python - Pandas :在组内正常化

假设我们有以下数据集:

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']

我想重新计算权重，以便它们在每组(示例中的苹果、番茄)中的总和为 1.0，并保持相关权重不变(例如，苹果/红色到苹果/绿色仍应为 155/102)。

最佳答案

使用 transform - 比 apply 和查找更快

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

或者，

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

时间

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

关于python - Pandas :在组内正常化，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/46419180/