Develop Reference

Mask an 2Darray row wise by another array

Consider the following 2d array:
>>> A = np.arange(2*3).reshape(2,3)
array([[0, 1, 2],
[3, 4, 5]])
>>> b = np.array([1, 2])
I would like to get the following mask from A as row wise condition from b as an upper index limit:
>>> mask
array([[True, False, False],
[True, True, False]])
Can numpy do this in a vectorized manner?

You can use array broadcasting:
mask = np.arange(A.shape[1]) < b[:,None]
output:
array([[ True, False, False],
[ True, True, False]])

Another possible solution, based on the idea that the wanted mask corresponds to a boolean lower triangular matrix:
mask = np.tril(np.ones(A.shape, dtype=bool))
Output:
array([[ True, False, False],
[ True, True, False]])

Get 2d np.array row indices depending on other 2d np.array

I want to fetch all row indicies of an 2d np.array on the basis of another 2d np.array.
Input:
all_elements = np.array([[1, 1], [2, 2], [3, 3]])
elements = np.array([[1, 1],[2, 2]])
Something like:
idx = np.row_idx(elements, all_elements, axis=0)
output:
[0, 1]
I have been trying to do this with some version of np.where(np.isin(.....)) but i can't get it to work properly.
Any1 have any tips?

IIUC, you might want to use:
np.where((all_elements==elements[:,None]).all(2).any(0))[0]
output: array([0, 1])
explanation:
# compare all elements using broadcasting
(all_elements==elements[:,None])
array([[[ True, True],
[False, False],
[False, False]],
[[False, False],
[ True, True],
[False, False]]])
# all True on the last dimension
(all_elements==elements[:,None]).all(2)
array([[ True, False, False],
[False, True, False]])
# any match per first dimension
(all_elements==elements[:,None]).all(2).any(0)
array([ True, True, False])

How can I use numpy array elements as indices to assign values for another numpy array

I have following problem, which I want to solve using numpy array elements.
The problem is:
Matrix = np.zeros((4*4), dtype = bool) which gives this 2D matrix.
Matrix = [[False, False, False, False],
[False, False, False, False],
[False, False, False, False],
[False, False, False, False]]
Les us suppose that we have an another array a = np.array([0,1], [2,1], [3,3])
a = [[0, 1],
[2, 1],
[3, 3]]
My question is: How to use the elements of the a array as indices to fill my matrix with True's. The output should seem like this
Matrix = [[False, True, False, False], # [0, 1]
[False, False, False, False],
[False, True, False, False], # [2, 1]
[False, False, False, True]] # [3, 3]

import numpy as np
Matrix = np.zeros((4*4), dtype = bool).reshape(4,4)
a = [[0, 1],
[2, 1],
[3, 3]]
Unroll them into a proper pair of indexing arrays for a 2d array
a = ([x[0] for x in a], [x[1] for x in a])
Matrix[a] = True
>>> Matrix
array([[False, True, False, False],
[False, False, False, False],
[False, True, False, False],
[False, False, False, True]])

Simple way to make the (4,4) bool array:
In [390]: arr = np.zeros((4,4), dtype = bool)
In [391]: arr
Out[391]:
array([[False, False, False, False],
[False, False, False, False],
[False, False, False, False],
[False, False, False, False]])
Proper syntax for making a:
In [392]: a = np.array([[0,1], [2,1], [3,3]])
In [393]: a
Out[393]:
array([[0, 1],
[2, 1],
[3, 3]])
Use the 2 columns of a as indices for the 2 dimensions of arr:
In [394]: arr[a[:,0],a[:,1]]=True
In [395]: arr
Out[395]:
array([[False, True, False, False],
[False, False, False, False],
[False, True, False, False],
[False, False, False, True]])

Compare columns of numpy matrix with array

I have a numpy matrix and want to compare every columns to a given array, like:
M = np.array([1,2,3,3,2,1,1,3,2]).reshape((3,3)).T
v = np.array([1,2,3])
Now I want to compare every columns of M with v, i.e. I want a matrix with the first column consisting of True, True, True. A second saying False, True, False. A third True, False, False.
How do I achieve this?
Thanks!

Use broadcasted comparison:
>>> M == v[:, None]
array([[ True, False, True],
[ True, True, False],
[ True, False, False]])

You might consider using np.equal column-wise:
np.array([np.equal(col, v) for col in M.T]).T
it compares the elements of two numpy arrays element-wise. The M.T makes for loop to pop your original M columns as one-dimensional arrays and the final transpose is needed to reverse it.
Here the equal/not_equal functions are described.

Alternatively, you could match each row in the matrix with the given vector using np.apply_along_axis
>>> M
array([[1, 3, 1],
[2, 2, 3],
[3, 1, 2]])
>>> v
array([1, 2, 3])
>>> np.apply_along_axis(lambda x: x==v, 1, M)
array([[ True, False, False],
[False, True, True],
[False, False, False]], dtype=bool)

Acces all off diagonal elements of boolean numpy matrix

Suppose there is a diagonal matrix M:
#import numpy as np
M = np.matrix(np.eye(5, dtype=bool))
Does anybody know a simple way to access all off diagonal elements, meaning all elements that are False? In R I can simply do this by executing
M[!M]
Unfortunately this is not valid in Python.

You need the bitwise not operator:
M[~M]

You might try np.extract combined with np.eye. For example:
M = np.array([[1, 2, 3], [4, 5, 6], [7, 8, 9]])
np.extract(1 - np.eye(3), M)
# result: array([2, 3, 4, 6, 7, 8])
In your example it's almost an identity:
M = np.matrix(np.eye(5, dtype=bool))
np.extract(1 - np.eye(5), M)
#result:
array([False, False, False, False, False, False, False, False, False,
False, False, False, False, False, False, False, False, False,
False, False], dtype=bool)