pandas：用另一個字符串替換字符串

Question

我有以下數據框

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

我想用responsive替換respon和r ，所以最終的數據框是

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

我嘗試了以下但沒有奏效：

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

Answer 1

用dictionary replace的解決方案：

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

如果需要將列中的所有值設置為某個string ：

df['prod_type'] = 'responsive' 

Answer 2

您不需要在此處傳遞regex=True ，因為這將查找部分匹配，因為您在精確匹配之后只需將參數作為單獨的參數傳遞：

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

其他解決方案，以防df['prod_type']所有項目都相同：

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，您可以使用帶有 lambda 語法的 apply 函數

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

Answer 5

按照 jezrael 的回答，您可以設置 inplace inplace=True<\/code>來更改數據框：

df = pd.DataFrame(
    {'prod_type':['responsive','responsive','respon','r','respon','r','responsive']},
    columns=['prod_type'])

df.replace({'prod_type': {'respon': 'responsive', 'r': 'responsive'}}, inplace=True)
df
    prod_type
0  responsive
1  responsive
2  responsive
3  responsive
4  responsive
5  responsive
6  responsive