Question

我有一些代码，它基本上将long numpy数组的某些元素设置为零。

import numpy as np
a = np.array([[[1, 2, 3, 4], [0, 1, 2, 3], [1, 2, 3, 4]], 
              [[0, 1, 2, 3], [0, 0, 1, 2], [0, 2, 3, 4]]])

aup = a + 2

b = np.array([[[0, 1, 2], [3, 4, 5], [0, 2, 3], [5, 6, 7], [0, 1, 2]],
              [[3, 4, 5], [0, 2, 3], [5, 6, 7], [0, 1, 2], [3, 4, 5]],
              [[0, 1, 2], [3, 4, 5], [0, 2, 3], [5, 6, 7], [0, 1, 2]], 
              [[3, 4, 5], [0, 2, 3], [5, 6, 7], [0, 1, 2], [3, 4, 5]],
              [[0, 1, 2], [3, 4, 5], [0, 2, 3], [5, 6, 7], [0, 1, 2]]])

for i in range(3):
    for j in range(4):
        for row_t in range(a[0, i, j], aup[0, i, j]):
            for col_t in range(a[1, i, j], aup[1, i, j]):
                row = row_t%5
                col = col_t%5
                b[row, col, i] = 0

主要问题是运行时间，因为我有4个for循环，所以可能很多。是否有任何笨拙的技巧来避免这些for循环并基本达到相同的目的？

更新：

我认为这一切归结为以下问题。首先在轴2上找到a和aup的最小值，即

min_a = np.min(a, axis=2)%5  \\ [[1 0 1], [0 0 0]]
max_aup = np.max(aup, axis=2)%5 \\ [[1 0 1], [0 4 1]]

Npw列出在min_a和min_aup之间形成的所有元素明智元组。与相应的索引元素：

tuples = [[1, 1, 0], [0, 0, 1], [0, 1, 1], [0, 2, 1], [0, 3, 1], 
          [0, 4, 1], [1, 0, 2], [1, 1, 2]]

并将这些元组的b元素设置为0：

b[tuples] = 0

因此，整个问题基本上是有效地找到这些tuples。

Answer 1

好的，这是你的循环的矢量化版本。它明确地使用了aup是a加上标量偏移的事实：

>>> import numpy as np
>>> a = np.array([[[1, 2, 3, 4], [0, 1, 2, 3], [1, 2, 3, 4]], 
...               [[0, 1, 2, 3], [0, 0, 1, 2], [0, 2, 3, 4]]])
>>> 
>>> aup = a + 2
>>> 
>>> b = np.array([[[0, 1, 2], [3, 4, 5], [0, 2, 3], [5, 6, 7], [0, 1, 2]],
...               [[3, 4, 5], [0, 2, 3], [5, 6, 7], [0, 1, 2], [3, 4, 5]],
...               [[0, 1, 2], [3, 4, 5], [0, 2, 3], [5, 6, 7], [0, 1, 2]], 
...               [[3, 4, 5], [0, 2, 3], [5, 6, 7], [0, 1, 2], [3, 4, 5]],
...               [[0, 1, 2], [3, 4, 5], [0, 2, 3], [5, 6, 7], [0, 1, 2]]])
>>> 
>>> b0 = b.copy()
>>> 
>>> 
>>> for i in range(3):
...     for j in range(4):
...         for row_t in range(a[0, i, j], aup[0, i, j]):
...             for col_t in range(a[1, i, j], aup[1, i, j]):
...                 row = row_t % 5
...                 col = col_t % 5
...                 b[row, col, i] = 0
... 
>>> b_loopy = b
>>> b = b0.copy()
>>> 
>>> i, j, k, _ = np.ogrid[:2, :2, :3, :0]
>>> 
>>> b[(a[0] + i) % 5, (a[1] + j) % 5, k] = 0
>>> 
>>> b_vect = b
>>> 
>>> np.all(b_vect == b_loopy)
True

对于任意aup > a，它有点毛茸茸。对于小问题大小，下面的代码比循环稍慢，但扩展得更好，请参阅本文末尾的时间。

import numpy as np

a = np.array([[[1, 2, 3, 4], [0, 1, 2, 3], [1, 2, 3, 4]], 
              [[0, 1, 2, 3], [0, 0, 1, 2], [0, 2, 3, 4]]])

aup = a + np.random.randint(2, 5, a.shape)

