问题描述
我有一批b
m x n
图像存储在数组x
中,还有一个大小为p x q
的卷积滤波器f
,我想将其应用于每个图像(然后使用求和池并批量存储在数组y
中,即all(np.allclose(y[i][j][k], (x[i, j:j+p, k:k+q] * f).sum()) for i in range(b) for j in range(m-p+1) for k in range(n-q+1))
为true.
I have a batch of b
m x n
images stored in an array x
, and a convolutional filter f
of size p x q
that I'd like to apply to each image (then use sum pooling and store in an array y
) in the batch, i.e. all(np.allclose(y[i][j][k], (x[i, j:j+p, k:k+q] * f).sum()) for i in range(b) for j in range(m-p+1) for k in range(n-q+1))
is true.
适应此答案,我可以编写以下内容:
Adapting this answer, I could write the following:
b, m, n, p, q = 6, 5, 4, 3, 2
x = np.arange(b*m*n).reshape((b, m, n))
f = np.arange(p*q).reshape((p, q))
y = []
for i in range(b):
shape = f.shape + tuple(np.subtract(x[i].shape, f.shape) + 1)
strides = x[i].strides * 2
M = np.lib.stride_tricks.as_strided(x[i], shape=shape, strides=strides)
y.append(np.einsum('ij,ijkl->kl', f, M))
assert all(np.allclose(y[i][j][k], (x[i, j:j+p, k:k+q] * f).sum()) for i in range(b) for j in range(m-p+1) for k in range(n-q+1))
但是我认为有一种方法可以只使用一个einsum
,这对我很有用,因为b
通常在100到1000之间.
but I think there's a way to do it with just one einsum
, which would be useful to me because b
is usually between 100 and 1000.
如何调整我的方法以仅使用一个einsum
?另外,出于我的目的,除了numpy
之外,我不能引入scipy
或任何其他依赖项.
How do I adapt my approach to use just one einsum
? Also, for my purposes, I can't bring in scipy
or any other dependencies besides numpy
.
推荐答案
只需将shape
设置为5d,然后获取strides
即可匹配shape
.
Just need to get shape
to be 5d and get the strides
to match the shape
.
shape = f.shape + (x.shape[0],) + tuple(np.subtract(x.shape[1:], f.shape) + 1)
strides = (x.strides * 2)[1:]
M = np.lib.stride_tricks.as_strided(x, shape=shape, strides=strides)
y = np.einsum('pq,pqbmn->bmn', f, M)
如果b
变大,现在M
可能会变大,但这可以解决玩具问题.
now M
might get really big if b
gets really big, but it works on your toy problem.
这篇关于批量卷积2d在numpy中没有秘诀?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!