Question

我有一个数据框，我想从中提取数字，如果'transfer'字是在8列上，它应该从位置13、15字符中提取出来，否则应该从第21位15个字符中提取

 SELECT IFNULL(SUM(ri.price),0) AS total, m.month
 FROM (
       SELECT 'Jan' AS MONTH
       UNION SELECT 'Feb' AS MONTH
       UNION SELECT 'Mar' AS MONTH
       UNION SELECT 'Apr' AS MONTH
       UNION SELECT 'May' AS MONTH
       UNION SELECT 'Jun' AS MONTH
       UNION SELECT 'Jul' AS MONTH
       UNION SELECT 'Aug' AS MONTH
       UNION SELECT 'Sep' AS MONTH
       UNION SELECT 'Oct' AS MONTH
       UNION SELECT 'Nov' AS MONTH
       UNION SELECT 'Dec' AS MONTH
      ) AS m
LEFT JOIN table_u pu 
ON MONTH(STR_TO_DATE(CONCAT(pu.created_date, '2019'),'%M %Y')) = MONTH(pu.created_date)
AND YEAR(pu.created_date) = '2019'
LEFT JOIN table ri 
ON ri.id = pu.id
GROUP BY m.month
ORDER by 1+1;

Answer 1

尝试一下：

import pandas as pd
import numpy as np

df['new_extract_column'] = np.nan

df.loc[ df['column8'].str.contains('transfer'), 'new_extract_column' ] = df[ df['column8'].str.contains('transfer') ].apply(lambda x: x[13:16])

df.loc[ ~df['column8'].str.contains('transfer'), 'new_extract_column' ] = df[ df['column8'].str.contains('transfer') ].apply(lambda x: x[15: 22])

Answer 2

如果您的数据后有空格，则可以使用分割字符串方法。否则您可以使用正则表达式

df = pd.DataFrame([['On-Line Transfer -  01901091900014'], ['On-Line Transfer -  02501091900004'], ['On-Line Transfer -  03601091900018'], ['Transfer -  03631081900095'], ['Transfer -  03829081900083']], columns=['Particular'])
df["Particular"].str.split().str[-1]

从熊猫中提取字符串

2 个答案: