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くもすけさんのbacktestをpandasからできるだけnumpyで速くしてみた

元記事はこちら

ドテン&ピラミッディングのコードの一部をnumpyで計算して検証してみました

# ポジション計算部(ドテン&ピラミッディング)
pyramiding=3
df['order'] = 0
df['order'] = df['order'].where(df['long']!=True,1)
df['order'] = df['order'].where(df['short']!=True,-1)
df['pos'] = df['order'].where(df['order']!=0,).fillna(method='ffill').fillna(0)
df['pos'] = df.groupby((df['pos']*df['pos'].shift(1)<0).cumsum().fillna(0))['order'].cumsum()
df['pos'] = df['pos'].where(df['pos']<=pyramiding,pyramiding ) 
df['pos'] = df['pos'].where(df['pos']>=-pyramiding,-pyramiding )
print( df )

numpyにはffill()とshift()がないのでこれらの関数を追加

def np_shift(arr, num=1, fill_value=np.nan):
   result = np.empty_like(arr)
   if num > 0:
       result[:num] = fill_value
       result[num:] = arr[:-num]
   elif num < 0:
       result[num:] = fill_value
       result[:num] = arr[-num:]
   else:
       result[:] = arr
   return result


def np_ffill(arr, axis=0):
   idx_shape = tuple([slice(None)] + [np.newaxis] * (len(arr.shape) - axis - 1))
   idx = np.where(~np.isnan(arr), np.arange(arr.shape[axis])[idx_shape], 0)
   np.maximum.accumulate(idx, axis=axis, out=idx)
   slc = [np.arange(k)[tuple([slice(None) if dim==i else np.newaxis
       for dim in range(len(arr.shape))])]
       for i, k in enumerate(arr.shape)]
   slc[axis] = idx
   return arr[tuple(slc)]

要素数108000のデータを用意

len(df)
# 108000

pandasとnumpyでの時間比較

pandas

%%timeit
df['order'] = 0
df['order'] = df['order'].where(df['long']!=True,1)
df['order'] = df['order'].where(df['short']!=True,-1)
df['pos'] = df['order'].where(df['order']!=0,).fillna(method='ffill').fillna(0)
df['pos'] = df.groupby((df['pos']*df['pos'].shift(1)<0).cumsum().fillna(0))['order'].cumsum()
pyramiding = 3
df['pos'] = df['pos'].where(df['pos']<=pyramiding,pyramiding ) 
df['pos'] = df['pos'].where(df['pos']>=-pyramiding,-pyramiding )

# 31.7 ms ± 383 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

numpy

%%timeit
df['order'] = 0
df['order'] = np.where(df['short'], -1, np.where(df['long'], 1, df['order']))
df['pos'] = np.nan_to_num(np_ffill(np.where(df['order']==0, np.nan, df['order'])))
df['pos'] = df.groupby(np.nan_to_num(np_ffill(np.cumsum(df['pos'].to_numpy()*np_shift(df['pos'].to_numpy())<0))))['order'].cumsum()
pyramiding = 3
df['pos'] = np.where(df['pos']<=-pyramiding,-pyramiding,np.where(df['pos']>=pyramiding,pyramiding,df['pos']))

# 24.4 ms ± 819 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

データに違いがないかチェック

def pd_arr(df):
   df['order'] = 0
   df['order'] = df['order'].where(df['long']!=True,1)
   df['order'] = df['order'].where(df['short']!=True,-1)
   df['pos'] = df['order'].where(df['order']!=0,).fillna(method='ffill').fillna(0)
   df['pos'] = df.groupby((df['pos']*df['pos'].shift(1)<0).cumsum().fillna(0))['order'].cumsum()
   pyramiding = 3
   df['pos'] = df['pos'].where(df['pos']<=pyramiding,pyramiding ) 
   df['pos'] = df['pos'].where(df['pos']>=-pyramiding,-pyramiding )
   return df
   
def np_arr(df):
   df['order'] = 0
   df['order'] = np.where(df['short'], -1, np.where(df['long'], 1, df['order']))
   df['pos'] = np.nan_to_num(np_ffill(np.where(df['order']==0, np.nan, df['order'])))
   df['pos'] = df.groupby(np.nan_to_num(np_ffill(np.cumsum(df['pos'].to_numpy()*np_shift(df['pos'].to_numpy())<0))))['order'].cumsum()
   pyramiding = 3
   df['pos'] = np.where(df['pos']<=-pyramiding,-pyramiding,np.where(df['pos']>=pyramiding,pyramiding,df['pos']))
   return df
   
np.isclose(pd_arr(df), np_arr(df)).all()
# True

31.7 ms ± 383 µs ---> 24.4 ms ± 819 µs

思ってたより速くならない結果となりました

groupbyの部分が一番時間掛かってたのでその部分を頑張ってnumpy化すれば速くなりそうですが今日はここまで


7/11 numpyのdf['pos']の式が間違っていたところを修正

pandasのwhere()の第2引数はFalseのときの値を返す。

numpyのwhere()の第2引数はTrueのときの値を返す。第3引数はFalse

passしたいときは元のデータを渡す

# before
df['pos'] = np.where(df['pos']>=-pyramiding,-pyramiding, np.where(df['pos']<=pyramiding,pyramiding,df['pos']))
# after
df['pos'] = np.where(df['pos']<=-pyramiding,-pyramiding,np.where(df['pos']>=pyramiding,pyramiding,df['pos']))

-------------------------------------  番外編  -------------------------------------

numpy-groupies & numbaを使った場合

pip installで簡単にインストールできます。

numpyのコードを以下のように変更

# numpy
df['pos'] = df.groupby(np.nan_to_num(np_ffill(np.cumsum(df['pos'].to_numpy()*np_shift(df['pos'].to_numpy())<0))))['order'].cumsum()​
# numpy-groupies & numba
import numpy_groupies as npg
group_index = np.nan_to_num(np_ffill(np.cumsum(df['pos'].to_numpy()*np_shift(df['pos'].to_numpy())<0)))
df['pos'] = npg.aggregate(group_index, df['order'].to_numpy(), func='cumsum')
%%timeit
df['order'] = 0
df['order'] = np.where(df['short'], -1, np.where(df['long'], 1, df['order']))
df['pos'] = np.nan_to_num(np_ffill(np.where(df['order']==0, np.nan, df['order'])))
group_index = np.nan_to_num(np_ffill(np.cumsum(df['pos'].to_numpy()*np_shift(df['pos'].to_numpy())<0)))
df['pos'] = npg.aggregate(group_index, df['order'].to_numpy(), func='cumsum')
pyramiding = 3
df['pos'] = np.where(df['pos']<=-pyramiding,-pyramiding,np.where(df['pos']>=pyramiding,pyramiding,df['pos']))
# 14.6 ms ± 150 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

pandas        --->         numpy     --->    numpy-groupies & numba

31.7 ms ± 383 µs ---> 24.4 ms ± 819 µs  --->  14.6 ms ± 150 µs


pandasに比べて半分以下で終わるようになりました

使い方も簡単で累積和が知りたい場合は

npg.aggregate(グループ化したい配列、統計取りたい配列、func='cumsum')

とするだけです。

いいなと思ったら応援しよう!