pypy

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Python pypy: Efficient sum of absolute array/vector difference

由 梦想的初衷 提交于 2020-08-12 04:33:25
问题 I am trying to reduce the computation time of my script,which is run with pypy. It has to calculate for a large number of lists/vectors/arrays the pairwise sums of absolute differences. The length of the input vectors is quite small, between 10 and 500. I tested three different approaches so far: 1) Naive approach, input as lists: def std_sum(v1, v2): distance = 0.0 for (a,b) in izip(v1, v2): distance += math.fabs(a-b) return distance 2) With lambdas and reduce, input as lists: lzi = lambda

Python pypy: Efficient sum of absolute array/vector difference

由 て烟熏妆下的殇ゞ 提交于 2020-08-12 04:32:09
问题 I am trying to reduce the computation time of my script,which is run with pypy. It has to calculate for a large number of lists/vectors/arrays the pairwise sums of absolute differences. The length of the input vectors is quite small, between 10 and 500. I tested three different approaches so far: 1) Naive approach, input as lists: def std_sum(v1, v2): distance = 0.0 for (a,b) in izip(v1, v2): distance += math.fabs(a-b) return distance 2) With lambdas and reduce, input as lists: lzi = lambda

Python pypy: Efficient sum of absolute array/vector difference

由 拜拜、爱过 提交于 2020-08-12 04:32:07
问题 I am trying to reduce the computation time of my script,which is run with pypy. It has to calculate for a large number of lists/vectors/arrays the pairwise sums of absolute differences. The length of the input vectors is quite small, between 10 and 500. I tested three different approaches so far: 1) Naive approach, input as lists: def std_sum(v1, v2): distance = 0.0 for (a,b) in izip(v1, v2): distance += math.fabs(a-b) return distance 2) With lambdas and reduce, input as lists: lzi = lambda
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