于通过 in 检查另一列值，axis=1 用于按行处理:

import numpy as np

m = df.apply(lambda x: x['company_name'].lower() in x['desc'].lower(), axis=1)
df['CompanyMatch'] = np.where(m, 'Match', np.nan)
print (df)
         company_name                              desc  year CompanyMatch
0  auckland suppliers    auckland has some good reviews  2000          nan
1            Octagone                          Octagone  2001        Match
2          SodaBottel               we shall update you  2003          nan
3        Shimla Mirch  we have varities of shimla mirch  2004        Match

编辑:

只比较第一个词:

m = df.apply(lambda x: x['company_name'].split()[0].lower() in x['desc'].lower(), axis=1)
df['CompanyMatch'] = np.where(m, 'Match', np.nan)
print (df)
         company_name                              desc  year CompanyMatch
0  auckland suppliers    auckland has some good reviews  2000        Match
1            Octagone                          Octagone  2001        Match
2          SodaBottel               we shall update you  2003          nan
3        Shimla Mirch  we have varities of shimla mirch  2004        Match