パンダのデータフレームで複雑なスライスをベクトル化する

配列内の範囲全体のベクトル化された標準偏差

def get_ranges_arr(starts,ends):
    # Taken from http://stackoverflow.com/a/37626057/3293881
    counts = ends - starts
    counts_csum = counts.cumsum()
    id_arr = np.ones(counts_csum[-1],dtype=int)
    id_arr[0] = starts[0]
    id_arr[counts_csum[:-1]] = starts[1:] - ends[:-1] + 1
    return id_arr.cumsum()

def ranged_std(arr,starts,ends):
    # Get all indices and the IDs corresponding to same groups
    idx = get_ranges_arr(starts,ends)
    id_arr = np.repeat(np.arange(starts.size),ends-starts)
    
    # Extract relevant data
    slice_arr = arr[idx]
    
    # Simulate standard deviation implementation for a number of groups
    # using id_arr as the basis to perform various mathematical operations
    # within each group. Since, std. deviation performs sum/mean reduction,
    # we can simply use np.bincount for an efficient implementation.
    # Std. deviation formula used :
    #https://github.com/numpy/numpy/blob/v1.11.0/numpy/core/fromnumeric.py#L2939
    grp_counts = np.bincount(id_arr)
    mean_vals = np.bincount(id_arr,slice_arr)/grp_counts
    abs_vals = np.abs(slice_arr - mean_vals[id_arr])**2
    return np.sqrt(np.bincount(id_arr,abs_vals)/grp_counts)

サンプル実行（ルーピーバージョンに対して検証）

In [173]: arr = np.random.randint(0,9,(20))

In [174]: starts = np.array([2,6,11])

In [175]: ends = np.array([8,9,15])

In [176]: [np.std(arr[i:j]) for i,j in zip(starts,ends)]
Out[176]: [1.9720265943665387, 0.81649658092772603, 0.82915619758884995]

In [177]: ranged_std(arr,starts,ends)
Out[177]: array([ 1.97202659,  0.81649658,  0.8291562 ])    

ランタイムテスト

ケース＃1：範囲の数が非常に少ない 3

In [21]: arr = np.random.randint(0,9,(20))

In [22]: starts = np.array([2,6,11])

In [23]: ends = np.array([8,9,15])

In [24]: %timeit [np.std(arr[i:j]) for i,j in zip(starts,ends)]
10000 loops, best of 3: 146 µs per loop

In [25]: %timeit ranged_std(arr,starts,ends)
10000 loops, best of 3: 45 µs per loop

ケース＃2：まともな範囲の数 1000

In [32]: arr = np.random.randint(0,9,(1010))

In [33]: starts = np.random.randint(0,9,(1000))

In [34]: ends = starts + np.random.randint(0,9,(1000))

In [35]: %timeit [np.std(arr[i:j]) for i,j in zip(starts,ends)]
10 loops, best of 3: 47.5 ms per loop

In [36]: %timeit ranged_std(arr,starts,ends)
1000 loops, best of 3: 217 µs per loop

ケース＃3：多数の範囲 10000

In [60]: arr = np.random.randint(0,9,(1010))

In [61]: arr = np.random.randint(0,9,(10010))

In [62]: starts = np.random.randint(0,9,(10000))

In [63]: ends = starts + np.random.randint(0,9,(10000))

In [64]: %timeit [np.std(arr[i:j]) for i,j in zip(starts,ends)]
1 loops, best of 3: 474 ms per loop

In [65]: %timeit ranged_std(arr,starts,ends)
100 loops, best of 3: 2.17 ms per loop

本当に驚くべきスピードアップ200x+！

ranged_std私たちのケースを解決するために使用する

# Get start, stop numeric indices as needed for getting ranges array later on
starts = asd_1.index.searchsorted(index_1)
ends = asd_1.index.searchsorted(index_2)

# Create final dataframe output using ranged_std func
df = pd.DataFrame(ranged_std(asd_1.values,starts,ends+1),index=index_1)

検証のためのサンプル実行-

In [17]: asd_1 = pd.Series(0.01 * np.random.randn(252), index=\
    ...:                   pd.date_range('2011-1-1', periods=252))
    ...: 
    ...: index_1 = pd.to_datetime(['2011-2-2', '2011-4-3', '2011-5-1',])
    ...: index_2 = pd.to_datetime(['2011-2-15', '2011-4-16', '2011-5-17',])
    ...: 
    ...: index_tot = list(zip(index_1,index_2))
    ...: aux_learning_std = pd.DataFrame([np.nanstd(asd_1.loc[i:j]) for i, j in \
    ...:                                                index_tot], index=index_1)
    ...: 

In [18]: starts = asd_1.index.searchsorted(index_1)
    ...: ends = asd_1.index.searchsorted(index_2)
    ...: df = pd.DataFrame(ranged_std(asd_1.values,starts,ends+1),index=index_1)
    ...: 

In [19]: aux_learning_std
Out[19]: 
                   0
2011-02-02  0.007244
2011-04-03  0.012862
2011-05-01  0.010155

In [20]: df
Out[20]: 
                   0
2011-02-02  0.007244
2011-04-03  0.012862
2011-05-01  0.010155