pandas：用另一个字符串替换字符串

Question

我有以下数据框

    prod_type
0   responsive
1   responsive
2   respon
3   r
4   respon
5   r
6   responsive

我想将respon和r替换为responsive，因此最终数据框为

    prod_type
0   responsive
1   responsive
2   responsive
3   responsive
4   responsive
5   responsive
6   responsive

我尝试了以下但是没有用：

df['prod_type'] = df['prod_type'].replace({'respon' : 'responsvie'}, regex=True)
df['prod_type'] = df['prod_type'].replace({'r' : 'responsive'}, regex=True)

Answer 1

dictionary df['prod_type'] = df['prod_type'].replace({'respon':'responsive', 'r':'responsive'}) print (df) prod_type 0 responsive 1 responsive 2 responsive 3 responsive 4 responsive 5 responsive 6 responsive 的解决方案：

string

如果需要将列中的所有值设置为某些df['prod_type'] = 'responsive' ：

{{1}}

Answer 2

你不需要在这里传递regex=True，因为这将寻找部分匹配，因为你在精确匹配之后只是将params作为单独的args传递：

In [7]:
df['prod_type'] = df['prod_type'].replace('respon' ,'responsvie')
df['prod_type'] = df['prod_type'].replace('r', 'responsive')
df

Out[7]:
    prod_type
0  responsive
1  responsive
2  responsvie
3  responsive
4  responsvie
5  responsive
6  responsive

Answer 3

如果df['prod_type']中的所有项目都相同，则为其他解决方案：

df['prod_type'] = ['responsive' for item in df['prod_type']]
In[0]: df
Out[0]:
prod_type
0  responsive
1  responsive
2  responsive
3  responsive
4  responsive
5  responsive
6  responsive

Answer 4

或者，您可以将应用功能与lambda语法一起使用

df['prod_type'] = df['prod_type'].apply(lambda x: x.replace('respon', 'responsvie'))