Question

在pandas中执行分组和聚合非常容易且快速。但是，由于存在lambda函数，执行pandas已经内置了C 而没有聚合的简单groupby-apply函数（至少以我的方式）要慢得多。

# Form data
>>> import numpy as np
>>> import pandas as pd
>>> df = pd.DataFrame(np.random.random((100,3)),columns=['a','b','c'])
>>> df['g'] = np.random.randint(0,3,100)
>>> df.head()

          a         b         c  g
0  0.901610  0.643869  0.094082  1
1  0.536437  0.836622  0.763244  1
2  0.647989  0.150460  0.476552  0
3  0.206455  0.319881  0.690032  2
4  0.153557  0.765174  0.377879  1

# groupby and apply and aggregate
>>> df.groupby('g')['a'].sum()

g
0    17.177280
1    15.395264
2    17.668056
Name: a, dtype: float64

# groupby and apply without aggregation
>>> df.groupby('g')['a'].transform(lambda x: x.sum())

0     15.395264
1     15.395264
2     17.177280
3     17.668056
4     15.395264

95    15.395264
96    17.668056
97    15.395264
98    17.668056
99    17.177280
Name: a, Length: 100, dtype: float64

因此，我具有lambda函数所需的功能，但是速度很慢。

>>> %timeit df.groupby('g')['a'].sum()

1.11 ms ± 143 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

>>> %timeit df.groupby('g')['a'].transform(lambda x:x.sum())

4.01 ms ± 699 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

这在较大的数据集中成为一个问题。我认为可以更快更有效地获得此功能。

Answer 1

可能您正在寻找

  df.groupby('g')['a'].transform('sum')

实际上比带有apply的版本要快

import numpy as np
import pandas as pd
import timeit
df = pd.DataFrame(np.random.random((100,3)),columns=['a','b','c'])
df['g'] = np.random.randint(0,3,100)
def groupby():
    df.groupby('g')['a'].sum()

def transform_apply():
    df.groupby('g')['a'].transform(lambda x: x.sum())

def transform():
    df.groupby('g')['a'].transform('sum')

print('groupby',timeit.timeit(groupby,number=10))

print('lambda transform',timeit.timeit(transform_apply,number=10))

print('transform',timeit.timeit(transform,number=10))

输出：

groupby 0.010655807999999989
lambda transform 0.029328375000000073
transform 0.01493376600000007

快速高效的熊猫Groupby sum / mean，无聚合

1 个答案: