如何加速非常慢的熊猫应用功能?

时间:2016-03-16 19:10:04

标签: python performance pandas

我有一个非常大的pandas数据集,在某些时候我需要使用以下函数

def proc_trader(data):
    data['_seq'] = np.nan
    # make every ending of a roundtrip with its index
    data.ix[data.cumq == 0,'tag'] = np.arange(1, (data.cumq == 0).sum() + 1)
    # backfill the roundtrip index until previous roundtrip;
    # then fill the rest with 0s (roundtrip incomplete for most recent trades)
    data['_seq'] =data['tag'].fillna(method = 'bfill').fillna(0)
    return data['_seq']
    # btw, why on earth this function returns a dataframe instead of the series `data['_seq']`??

我使用了申请

reshaped['_spell']=reshaped.groupby(['trader','stock'])[['cumq']].apply(proc_trader)

显然,我不能在这里分享数据,但你看到我的代码中存在瓶颈吗?可能是arange的事吗?数据中有许多name-productid个组合。

最小工作示例:

import pandas as pd
import numpy as np

reshaped= pd.DataFrame({'trader' : ['a','a','a','a','a','a','a'],'stock' : ['a','a','a','a','a','a','b'], 'day' :[0,1,2,4,5,10,1],'delta':[10,-10,15,-10,-5,5,0] ,'out': [1,1,2,2,2,0,1]})


reshaped.sort_values(by=['trader', 'stock','day'], inplace=True)
reshaped['cumq']=reshaped.groupby(['trader', 'stock']).delta.transform('cumsum')
reshaped['_spell']=reshaped.groupby(['trader','stock'])[['cumq']].apply(proc_trader).reset_index()['_seq']

1 个答案:

答案 0 :(得分:0)

这里没什么好看的,只是在几个地方调整过。实际上没有必要输入功能,所以我没有。在这个微小的样本数据上,它的速度大约是原始数据的两倍。

reshaped.sort_values(by=['trader', 'stock','day'], inplace=True)
reshaped['cumq']=reshaped.groupby(['trader', 'stock']).delta.cumsum()
reshaped.loc[ reshaped.cumq == 0, '_spell' ] = 1
reshaped['_spell'] = reshaped.groupby(['trader','stock'])['_spell'].cumsum()
reshaped['_spell'] = reshaped.groupby(['trader','stock'])['_spell'].bfill().fillna(0)

结果:

   day  delta  out stock trader  cumq  _spell
0    0     10    1     a      a    10     1.0
1    1    -10    1     a      a     0     1.0
2    2     15    2     a      a    15     2.0
3    4    -10    2     a      a     5     2.0
4    5     -5    2     a      a     0     2.0
5   10      5    0     a      a     5     0.0
6    1      0    1     b      a     0     1.0
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