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 Slicing using vectors
#1
Hi all

Does somebedoy know how to use vectors for slicing (see code herebellow)?

I got the following error "only integer scalar arrays can be converted to a scalar index" but I do not understand since I'm using a scalar (numpy) array, or I'm missing something

Thanks

Paul

n = 100
m = 2
A = np.random.randint(66, size=(n,m), dtype=np.int32)
i = np.random.randint(n-4, size=int(0.5*n), dtype=np.int32)
j = i + 4*np.ones(int(0.5*n), dtype=np.int32)
extract1_A = A[i,:]     # as usual = OK
#extract2_A = A[i:i+4,:] # fails
extract3_A = A[i:j,:] # fails
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#2
Try this:

import numpy as np
n = 100
m = 2
A = np.array(np.random.randint(66, size=(n,m), dtype=np.int32))
i = np.array(np.random.randint(n-4, size=int(0.5*n), dtype=np.int32))
j = i + 4*np.ones(int(0.5*n), dtype=np.int32)
print(i)
print(j)
print(i.shape)
print(j.shape)
extract1_A = np.array(A[i,:])
extract2_A = np.array([ A[x:x+4,:] for x in i])
extract3_A = np.array([ A[x:y,:] for x in i for y in j])
print (extract1_A.shape)
print (extract2_A.shape)
print (extract3_A.shape)
Smile wish everyone happy coding Smile
Quote
#3
thanks for the interest, but the goal has ever been to avoid the use of loops.

Be carefull with the dimensions of your matrixes

Paul
Quote
#4
It's generally helpful if you post runnable code (yours lacks at least one import) and the full, verbatim error message (ideally in error tags). Here's what I get when I run your code after adding the import:
Error:
Traceback (most recent call last): File "doit.py", line 10, in <module> extract3_A = A[i:j,:] # fails TypeError: only integer scalar arrays can be converted to a scalar index
(Nov-14-2019, 10:06 AM)paul18fr Wrote: I got the following error "only integer scalar arrays can be converted to a scalar index" but I do not understand since I'm using a scalar (numpy) array, or I'm missing something
So I tried printing your object and I got something like this:
Output:
[[22 45] [24 48] [51 24] [23 63] [ 9 29] .../
That... looks like a collection of non-scalars to me. I don't usually link to SO, but this might be useful.
Feel like you're not getting the answers you want? Checkout the help/rules for things like what to include/not include in a post, how to use code tags, how to ask smart questions, and more.

Pro-tip - there's an inverse correlation between the number of lines of code posted and my enthusiasm for helping with a question :)
Quote
#5
(Nov-16-2019, 12:00 AM)micseydel Wrote: It's generally helpful if you post runnable code (yours lacks at least one import) and the full, verbatim error message (ideally in error tags). Here's what I get when I run your code after adding the import:
Error:
Traceback (most recent call last): File "doit.py", line 10, in <module> extract3_A = A[i:j,:] # fails TypeError: only integer scalar arrays can be converted to a scalar index

The code has been added as it stands to highlight the issue I got.

Finally I found a way that answers to my need without using any loop but the Kronecker product; it has been checked on a small size matrix, but it quite interesting with million of lines (tested with 10 million on my old laptop).

Paul
import time
import numpy as np

#n = 1_000_000
n = 10
m = 2
A = np.array(np.random.randint(66, size=(n,m), dtype=np.int32))
i = np.array(np.random.randint(n-4, size=int(0.5*n), dtype=np.int32))
j = i + 4*np.ones(int(0.5*n), dtype=np.int32)

## the i vector gives us the first index of values we want to get from A
## in the current case we want to get values from i to (i+4)
## with only 1 index, slicing is traditionnally used as A[100:104,4] for example

## the "trick" or the solution I've been using is to specify each index I want to extract
## using the Kronecker product as follow:
t0 = time.time()
k1 = np.arange(4, dtype=np.int32)
k2 = np.ones(int(0.5*n), dtype=np.int32)
k3 = np.ones(4, dtype=np.int32)
kron1 = np.kron(k2,k1)  # here [0 1 2 3] is repeated (0.5*n) times => from j vector
kron2 = np.kron(i,k3)   # here each index is repeated (0.5*n) times => from i vector
index = kron1 + kron2   # then each index varies from its initial value to (initial+4)
Extract_A = np.copy(A[index,:]) # all the indexes have been explicitly expressed and we can extract the values as usually
t1 = time.time()
print("The new solution took {} seconds".format(t1-t0))
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