Develop Reference

How can I retrieve indptr, indices from scipy csr_matrix?

According to the documentation of scipy.sparse.csr_matrix, there are many ways to create a csr_matrix. One of the ways is to give data, indptr, and indices as inputs. I wonder, if there is a way to retrieve them in the opposite direction, i.e., assume that I have my_csr_matrix created as below:
>>> indptr = np.array([0, 2, 3, 6])
>>> indices = np.array([0, 2, 2, 0, 1, 2])
>>> data = np.array([1, 2, 3, 4, 5, 6])
>>> my_csr_matrix = csr_matrix((data, indices, indptr), shape=(3, 3))
>>> my_csr_matrix.toarray()
array([[1, 0, 2],
[0, 0, 3],
[4, 5, 6]])
The question is, how can I retrieve indptr and indices from my_csr_matrix without knowing the information about them or data before-hand explicitly?

Literally just my_csr_matrix.indptr and my_csr_matrix.indices. You can also get the data array with my_csr_matrix.data.
These attributes are documented further down the page, under the "Attributes" heading.
Note that these are the actual underlying arrays used by the sparse matrix representation. Modifying these arrays will modify the sparse matrix, unless you do something that causes the CSR matrix to allocate new underlying arrays. (For example, my_csr_matrix[1, 1] = 7 would change the sparsity structure of the matrix and require allocating new arrays.)

How does slicing numpy arrays with other arrays work?

I have a numpy array of shape [batch_size, timesteps_per_samples, width, height], where width and height refer to a 2D grid. The values in this array can be interpreted as an elevation at a certain location that changes over time.
I want to know the elevation over time for various paths within this array. Therefore i have a second array of shape [batch_size, paths_per_batch_sample, timesteps_per_path, coordinates] (coordinates = 2, for x and y in the 2D plane).
The resulting array should be of shape [batch_size, paths_per_batch_sample, timesteps_per_path] containing the elevation over time for each sample within the batch.
The following two examples work. The first one is very slow and just serves for understanding what I am trying to do. I think the second one does what I want but I have no idea why this works nor if it may crash under certain circumstances.
Code for the problem setup:
import numpy as np
batch_size=32
paths_per_batch_sample=10
timesteps_per_path=4
width=64
height=64
elevation = np.arange(0, batch_size*timesteps_per_path*width*height, 1)
elevation = elevation.reshape(batch_size, timesteps_per_path, width, height)
paths = np.random.randint(0, high=width-1, size=(batch_size, paths_per_batch_sample, timesteps_per_path, 2))
range_batch = range(batch_size)
range_paths = range(paths_per_batch_sample)
range_timesteps = range(timesteps_per_path)
The following code works but is very slow:
elevation_per_time = np.zeros((batch_size, paths_per_batch_sample, timesteps_per_path))
for s in range_batch:
for k in range_paths:
for t in range_timesteps:
x_co, y_co = paths[s,k,t,:].astype(int)
elevation_per_time[s,k,t] = elevation[s,t,x_co,y_co]
The following code works (even fast) but I can't understand why and how o.0
elevation_per_time_fast = elevation[
:,
range_timesteps,
paths[:, :, range_timesteps, 0].astype(int),
paths[:, :, range_timesteps, 1].astype(int),
][range_batch, range_batch, :, :]
Prove that the results are equal
check = (elevation_per_time == elevation_per_time_fast)
print(np.all(check))
Can somebody explain how I can slice an nd-array by multiple other arrays?
Especially, I don't understand how the numpy knows that 'range_timesteps' has to run in step (for the index in axis 1,2,3).
Thanks in advance!

Lets take a quick look at slicing numpy array first:
a = np.arange(0,9,1).reshape([3,3])
array([[0, 1, 2],
[3, 4, 5],
[6, 7, 8]])
Numpy has 2 ways of slicing array, full sections start:stop and by index from a list [index1, index2 ...]. The output will still be an array with the shape of your slice:
a[0:2,:]
array([[0, 1, 2],
[3, 4, 5]])
a[:,[0,2]]
array([[0, 2],
[3, 5],
[6, 8]])
The second part is that since you get a returned array with the same amount of dimensions you can easily stack any number of slices as long as you dont try to directly access an index outside of the array.
a[:][:][:][:][:][:][:][[0,2]][:,[0,2]]
array([[0, 2],
[6, 8]])

Fill numpy array with other numpy array

I have following numpy arrays:
whole = np.array(
[1, 0, 3, 0, 6]
)
sparse = np.array(
[9, 8]
)
Now I want to replace every zero in the whole array in chronological order with the items in the sparse array. In the example my desired array would look like:
merged = np.array(
[1, 9, 3, 8, 6]
)
I could write a small algorithm by myself to fix this but if someone knows a time efficient way to solve this I would be very grateful for you help!

Do you assume that sparse has the same length as there is zeros in whole ?
If so, you can do:
import numpy as np
from copy import copy
whole = np.array([1, 0, 3, 0, 6])
sparse = np.array([9, 8])
merge = copy(whole)
merge[whole == 0] = sparse
if the lengths mismatch, you have to restrict to the correct length using len(...) and slicing.

Need help converting Matlab's bsxfun to numpy

I'm trying to convert a piece of MATLAB code, and this is a line I'm struggling with:
f = 0
wlab = reshape(bsxfun(#times,cat(3,1-f,f/2,f/2),lab),[],3)
I've come up with
wlab = lab*(np.concatenate((3,1-f,f/2,f/2)))
How do I reshape it now?

