如何使用st.contains将文本从旧列移动到新创建的列

下面是一种使用带有命名捕获组的正则表达式的简单方法:

df = pd.DataFrame({'Desc': ['The table is long', 'The fan is nice', 'The fan is cheap', 'The chair is brown']})
words = ['table', 'fan', 'chair']

regex = '|'.join(f'(?P<{w}>.*{w}.*)' for w in words)
df.join(df['Desc'].str.extract(regex, expand=False).add_prefix('keyword_'))

注意。命名的捕获组不能有特殊字符或空格。如果是这种情况,请告诉我,并且可以更改捕获组的名称。 输出:

                 Desc      keyword_table       keyword_fan       keyword_chair
0   The table is long  The table is long               NaN                 NaN
1     The fan is nice                NaN   The fan is nice                 NaN
2    The fan is cheap                NaN  The fan is cheap                 NaN
3  The chair is brown                NaN               NaN  The chair is brown

其他选择 get_dummies

df = pd.DataFrame({'Desc': ['The table is long', 'The fan is nice', 'The fan is cheap', 'The chair is brown']})
words = ['table', 'fan', 'chair']

regex = '(%s)' % '|'.join(words)
df.join(pd.get_dummies(df['Desc'].str.extract(regex, expand=False))
          .mul(df['Desc'], axis=0)
          .add_prefix('keyword_')
        )

您的布尔数列可以用作索引来分割数据帧,如下所示:

df['Keyword Table'] = df[df['Description'].str.contains('Table', na = False)]['Description']

要查看关键字列表,可以使用apply:

keywords = ['Table', 'Fan', 'Chair']

df['Keywords'] = df[df['Description'].apply(lambda x: any(k in x for k in keywords))]['Description']

这段代码有用吗?

df = pd.DataFrame({'Desc':['cat is black','dog is white']})
kw = ['cat','dog']
for k in kw:
   df[k + ' col'] = df.Desc.map(lambda s: s if k in s else '' )

输出是