如何將第一個字符串與列匹配並打印匹配項？

Question

我的數據框是

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)

我嘗試了這段代碼

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

如果company_name列的第一個單詞與desc列匹配，我想打印“匹配”。我對放置索引[0]的位置感到困惑，因此它的打印方式如下：

> 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

您可以將numpy.where與apply一起apply ，以in檢查一個列的值， axis=1用於按行處理：

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

編輯：

僅用於比較第一個單詞：

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