Not going to do it for your code, but more as a general knowledge:
bsxfun is a function that fills a gap in MATLAB that python doesn't need to fill: broadcasting.
Broadcasting is a thing where if a matrix that is being multiplied/added/whatever similar is not the same size as the other one being used, the matrix will be repeated.
So in python, if you have a 3D matrix A and you want to multiply every 2D slice of it with a matrix B that is 2D, you dont need anything else, python will broadcast B for you, it will repeat the matrix again and again. A*B will suffice. However, in MATLAB that will raise an error Matrix dimension mismatch. To overcome that, you'd use bsxfun as bsxfun(#times,A,B) and this will broadcast (repeat) B over the 3rd dimension of A.
This means that converting bsxfun to python generally requires nothing.

MATLAB
reshape(x,[],3)
is the equivalent of numpy
np.reshape(x,(-1,3))
the [] and -1 are place holders for 'fill in the correct shape here'.
===============
I just tried the MATLAB expression is Octave - it's on a different machine, so I'll just summarize the action.
For lab=1:6 (6 elements) the bsxfun produces a (1,6,3) matrix; the reshape turns it into (6,3), i.e. just removes the first dimension. The cat produces a (1,1,3) matrix.
np.reshape(np.array([1-f,f/2,f/2])[None,None,:]*lab[None,:,None],(-1,3))
For lab with shape (n,m), the bsxfun produces a (n,m,3) matrix; the reshape would make it (n*m,3)
So for a 2d lab, the numpy needs to be
np.array([1-f,f/2,f/2])[None,None,:]*lab[:,:,None]
(In MATLAB the lab will always be 2d (or larger), so this 2nd case it closer to its action even if n is 1).
=======================
np.array([1-f,f/2,f/2])*lab[...,None]
would handle any shaped lab
If I make the Octave lab (4,2,3), the `bsxfun is also (4,2,3)
The matching numpy expression would be
In [94]: (np.array([1-f,f/2,f/2])*lab).shape
Out[94]: (4, 2, 3)
numpy adds dimensions to the start of the (3,) array to match the dimensions of lab, effectively
(np.array([1-f,f/2,f/2])[None,None,:]*lab) # for 3d lab
If f=0, then the array is [1,0,0], so this has the effect of zeroing values on the last dimension of lab. In effect, changing the 'color'.

It is equivalent to
import numpy as np
wlab = np.kron([1-f,f/2,f/2],lab.reshape(-1,1))

In Python, if you use numpy you do not need to do any broadcasting, as this is done automatically for you.
For instance, looking at the following code should make it clearer:
>>> import numpy as np
>>> a = np.array([[1, 2, 3], [3, 4, 5], [6, 7, 8], [9, 10, 100]])
>>> b = np.array([1, 2, 3])
>>>
>>> a
array([[ 1, 2, 3],
[ 3, 4, 5],
[ 6, 7, 8],
[ 9, 10, 100]])
>>> b
array([1, 2, 3])
>>>
>>> a - b
array([[ 0, 0, 0],
[ 2, 2, 2],
[ 5, 5, 5],
[ 8, 8, 97]])
>>>

numpy.square returns incorrect result for sparse matrices

numpy.square seems to give incorrect output when scipy.sparse matrices are passed to it:
import numpy as np
import scipy.sparse as S
a = np.array([np.arange(5), np.arange(5), np.arange(5), np.arange(5), np.arange(5)])
a
# array([[0, 1, 2, 3, 4],
# [0, 1, 2, 3, 4],
# [0, 1, 2, 3, 4],
# [0, 1, 2, 3, 4],
# [0, 1, 2, 3, 4]])
np.square(a)
# array([[ 0, 1, 4, 9, 16],
# [ 0, 1, 4, 9, 16],
# [ 0, 1, 4, 9, 16],
# [ 0, 1, 4, 9, 16],
# [ 0, 1, 4, 9, 16]])
b = S.lil_matrix(a)
c = np.square(b)
c
# <5x5 sparse matrix of type '<class 'numpy.int64'>'
# with 20 stored elements in Compressed Sparse Row format>
c[2,2]
# 20
# Expected output is 4, as in np.square(a) output above.
Is this a bug?

In general, passing in scipy.sparse matrices into numpy functions that take arrays ("array_like") as input, results to undefined/unintended behavior.
There is no automatic sparse -> dense cast.
Numpy does not know anything about Scipy's sparse matrices.
Sparse matrices are not "array_like" in the sense understood by Numpy.
What the numpy functions then do is to treat the sparse matrices as just some Python objects of an unknown type --- in general resulting to putting them to 1-element object arrays, and working on from there. For returning scalar results, the temporary object array is discarded and just the object contained inside it is returned, so it's easy to miss that something strange was actually done.
Object arrays have some fallbacks for performing arithmetic etc operations on their elements (unknown Python objects), including calling operator.mul of the element if * needs to be performed and so on. This then combined with the above results to the behavior you see.

Update: As pointed out by hpaulj, the reason is probably a bit more involved. np.square is able to detect np.matrix and is able to square the elements. However, it falters on sp.sparse.*matrix.
This is not a bug; this is the subtle difference between how numpy and scipy implement the __mul__ operator. By default, * for numpy.ndarray performs element-wise multiplication whereas for numpy.matrix (and consequently, for scipy.sparse.*matrix), it performs matrix multiplication (from PEP 465):
numpy provides two different types with different __mul__ methods. For
numpy.ndarray objects, * performs elementwise multiplication, and
matrix multiplication must use a function call (numpy.dot). For
numpy.matrix objects, * performs matrix multiplication, and
elementwise multiplication requires function syntax.
Internally, numpy.square uses the provided argument's __mul__ method, which is different for ndarrays and matrixes.