How does one test if a matrix in Python has only 1

Question

Let's say I've got a matrix like this:

mat1 = np.array([1,0,1], [1,1,0], [0,0,0]);

And I've got another one like this:

mat2 = np.array([0,1,0], [0,0,1], [1,1,1]);

I want to detect if something like

np.add(mat1, mat2);

has only 1's or 0's, namely some 1's and some 0's, all 0's, or all 1's.

---

n.b. - Comment your code.

Answer 1

You can use unique

import numpy as np
mat1 = np.array([[1,0,1], [1,1,0], [0,0,0]])

np.unique(mat1)
# array([0, 1])
1 in np.unique(mat1)
# True
0 in np.unique(mat1)
# True
np.unique(mat1) == [0, 1]
# array([ True,  True], dtype=bool)

You can also use setdiff1d

np.setdiff1d(mat1, [0, 1])
# array([], dtype=int64)
np.setdiff1d(mat1, [0, 1]).size
# 0

Answer 2

How about this:

>>> def check(matrix):
...     # flatten up the matrix into one single list
...     # and set on the list it should be [0,1] if it
...     # contains only 0 and 1. Then do sum on that will 
...     # return 1
...     if sum(set(sum(matrix,[]))) > 1:
...         return False
...     return True
...
>>>
>>> check([[1,0,1], [1,1,0], [0,0,0]])
True
>>> check([[1,0,1], [1,1,0], [0,0,2]])
False
>>> check([[1,0,1], [1,1,0], [0,0,3]])
False
>>>

Answer 3

Simply:

In [6]:

set((mat1+mat2).ravel()).issubset(set((1,0)))
Out[6]:
True

In [7]:

mat3 = np.array([[0,5,0], [0,0,1], [1,1,1]])
set((mat1+mat3).ravel()).issubset(set((1,0)))
Out[7]:
False

Answer 4

Use numpy.all, numpy.any:

• all 0: np.all(mat == 0)
• all 1: np.all(mat == 1)
• some 0: np.any(mat == 0)
• some 1: np.any(mat == 1)

---

>>> mat1 = np.array([[1,0,1], [1,1,0], [0,0,0]])
>>> mat2 = np.array([[0,1,0], [0,0,1], [1,1,1]])

>>> np.all(mat1 == 0)
False
>>> np.any(mat1 == 0)
True
>>> np.all(mat1 == 1)
False
>>> np.any(mat1 == 1)
True

>>> mat3 = mat1 + mat2
>>> mat3
array([[1, 1, 1],
       [1, 1, 1],
       [1, 1, 1]])
>>> np.all(mat3 == 1)
True

UPDATE

To check whether the array contains only 1 or 0, nothing else, use following:

>>> mat1 = np.array([[1,0,1], [1,1,0], [0,0,0]])
>>> mat2 = np.array([[0,1,2], [3,4,5], [6,7,8]])
>>> np.all((mat1 == 0) | (mat1 == 1))
True
>>> np.all((mat2 == 0) | (mat2 == 1))
False

Answer 5

Check this out: np.sum(np.unique(mat0.ravel()))

---

So, mat0.ravel() does this:

[[1,0,0],[0,0,0],[1,1,0]] ---> [1,0,0,0,0,0,1,1,0]

This new object is an array, namely the [1,0,0,0,0,0,1,1,0] object above. Now, np.unique(mat0.ravel()) finds all the unique elements and sorts them and puts them in a set, like this:

[1,0,0,0,0,0,1,1,0] ---> {0,1}

From here if one applies np.sum on this, namely np.sum(np.unique(mat0.ravel())) we get the sum of the contents of the set, so a good condition to check if only a 0 or 1 in each and every cell in matrix is the following:

np.sum(np.unique(mat0.ravel())) > 1

---

n.b. - This is only for non-negative integers.

Answer 6

If you know it's int dtype, then (suprisingly) it's faster to check the max and min (even without doing these operations simultaneously):

In [11]: m = np.random.randint(0, 2, (10, 10))

In [12]: %timeit np.all((m == 0) | (m == 1))
10000 loops, best of 3: 33.7 µs per loop

In [13]: %timeit m.dtype == int and m.min() == 0 and m.max() == 1
10000 loops, best of 3: 29.8 µs per loop

In [21]: m = np.random.randint(0, 2, (10000, 10000))

In [22]: %timeit np.all((m == 0) | (m == 1))
1 loops, best of 3: 705 ms per loop

In [23]: %timeit m.dtype == int and m.min() == 0 and m.max() == 1
1 loops, best of 3: 481 ms per loop