从数据框中提取与列表进行比较的字符串

你可以用

extract

join

List

|

or

regex

List1 = ['Three has' , 'Mail' , 'Done' , 'Game' , 'Time has come']df['Col 3'] = df['Col 2'].str.extract("(" + "|".join(List1) +")", expand=False)print (df)   Col 1Col 2      Col 30      1        The date        NaN1      2  Three has come  Three has2      3       Mail Sent       Mail3      4       Done Deal       Done

另一个解决方案：

List1 = ['Three has' , 'Mail' , 'Done' , 'Game' , 'Time has come']df['Col 3'] = df['Col 2'].apply(lambda x: ''.join([L for L in List1 if L in x]))df['Col 3'] = df['Col 3'].mask(df['Col 3'] == '')print (df)   Col 1Col 2      Col 30      1        The date        NaN1      2  Three has come  Three has2      3       Mail Sent       Mail3      4       Done Deal       Done