是否有一个函数可以将NumPy的广播规则应用于形状列表并返回最终形状?

下面是另一个直接实现,它恰好在示例中击败了其他实现。值得一提的是@hpaulj和@Warren Weckesser的hack,它几乎同样快,而且更简洁:

def bs_pp(*shapes):
    ml = max(shapes, key=len)
    out = list(ml)
    for l in shapes:
        if l is ml:
            continue
        for i, x in enumerate(l, -len(l)):
            if x != 1 and x != out[i]:
                if out[i] != 1:
                    raise ValueError
                out[i] = x
    return (*out,)

def bs_mq1(*shapes):
    max_rank = max([len(shape) for shape in shapes])
    shapes = [[1] * (max_rank - len(shape)) + shape for shape in shapes]
    final_shape = [1] * max_rank
    for shape in shapes:
        for dim, size in enumerate(shape):
            if size != 1:
                final_size = final_shape[dim]
                if final_size == 1:
                    final_shape[dim] = size
                elif final_size != size:
                    raise ValueError("Cannot broadcast these shapes")
    return (*final_shape,)

import numpy as np

def bs_mq2(*shapes):
    max_rank = max([len(shape) for shape in shapes])
    shapes = np.array([[1] * (max_rank - len(shape)) + shape
                      for shape in shapes])
    shapes[shapes==1] = -1
    final_shape = shapes.max(axis=0)
    final_shape[final_shape==-1] = 1
    return (*final_shape,)

def bs_hp_ww(*shapes):
    return np.broadcast(*[np.empty(shape + [0,], int) for shape in shapes]).shape[:-1]

L = [6], [4, 2, 3, 1], [2, 1, 1]

from timeit import timeit

print('pp:       ', timeit(lambda: bs_pp(*L), number=10_000)/10)
print('mq 1:     ', timeit(lambda: bs_mq1(*L), number=10_000)/10)
print('mq 2:     ', timeit(lambda: bs_mq2(*L), number=10_000)/10)
print('hpaulj/ww:', timeit(lambda: bs_hp_ww(*L), number=10_000)/10)

assert bs_pp(*L) == bs_mq1(*L) and bs_pp(*L) == bs_mq2(*L) and bs_pp(*L) == bs_hp_ww(*L)

样本运行:

pp:        0.0021552839782088993
mq 1:      0.00398325570859015
mq 2:      0.01497043427079916
hpaulj/ww: 0.003267909213900566

我在NumPy文档中没有看到任何与此相关的内容。可以向每个目标形状广播单个0维数组,然后相互广播所有结果:

def broadcast_shapes(*shapes):
    base = numpy.array(0)
    broadcast1 = [numpy.broadcast_to(base, shape) for shape in shapes]
    return numpy.broadcast(*broadcast1).shape

这样可以避免为大型形状分配大量内存。不过,需要创建阵列感觉有点傻。

下面是一个简单的实现,以防有人需要它(它可能有助于理解广播)。不过我更喜欢使用NumPy函数。

def broadcast_shapes(*shapes):
    max_rank = max([len(shape) for shape in shapes])
    shapes = [[1] * (max_rank - len(shape)) + shape for shape in shapes]
    final_shape = [1] * max_rank
    for shape in shapes:
        for dim, size in enumerate(shape):
            if size != 1:
                final_size = final_shape[dim]
                if final_size == 1:
                    final_shape[dim] = size
                elif final_size != size:
                    raise ValueError("Cannot broadcast these shapes")
    return final_shape

编辑

我将这个函数与其他几个答案进行了比较,结果证明它是最快的( 编辑 ,Paul Panzer编写了一个更快的函数,请参见他的答案,我将其添加到下面的列表中):

%timeit bs_pp(*shapes) # Peter Panzer's answer
2.33 µs ± 10.2 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

%timeit broadcast_shapes1(*shapes)  # this answer
4.21 µs ± 11.2 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

%timeit broadcast_shapes2(*shapes) # my other answer with shapes.max(axis=0)
12.8 µs ± 67.3 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

%timeit broadcast_shapes3(*shapes) # user2357112's answer
18 µs ± 26.9 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

%timeit broadcast_shapes4(*shapes) # hpaulj's answer
18.1 µs ± 263 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

In [120]: shapes = [6], [4, 2, 3, 1], [2, 1, 1]                                 
In [121]: arrs = np.broadcast_arrays(*[np.empty(shape,int) for shape in shapes])
     ...:                                                                       
In [122]: [a.shape for a in arrs]                                               
Out[122]: [(4, 2, 3, 6), (4, 2, 3, 6), (4, 2, 3, 6)]

In [124]: np.lib.stride_tricks._broadcast_shape(*[np.empty(shape,int) for shape 
     ...: in shapes])                                                           
Out[124]: (4, 2, 3, 6)

In [131]: np.broadcast(*[np.empty(shape,int) for shape in shapes]).shape        
Out[131]: (4, 2, 3, 6)

第二个速度快了很多,4.79秒对42.4秒。第三个速度快了一点。

正如我第一次评论的那样,我从 broadcast_arrays ,然后查看代码。那是给我的吗 _broadcast_shape ,然后 np.broadcast .

假设这些形状实际上可以被广播,那么这是可行的:

def broadcast_shapes(*shapes):
    max_rank = max([len(shape) for shape in shapes])
    shapes = np.array([[1] * (max_rank - len(shape)) + shape
                      for shape in shapes])
    shapes[shapes==1] = -1
    final_shape = shapes.max(axis=0)
    final_shape[final_shape==-1] = 1
    return final_shape

如果假设没有空维度,那么 -1 黑客是没有必要的:

def broadcast_shapes(*shapes):
    max_rank = max([len(shape) for shape in shapes])
    shapes = np.array([[1] * (max_rank - len(shape)) + shape
                      for shape in shapes])
    return shapes.max(axis=0)