I use numpy to do image processing, I wanted to switch the image to black and white and for that I did the calculation in each cell to see the luminosity, but if i want to show it i have to transform a 2d array into 2d array with 3 times the same value
for exemple i have this:
a = np.array([[255,0][0,255]])
#into
b = np.array([[[255,255,255],[0,0,0]],[[0,0,0],[255,255,255]]])
I've been searching for a while but i don't find anything to help
PS: sorry if i have made some mistake with my English.
You'll want to us an explicit broadcast: https://numpy.org/doc/stable/reference/generated/numpy.broadcast_to.html#numpy.broadcast_to
b = np.broadcast_to(a[..., np.newaxis], (2, 2, 3))
Usually you don't need to do it explicitly, maybe try and see if just a[..., np.newaxis] and the standard broadcasting rules are enough.
Another way to do it
np.einsum('ij,k->ijk', a, [1,1,1])
It's a way to create a 3 dimensional array (hence the ijk), from a 2d array (ij) and a 1d array (k). Whose result is for all i,j,k being indices of a and of [1,1,1], the 3d matrix of a[i,j]×[1,1,1][k].
Related
Let a be a numpy array of shape (n,m,k) and a_msk is an array of shape (n,m) containing that masks elements from a through multiplication.
Up to my knowledge, I had to create a new axis in a_msk in order to make it compatible with a for multiplication.
b = a * a_msk[:,:,np.newaxis]
Unfortunately, my Google Colab runtime is running out of memory at this very operation given the large size of the arrays.
My question is whether I can achieve the same thing without creating that new axis for the mask array.
As #hpaulj commented adding an axis to make the two arrays "compatible" for broadcasting is the most straightforward way to do your multiplication.
Alternatively, you can move the last axis of your array a to the front which would also make the two arrays compatible (I wonder though whether this would solve your memory issue):
a = np.moveaxis(a, -1, 0)
Then you can simply multiply:
b = a * a_msk
However, to get your result you have to move the axis back:
b = np.moveaxis(b, 0, -1)
Example: both solutions return the same answer:
import numpy as np
a = np.arange(24).reshape(2, 3, 4)
a_msk = np.arange(6).reshape(2, 3)
print(f'newaxis solution:\n {a * a_msk[..., np.newaxis]}')
print()
print(f'moveaxis solution:\n {np.moveaxis((np.moveaxis(a, -1, 0) * a_msk), 0, -1)}')
Suppose that I have a 3d array A and a 2d array B. A has dimension (s,m,m) while B has dimension (m,s).
I want to write code for a 2d array C with dimension (m,s) such that C[:,i] = A[i,:,:] # B[:,i].
Is there a way to do this elegantly without using a for loop in numpy?
One solution I thought of was to reshape B into a 3d array with dimension (m,s,1), multiply A and B via A#B, then reshape the resulting 3d array into a 2d array. This sounds a bit tedious and was wondering if tensordot or einsum can be applied here.
Suggestions appreciated. Thanks!
Using einsum is straight forward here:
A = np.arange(18).reshape(2,3,3)
B = np.arange(6).reshape(3,2)
C = np.einsum("ijk,ki->ji",A,B)
for i in range(2):
A[i]#B[:,i]==C[:,i]
# array([ True, True, True])
# array([ True, True, True])
I am trying to perform non-consectuitive slicing of a multidimensional array like this (Matlab peudo code)
A = B(:,:,[1,3],[2,4,6]) %A and B are two 4D matrices
But when I try to write this code in Python:
A = B[:,:,np.array([0,2]),np.array([1,3,5])] #A and B are two 4D arrays
it gives an error: IndexError: shape mismatch: indexing arrays could not be broadcast...
It should be noted that slicing for one dimension each time works fine!
In numpy, if you use more than one fancy index (i.e. array) to index different dimension of the same array at the same time, they must broadcast. This is designed such that indexing can be more powerful. For your situation, the simplest way to solve the problem is indexing twice:
B[:, :, [0,2]] [..., [1,3,5]]
where ... stands for as many : as possible.
Indexing twice this way would generate some extra data moving time. If you want to index only once, make sure they broadcast (i.e. put fancy indices on different dimension):
B[:, :, np.array([0,2])[:,None], [1,3,5]]
which will result in a X by Y by 2 by 3 array. On the other hand, you can also do
B[:, :, [0,2], np.array([1,3,5])[:,None]]
which will result in a X by Y by 3 by 2 array. The [1,3,5] axis is transposed before the [0,2] axis.
Yon don't have to use np.array([0,2]) if you don't need to do fancy operation with it. Simply [0,2] is fine.
np.array([0,2])[:,None] is equivalent to [[0],[2]], where the point of [:,None] is to create an extra dimension such that the shape becomes (2,1). Shape (2,) and (3,) cannot broadcast, while shape (2,1) and (3,) can, which becomes (2,3).
I have an array with shape (64,64) in Python and I want to repeat these elements (three times) in the way that I could have an array with the shape (64,64,3). Any idea?
Probably the most simplest way to accomplish this here is by using numpy.dstack:
import numpy as np
b = np.dstack((a, a, a))
where a was the original array (shape 64×64), and b is the new array (shape 64×64×3).
I am trying to create 3D array in python using Numpy and by multiplying 2D array in to 3rd dimension. I am quite new in Numpy multidimensional arrays and basically I am missing something important here.
In this example I am trying to make 10x10x20 3D array using base 2D array(10x10) by copying it 20 times.
My starting 2D array:
a = zeros(10,10)
for i in range(0,9):
a[i+1, i] = 1
What I tried to create 3D array:
b = zeros(20)
for i in range(0,19):
b[i]=a
This approach is probably stupid. So what is correct way to approach construction of 3D arrays from base 2D arrays?
Cheers.
Edit
Well I was doing things wrongly probably because of my R background.
Here is how I did it finally
b = zeros(20*10*10)
b = b.reshape((20,10,10))
for i in b:
for m in range(0, 9):
i[m+1, m] = 1
Are there any other ways to do the same?
There are many ways how to construct multidimensional arrays.
If you want to construct a 3D array from given 2D arrays you can do something like
import numpy
# just some 2D arrays with shape (10,20)
a1 = numpy.ones((10,20))
a2 = 2* numpy.ones((10,20))
a3 = 3* numpy.ones((10,20))
# creating 3D array with shape (3,10,20)
b = numpy.array((a1,a2,a3))
Depending on the situation there are other ways which are faster. However, as long as you use built-in constructors instead of loops you are on the fast side.
For your concrete example in Edit I would use numpy.tri
c = numpy.zeros((20,10,10))
c[:] = numpy.tri(10,10,-1) - numpy.tri(10,10,-2)
Came across similar problem...
I needed to modify 2D array into 3D array like so:
(y, x) -> (y, x, 3).
Here is couple solutions for this problem.
Solution 1
Using python tool set
array_3d = numpy.zeros(list(array_2d.shape) + [3], 'f')
for z in range(3):
array_3d[:, :, z] = array_2d.copy()
Solution 2
Using numpy tool set
array_3d = numpy.stack([array_2d.copy(), ]*3, axis=2)
That is what I came up with. If someone knows numpy to give a better solution I would love to see it! This works but I suspect there is a better way performance-wise.