[Numpy-discussion] Tensordot and memory consumption
Nicolas Rougier
Nicolas.Rougier@loria...
Wed Feb 18 01:23:44 CST 2009
Hello,
I'm using tensordot in some computation and while I've been amazed by
the speed, I'm now trying to reduce memory consumption in some very
particular cases:
Let S be a 2 dims array of size (s1,s2)
Let D be a 2 dims array of size (d1,d2)
Let W be a 4 dims array of size (d1,d2,s1,s2)
Currently, I'm computing D as tensordot(W,S,2) and it works really
fine and fast.
However, in some cases, W is based solely on a single 2 dims array K and
each of the W[i,j] is a slice of K. I would like to know if there is a
way to build W such that memory footprint is reduced ? Or maybe there is
other ways to perform the same computation ? For example, I know that
when S and D are of same shape, I can use 2d convolution, but I would
need the general case.
Since I imagine my explanations are not sot clear, I join a small
example.
Nicolas
---
import numpy
n = 3
# Source
S = numpy.random.random((n,n))
# Destination (at this stage, only D shape matters)
D = numpy.zeros((2*n,2*n))
# Kernel
K = numpy.zeros((S.shape[0]*n+1,S.shape[1]*n+1))
K[S.shape[0],S.shape[1]] = 1
# Kernel decomposition for computing tensordot
W = numpy.zeros(D.shape + S.shape)
for i in range(W.shape[0]):
for j in range(W.shape[1]):
x = int(i/float(W.shape[0])*float(S.shape[0]))
y = int(j/float(W.shape[1])*float(S.shape[1]))
W[i,j] = K[n-x:n-x+S.shape[0],n-y:n-y+S.shape[1]]
D = numpy.tensordot(W,S,2)
print S
print
print D
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