Pseudocode:
if (all elements in c == 0) or (all elements in c == 2):
# all elements in 'c' are either 0 or 2.
. . .
If c = numpy.array[0,0,2] the condition is true,
but if c = numpy.array[0,1,2] it is false.
How to test this condition in Numpy efficiently?
numpy.isin is designed for this:
import numpy as np
arr1 = np.array([0, 0, 2])
arr2 = np.array([0, 1, 2])
np.isin(arr1, [0, 2]).all()
# True
np.isin(arr2, [0, 2]).all()
# False
This, of course, works regardless of ndim:
arr3 = np.random.randint(0, 3, (100, 100))
arr4 = np.random.choice([0,2], (100, 100))
np.isin(arr3, [0, 2]).all()
# False
np.isin(arr4, [0, 2]).all()
# True
You could use binary operators as logical ones:
((x == 0) | (x == 2)).all()
This is slightly faster (~20-30%) than "np.isin" solution.
Simple method : Just count the number of 0s and 2s , check if it's count is equal to length of array:
def check(array):
return array.count(0) + array.count(2) == len(array)
quick visual test is doing
np.unique(arr1)
This lists all the unique elements in arr1. So if you get anything that doens't contain only 0 or 2 you can visually know immediately. Just a tip.
arr1, thus it will be significantly slower than other solutions