熊猫：在群体内正常化

Question

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']

我想重新计算重量，这样它们在每组中总计达到1.0（例子中的苹果，番茄）并保持相关的重量（例如苹果/红色到苹果/绿色仍然应该是155/102）

Answer 1

使用transform - 快于apply并查找

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

<强>计时

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

Answer 2

您可以使用groupby计算每个组的总权重，然后apply每行的标准化lambda函数：

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

输出：

    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

Answer 3

使用np.bincount＆amp; pd.factorize
这应该是非常快速和可扩展的

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