我正在使用Theano 0.7创建一个使用 max-pooling 的convolutional neural net（即通过仅保留局部最大值来缩小矩阵） .
为了“撤消”或“反转”最大池化步骤，一种方法是将最大值的位置存储为辅助数据，然后通过制作大量零和使用这些辅助位置来简单地重新创建未合并的数据将最大值放在适当的位置 .
这是我目前正在做的事情：
import numpy as np
import theano
import theano.tensor as T
minibatchsize = 2
numfilters = 3
numsamples = 4
upsampfactor = 5
# HERE is the function that I hope could be improved
def upsamplecode(encoded, auxpos):
shp = encoded.shape
upsampled = T.zeros((shp[0], shp[1], shp[2] * upsampfactor))
for whichitem in range(minibatchsize):
for whichfilt in range(numfilters):
upsampled = T.set_subtensor(upsampled[whichitem, whichfilt, auxpos[whichitem, whichfilt, :]], encoded[whichitem, whichfilt, :])
return upsampled
totalitems = minibatchsize * numfilters * numsamples
code = theano.shared(np.arange(totalitems).reshape((minibatchsize, numfilters, numsamples)))
auxpos = np.arange(totalitems).reshape((minibatchsize, numfilters, numsamples)) % upsampfactor # arbitrary positions within a bin
auxpos += (np.arange(4) * 5).reshape((1,1,-1)) # shifted to the actual temporal bin location
auxpos = theano.shared(auxpos.astype(np.int))
print "code:"
print code.get_value()
print "locations:"
print auxpos.get_value()
get_upsampled = theano.function([], upsamplecode(code, auxpos))
print "the un-pooled data:"
print get_upsampled()
（顺便说一下，在这种情况下，我有一个3D张量，它只是第三个轴得到最大池 . 使用图像数据的人可能会看到两个维度得到最大化 . ）
输出是：
code:
[[[ 0 1 2 3]
[ 4 5 6 7]
[ 8 9 10 11]]
[[12 13 14 15]
[16 17 18 19]
[20 21 22 23]]]
locations:
[[[ 0 6 12 18]
[ 4 5 11 17]
[ 3 9 10 16]]
[[ 2 8 14 15]
[ 1 7 13 19]
[ 0 6 12 18]]]
the un-pooled data:
[[[ 0. 0. 0. 0. 0. 0. 1. 0. 0. 0. 0. 0. 2. 0.
0. 0. 0. 0. 3. 0.]
[ 0. 0. 0. 0. 4. 5. 0. 0. 0. 0. 0. 6. 0. 0.
0. 0. 0. 7. 0. 0.]
[ 0. 0. 0. 8. 0. 0. 0. 0. 0. 9. 10. 0. 0. 0.
0. 0. 11. 0. 0. 0.]]
[[ 0. 0. 12. 0. 0. 0. 0. 0. 13. 0. 0. 0. 0. 0.
14. 15. 0. 0. 0. 0.]
[ 0. 16. 0. 0. 0. 0. 0. 17. 0. 0. 0. 0. 0. 18.
0. 0. 0. 0. 0. 19.]
[ 20. 0. 0. 0. 0. 0. 21. 0. 0. 0. 0. 0. 22. 0.
0. 0. 0. 0. 23. 0.]]]
这个方法 works 但它是一个 bottleneck ，占用了我计算机的大部分时间（我认为set_subtensor调用可能意味着cpu < - > gpu数据复制） . 那么：这可以更有效地实施吗？
我怀疑有一种方法可以将此表达为单个 set_subtensor() 调用，这可能会更快，但我不知道如何让张量索引正确播放 .
UPDATE: 我在一次通话中想到了一种方法，通过处理扁平化的张量：
def upsamplecode2(encoded, auxpos):
shp = encoded.shape
upsampled = T.zeros((shp[0], shp[1], shp[2] * upsampfactor))
add_to_flattened_indices = theano.shared(np.array([ [[(y + z * numfilters) * numsamples * upsampfactor for x in range(numsamples)] for y in range(numfilters)] for z in range(minibatchsize)], dtype=theano.config.floatX).flatten(), name="add_to_flattened_indices")
upsampled = T.set_subtensor(upsampled.flatten()[T.cast(auxpos.flatten() + add_to_flattened_indices, 'int32')], encoded.flatten()).reshape(upsampled.shape)
return upsampled
get_upsampled2 = theano.function([], upsamplecode2(code, auxpos))
print "the un-pooled data v2:"
ups2 = get_upsampled2()
print ups2
但是，这仍然不是很好的效率，因为当我运行它（添加到上面的脚本的末尾）时，我发现Cuda库目前无法有效地执行整数索引操作：
ERROR (theano.gof.opt): Optimization failure due to: local_gpu_advanced_incsubtensor1
ERROR (theano.gof.opt): TRACEBACK:
ERROR (theano.gof.opt): Traceback (most recent call last):
File "/usr/local/lib/python2.7/dist-packages/theano/gof/opt.py", line 1493, in process_node
replacements = lopt.transform(node)
File "/usr/local/lib/python2.7/dist-packages/theano/sandbox/cuda/opt.py", line 952, in local_gpu_advanced_incsubtensor1
gpu_y = gpu_from_host(y)
File "/usr/local/lib/python2.7/dist-packages/theano/gof/op.py", line 507, in __call__
node = self.make_node(*inputs, **kwargs)
File "/usr/local/lib/python2.7/dist-packages/theano/sandbox/cuda/basic_ops.py", line 133, in make_node
dtype=x.dtype)()])
File "/usr/local/lib/python2.7/dist-packages/theano/sandbox/cuda/type.py", line 69, in __init__
(self.__class__.__name__, dtype, name))
TypeError: CudaNdarrayType only supports dtype float32 for now. Tried using dtype int64 for variable None
1 回答
我不知道这是否更快，但它可能更简洁一些 . 看看它对你的情况是否有用 .