Python 熊猫：用另一个字符串替换字符串

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

，因此最终的数据帧是

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

我尝试了以下方法，但无效：

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

---

您不需要在此处传递

regex=True

，因为这将查找部分匹配，因为您需要精确匹配，只需将参数作为单独的参数传递：

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

---

您不需要在此处传递

regex=True

，因为这将查找部分匹配，因为您需要精确匹配，只需将参数作为单独的参数传递：

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

通过

字典

解决方案：

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

如果需要将列中的所有值设置为某个

字符串

：

df['prod_type'] = 'responsive' 

通过

字典

解决方案：

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

如果需要将列中的所有值设置为某个

字符串

：

df['prod_type'] = 'responsive' 

---

如果

df['prod_type']

中的所有项目都相同，则其他解决方案如下：

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

---

如果

df['prod_type']

中的所有项目都相同，则其他解决方案如下：

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

---

或者，您可以将apply函数与lambda语法一起使用

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

---

或者，您可以将apply函数与lambda语法一起使用

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