pandas apply custom function to every row of one column grouped by another column - python-3.x

Question

I have a dataframe containing two columns: id and val.
df = pd.DataFrame ({'id': [1,1,1,2,2,2,3,3,3,3], 'val' : np.random.randn(10)})
id val
0 1 2.644347
1 1 0.378770
2 1 -2.107230
3 2 -0.043051
4 2 0.115948
5 2 0.054485
6 3 0.574845
7 3 -0.228612
8 3 -2.648036
9 3 0.569929
And I want to apply a custom function to every val according to id. Let's say I want to apply min-max scaling. This is how I would do it using a for loop:
df['scaled']=0
ids = df.id.drop_duplicates()
for i in range(len(ids)):
df1 = df[df.id==ids.iloc[i]]
df1['scaled'] = (df1.val-df1.val.min())/(df1.val.max()-df1.val.min())
df.loc[df.id==ids.iloc[i],'scaled'] = df1['scaled']
And the result is:
id val scaled
0 1 0.457713 1.000000
1 1 -0.464513 0.000000
2 1 0.216352 0.738285
3 2 0.633652 0.990656
4 2 -1.099065 0.000000
5 2 0.649995 1.000000
6 3 -0.251099 0.306631
7 3 -1.003295 0.081387
8 3 2.064389 1.000000
9 3 -1.275086 0.000000
How can I do this faster without a loop?

Answer 1

You can do this with groupby:
In [6]: def minmaxscale(s): return (s - s.min()) / (s.max() - s.min())
In [7]: df.groupby('id')['val'].apply(minmaxscale)
Out[7]:
0 0.000000
1 1.000000
2 0.654490
3 1.000000
4 0.524256
5 0.000000
6 0.000000
7 0.100238
8 0.014697
9 1.000000
Name: val, dtype: float64
(Note that np.ptp() / peak-to-peak can be used in placed of s.max() - s.min().)
This applies the function minmaxscale() to each smaller-sized Series of val, grouped by id.
Taking the first group, for example:
In [11]: s = df[df.id == 1]['val']
In [12]: s
Out[12]:
0 0.002722
1 0.656233
2 0.430438
Name: val, dtype: float64
In [13]: s.max() - s.min()
Out[13]: 0.6535106879021447
In [14]: (s - s.min()) / (s.max() - s.min())
Out[14]:
0 0.00000
1 1.00000
2 0.65449
Name: val, dtype: float64

Answer 2

Solution from sklearn MinMaxScaler
from sklearn.preprocessing import MinMaxScaler
scaler = MinMaxScaler()
df['new']=np.concatenate([scaler.fit_transform(x.values.reshape(-1,1)) for y, x in df.groupby('id').val])
df
Out[271]:
id val scaled new
0 1 0.457713 1.000000 1.000000
1 1 -0.464513 0.000000 0.000000
2 1 0.216352 0.738285 0.738284
3 2 0.633652 0.990656 0.990656
4 2 -1.099065 0.000000 0.000000
5 2 0.649995 1.000000 1.000000
6 3 -0.251099 0.306631 0.306631
7 3 -1.003295 0.081387 0.081387
8 3 2.064389 1.000000 1.000000
9 3 -1.275086 0.000000 0.000000