將列值的字符替換為Pandas中另一列的字符串

Question

name value_1
dd3   what, _ is
dd4   what, _ is

如何用名稱列中的整個字符串替換value_1列中的'_'？

value_1列的所需輸出

value_1
what, dd3 is
what, dd4 is

我已經嘗試過：

df['value_1'] = df['value_1'].apply(lambda x:x.replace("_", df['name']))

我得到了這個error :expected a string or other character buffer object

Answer 1

將apply with axis=1用於按行處理：

df['value_1'] = df.apply(lambda x:x['value_1'].replace("_", x['name']), axis=1)
print (df)
  name       value_1
0  dd3  what, dd3 is
1  dd4  what, dd4 is

Answer 2

更新：類似於@jezrael的解決方案，但對於較大的數據集（矢量化方法）應該更快一些：

In [221]: df['value_1'] = (df.groupby('name')['value_1']
                             .transform(lambda x: x.str.replace('_', x.name)))

In [222]: df
Out[222]:
  name       value_1
0  dd3  what, dd3 is
1  dd4  what, dd4 is

舊答案：

您可以創建一個幫助DF：

In [181]: x = df.value_1.str.split('_', expand=True)

In [192]: x
Out[192]:
        0    1
0  what,    is
1  what,    is

然后在其中插入新列：

In [182]: x.insert(1, 'name', df['name'])

產生：

In [194]: x
Out[194]:
        0 name    1
0  what,   dd3   is
1  what,   dd4   is

並替換原始列：

In [183]: df['value_1'] = x.sum(1)

In [184]: df
Out[184]:
  name       value_1
0  dd3  what, dd3 is
1  dd4  what, dd4 is