Issue
I was wondering if it is possible to write an iterative algorithm without using a for loop using as_strided and some operation that edits the memory in place.
For example, if I want to write an algorithm that replaces a number in an array with the sum of its neighbors. I came up with this abomination (yep its summing an element with 2 right neighbors but its just to get an idea):
import numpy as np
a = np.arange(10)
ops = 2
a_view_window = np.lib.stride_tricks.as_strided(a, shape = (ops,a.size - 2, 3), strides=(0,) + 2*a.strides)
a_view = np.lib.stride_tricks.as_strided(a, shape = (ops,a.size - 2), strides=(0,) + a.strides)
np.add.reduce(a_view_window, axis = -1, out=a_view)
print(a)
So I am taking an array of 10 numbers and creating this strange view which increases dimensionality without changing the strides. Thus my thinking is the reduction it will run over the fake new dimension and write over the previous values thus when it gets to the next major dimension it will have to read from the data it overwrote and thus iteratively perform the addition.
Sadly this does not work :(
(yes I know this is a terrible way to do things but I am curious about how the underlying numpy stuff works and if it can be hacked in this way)
Solution
This code results in an undefined behavior prior to Numpy 1.13 and works out-of-place in newer versions so to avoid overlapping/aliasing issues. Indeed, you cannot assume Numpy iterate in a given order on the input/output array view. In fact, Numpy often use SIMD instructions to speed up the code and sometimes tell compilers that views are not overlapping/aliasing each other (using the restrict
keyword) to they can generate a much more efficient code. For more information you can read the doc on ufuncs (and this issue):
Operations where ufunc input and output operands have memory overlap produced undefined results in previous NumPy versions, due to data dependency issues. In NumPy 1.13.0, results from such operations are now defined to be the same as for equivalent operations where there is no memory overlap.
Operations affected now make temporary copies, as needed to eliminate data dependency. As detecting these cases is computationally expensive, a heuristic is used, which may in rare cases result to needless temporary copies. For operations where the data dependency is simple enough for the heuristic to analyze, temporary copies will not be made even if the arrays overlap, if it can be deduced copies are not necessary.
Answered By - Jérôme Richard
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