b = np.array([[[0, 1, 2], [3, 4, 5], [0, 2, 3], [5, 6, 7], [0, 1, 2]],
              [[3, 4, 5], [0, 2, 3], [5, 6, 7], [0, 1, 2], [3, 4, 5]],
              [[0, 1, 2], [3, 4, 5], [0, 2, 3], [5, 6, 7], [0, 1, 2]], 
              [[3, 4, 5], [0, 2, 3], [5, 6, 7], [0, 1, 2], [3, 4, 5]],
              [[0, 1, 2], [3, 4, 5], [0, 2, 3], [5, 6, 7], [0, 1, 2]]])

def setup(i, j, k, l):
    b = np.random.randint(1, 10, (i, j, k))
    a = np.array(np.unravel_index(np.random.randint(0, i*j, (k, l)), (i, j)))
    aup = a + 1 + np.array(np.unravel_index(np.random.randint(
        0, (i-2)*(j-2), (k, l)), (i-2, j-2)))
    return b, a, aup

def f_loopy(b, a, aup):
    b = b.copy()
    for i in range(b.shape[-1]):
        for j in range(a.shape[-1]):
            for row_t in range(a[0, i, j], aup[0, i, j]):
                for col_t in range(a[1, i, j], aup[1, i, j]):
                    row = row_t % b.shape[0]
                    col = col_t % b.shape[1]
                    b[row, col, i] = 0
    return b

def unwrap_indices(a, aup, shp):
    i, j, k = shp
    _, k, m = a.shape
    ij = np.array((i, j))[:, None]
    m *= k
    a = a.reshape(2, -1)
    aup = aup.reshape(2, -1)
    fst, snd = mask = aup > ij
    fsti = np.flatnonzero(fst)
    sndi = np.flatnonzero(snd)
    bthi = fsti[snd[fsti]]
    m2 = m + len(fsti)
    m3 = m2 + len(sndi)
    d = np.empty((2, m3 + len(bthi)), dtype=int)
    d[0, m:m2] = aup[0, fsti] - i
    d[1, m2:m3] = aup[1, sndi] - j
    d[:, m3:] = aup[:, bthi] - ij
    d[:, :m] = np.where(mask, ij, aup) - a
    d[1, m:m2] = d[1, fsti]
    d[0, m2:m3] = d[0, sndi]
    aa = np.empty_like(d)
    aa[:, :m] = a
    aa[1, m:m2] = a[1, fsti]
    aa[0, m2:m3] = a[0, sndi]
    aa[0, m:m2] = 0
    aa[1, m2:m3] = 0
    aa[:, m3:] = 0
    z = np.empty(aa.shape[1:], dtype=int)
    z[:m].reshape(k, -1)[...] = np.arange(k)[:, None]
    z[m:m2] = z[fsti]
    z[m2:m3] = z[sndi]
    z[m3:] = z[bthi]
    return aa, d, z

def embed_indices_flat(aa, d, z, shp):
    i, j, k = shp
    _, m = aa.shape
    A = np.ravel_multi_index((aa[0], aa[1], z), shp)
    A[1:] -= A[:-1] + np.einsum('ij,i->j', d[:, :-1]-1, (j*k, k))
    A1 = (((j+1)-d[1]) * k).repeat(d[0]) 
    A1[0] = A[0]
    A1[d[0, :-1].cumsum()] = A[1:]
    idx = d[1].repeat(d[0]).cumsum()
    A2 = np.full(idx[-1:], k)
    A2[0] = A1[0]
    A2[idx[:-1]] = A1[1:]
    return A2.cumsum()

def f_vect(b, a, aup, switch_strat=20):
    b = b.copy()
    A, D, Z = unwrap_indices(a, aup, b.shape)
    if D.sum() > switch_strat * D.size:
        for ai, di, zi in zip(A.T, D.T, Z):
            b[ai[0]:ai[0]+di[0], ai[1]:ai[1]+di[1], zi] = 0
    else:
        b.ravel()[embed_indices_flat(A, D, Z, b.shape)] = 0
    return b

from timeit import timeit

print(np.all(f_vect(b, a, aup) == f_loopy(b, a, aup)))
print(timeit(lambda: f_vect(b, a, aup), number=1000))
print(timeit(lambda: f_loopy(b, a, aup), number=1000))

b, a, aup = setup(40, 30, 10, 8)

print(np.all(f_vect(b, a, aup) == f_loopy(b, a, aup)))
print(timeit(lambda: f_vect(b, a, aup), number=1000))
print(timeit(lambda: f_loopy(b, a, aup), number=1000))

示例输出：

True                      # <- results equal
0.08376670675352216       # <- vectorized
0.062134379986673594      # <- loopy
True                      # same for larger problem size 
0.3771689278073609        #
8.375985411927104         #

在这段代码中是否存在任何避免循环的麻烦技巧？

1 个答案: