在 Pandas 中用平均值转换组的更快方法
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Faster way to transform group with mean value in Pandas
提问by YXD
I have a Pandas dataframe where I am trying to replace the values in each group by the mean of the group. On my machine, the line df["signal"].groupby(g).transform(np.mean)takes about 10 seconds to run with Nand N_TRANSITIONSset to the numbers below.
我有一个 Pandas 数据框，我试图用该组的平均值替换每个组中的值。在我的机器上，这条线df["signal"].groupby(g).transform(np.mean)大约需要 10 秒才能运行N并N_TRANSITIONS设置为下面的数字。
Is there any faster way to achieve the same result?
有没有更快的方法来达到同样的结果？
import pandas as pd
import numpy as np
from time import time
np.random.seed(0)
N = 120000
N_TRANSITIONS = 1400
# generate groups
transition_points = np.random.permutation(np.arange(N))[:N_TRANSITIONS]
transition_points.sort()
transitions = np.zeros((N,), dtype=np.bool)
transitions[transition_points] = True
g = transitions.cumsum()
df = pd.DataFrame({ "signal" : np.random.rand(N)})
# here is my bottleneck for large N
tic = time()
result = df["signal"].groupby(g).transform(np.mean)
toc = time()
print toc - tic
回答by Jeff
Current method, using transform
当前方法，使用变换
In [44]: grp = df["signal"].groupby(g)
In [45]: result2 = df["signal"].groupby(g).transform(np.mean)
In [47]: %timeit df["signal"].groupby(g).transform(np.mean)
1 loops, best of 3: 535 ms per loop
Using 'broadcasting' of the results
使用结果的“广播”
In [43]: result = pd.concat([ Series([r]*len(grp.groups[i])) for i, r in enumerate(grp.mean().values) ],ignore_index=True)
In [42]: %timeit pd.concat([ Series([r]*len(grp.groups[i])) for i, r in enumerate(grp.mean().values) ],ignore_index=True)
10 loops, best of 3: 119 ms per loop
In [46]: result.equals(result2)
Out[46]: True
I think you might need to set the index of the returned on the broadcast result (it happens to work here because its a default index
我认为您可能需要设置广播结果返回的索引（它碰巧在这里工作，因为它是默认索引
result = pd.concat([ Series([r]*len(grp.groups[i])) for i, r in enumerate(grp.mean().values) ],ignore_index=True)
result.index = df.index
回答by YXD
Inspired by Jeff's answer. This is the fastest method on my machine:
受到杰夫回答的启发。这是我机器上最快的方法：
pd.Series(np.repeat(grp.mean().values, grp.count().values))
