Pandas: normalize within the group

user_1065145 0
Viewed 9482

Let's say we have the following dataset:

import pandas as pd

data = [('apple', 'red', 155), ('apple', 'green', 102), ('apple', 'iphone', 48),
         ('tomato', 'red', 175), ('tomato', 'ketchup', 96), ('tomato', 'gun', 12)]

df = pd.DataFrame(data)
df.columns = ['word', 'rel_word', 'weight']

enter image description here

I would like to recompute weights, so that they sum up to 1.0 within each group (apple, tomato in the example) and keep related weights as is (e.g. apple/red to apple/green still should be 155/102).

pandas python
3 Answers

Use transform - faster than apply and lookup

In [3849]: df['weight'] / df.groupby('word')['weight'].transform('sum')
Out[3849]:
0    0.508197
1    0.334426
2    0.157377
3    0.618375
4    0.339223
5    0.042403
Name: weight, dtype: float64

In [3850]: df['norm_w'] = df['weight'] / df.groupby('word')['weight'].transform('sum')

In [3851]: df
Out[3851]:
     word rel_word  weight    norm_w
0   apple      red     155  0.508197
1   apple    green     102  0.334426
2   apple   iphone      48  0.157377
3  tomato      red     175  0.618375
4  tomato  ketchup      96  0.339223
5  tomato      gun      12  0.042403

Or,

In [3852]: df.groupby('word')['weight'].transform(lambda x: x/x.sum())
Out[3852]:
0    0.508197
1    0.334426
2    0.157377
3    0.618375
4    0.339223
5    0.042403
Name: weight, dtype: float64

Timings

In [3862]: df.shape
Out[3862]: (12000, 4)

In [3864]: %timeit df['weight'] / df.groupby('word')['weight'].transform('sum')
100 loops, best of 3: 2.44 ms per loop

In [3866]: %timeit df.groupby('word')['weight'].transform(lambda x: x/x.sum())
100 loops, best of 3: 5.16 ms per loop

In [3868]: %%timeit
      ...: group_weights = df.groupby('word').aggregate(sum)
      ...: df.apply(lambda row: row['weight']/group_weights.loc[row['word']][0],axis=1)
1 loop, best of 3: 2.5 s per loop
user_21372550

You can use groupby to calculate the total weight of each group, and then apply a normalization lambda function to each row:

group_weights = df.groupby('word').aggregate(sum)
df['normalized_weights'] = df.apply(lambda row: row['weight']/group_weights.loc[row['word']][0],axis=1)

Output:

    word    rel_word    weight  normalized_weights
0   apple   red         155     0.508197
1   apple   green       102     0.334426
2   apple   iphone      48      0.157377
3   tomato  red         175     0.618375
4   tomato  ketchup     96      0.339223
user_81188410

Using np.bincount & pd.factorize
This should be very fast and scalable 

f, u = pd.factorize(df.word.values)
w = df.weight.values

df.assign(norm_w=w / np.bincount(f, w)[f])

     word rel_word  weight    norm_w
0   apple      red     155  0.508197
1   apple    green     102  0.334426
2   apple   iphone      48  0.157377
3  tomato      red     175  0.618375
4  tomato  ketchup      96  0.339223
5  tomato      gun      12  0.042403
user_23366540
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