Add 2-d array to 3-d array with constantly changing index fast
Add 2-d array to 3-d array with constantly changing index fast
I'm trying to add a 2-d array to a 3-d array with constantly changing index , I come up with following code:
import numpy as np
a = np.zeros([8, 3, 5])
k = 0
for i in range(2):
for j in range(4):
a[k, i: i + 2, j: j + 2] += np.ones([2, 2], dtype=int)
k += 1
print(a)
which will give exactly what i want:
[[[1. 1. 0. 0. 0.]
[1. 1. 0. 0. 0.]
[0. 0. 0. 0. 0.]]
[[0. 1. 1. 0. 0.]
[0. 1. 1. 0. 0.]
[0. 0. 0. 0. 0.]]
[[0. 0. 1. 1. 0.]
[0. 0. 1. 1. 0.]
[0. 0. 0. 0. 0.]]
[[0. 0. 0. 1. 1.]
[0. 0. 0. 1. 1.]
[0. 0. 0. 0. 0.]]
[[0. 0. 0. 0. 0.]
[1. 1. 0. 0. 0.]
[1. 1. 0. 0. 0.]]
[[0. 0. 0. 0. 0.]
[0. 1. 1. 0. 0.]
[0. 1. 1. 0. 0.]]
[[0. 0. 0. 0. 0.]
[0. 0. 1. 1. 0.]
[0. 0. 1. 1. 0.]]
[[0. 0. 0. 0. 0.]
[0. 0. 0. 1. 1.]
[0. 0. 0. 1. 1.]]]
I wish it can be faster so I create an array for index and trying to use np.vectorize. But as manual described, vectorize is not for performance. And my goal is running through an array with shape of (10^6, 15, 15) which end up with 10^6 iteration. I hope there are some cleaner solution can get rid of all the for-loop.
This is the first time I using stack overflow, any suggestion are appreciated.
Thank you.
@NilsWerner My ultimate goal is to find the best next step in board games. More precisely, I want to solve a game called 1010! (like Tetris), because it has three blocks that can be used as the next step, and each block may have up to 100 positions that can be placed in plate, so consider each possibility, and there may be up to 100*100*100 choices. I want to list all possible positions and scores and figure out the best next step.
– Waffle
Aug 31 at 15:33
It looks like the kind of moving window that can be produced with
as_strided
, but working out the details could take some time. I'd start with a
, or a simplification of it, and try to find a way to index all the 1s (as a (8,2,2) array). Then use that indexing in an assignment expression.– hpaulj
Aug 31 at 16:28
as_strided
a
1 Answer
1
A efficient solution using numpy.lib.stride_tricks, which can "view" all the possibilities.
N=4 #tray size #(square)
P=3 # chunk size
R=N-P
from numpy.lib.stride_tricks import as_strided
tray = zeros((N,N),numpy.int32)
chunk = ones((P,P),numpy.int32)
tray[R:,R:] = chunk
tray = np.vstack((tray,tray))
view = as_strided(tray,shape=(R+1,R+1,N,N),strides=(4*N,4,4*N,4))
a_view = view.reshape(-1,N,N)
a_hard = a_view.copy()
Here is the result :
In [3]: a_view
Out[3]:
array([[[0, 0, 0, 0],
[0, 1, 1, 1],
[0, 1, 1, 1],
[0, 1, 1, 1]],
[[0, 0, 0, 0],
[1, 1, 1, 0],
[1, 1, 1, 0],
[1, 1, 1, 0]],
[[0, 1, 1, 1],
[0, 1, 1, 1],
[0, 1, 1, 1],
[0, 0, 0, 0]],
[[1, 1, 1, 0],
[1, 1, 1, 0],
[1, 1, 1, 0],
[0, 0, 0, 0]]])
a_view
is just a view on possible positions of a chunk on the tray. It doesn't cost any computation, and it just uses twice the tray space.
a_hard
is a hard copy, necessary if you need to modify it.
a_view
a_hard
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Are you treally trying to add ones to a, or are you eventually trying to solve something else, like a convolution?
– Nils Werner
Aug 31 at 15:11