name value_1
dd3 what, _ is
dd4 what, _ is
How to replace '_' from value_1 column with whole string from name column?
Desired output for value_1 column
value_1
what, dd3 is
what, dd4 is
I have tried with this:
df['value_1'] = df['value_1'].apply(lambda x:x.replace("_", df['name']))
And I got this error :expected a string or other character buffer object
Use apply with axis=1 for process by rows:
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
UPDATE: similar to @jezrael's solution, but it should be bit faster for larger data sets (vectorized approach):
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
Old answer:
you can create a helper DF:
In [181]: x = df.value_1.str.split('_', expand=True)
In [192]: x
Out[192]:
0 1
0 what, is
1 what, is
then insert a new column into it:
In [182]: x.insert(1, 'name', df['name'])
which yields:
In [194]: x
Out[194]:
0 name 1
0 what, dd3 is
1 what, dd4 is
and replace the original column:
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
