How to check if all elements of a numpy array have integer values?

Question

As the title says, it is not about type of elements. I need to be sure that the values of elements are integers, i.e.

np.array([1, 2, 3])
np.array([1., 2.0, 9/3])

must both give [True, True, True] after the 'Are they integers?'-checking.

Is there a clean and pythonic/numpyic way of doing this?

I've tried some many-lines-combinations such as:

isinstance(x, (int, np.integer)) 
#or
(1.0).is_integer()

but they are cumbersome and ugly.

I've made some progress in this regard. There is a standalone numpy function now available. Let's see if it makes it to numpy. — Mad Physicist, Dec 31 '21 at 01:24

score 3 · Accepted Answer · answered Sep 24 '20 at 10:10

3

What I use is this quantity % int(quantity) == 0

answered Sep 24 '20 at 10:10

this doesn't work for `v` such that `int(v) == 0` though – Kukuster Aug 21 '22 at 20:41

score 2 · Answer 2 · answered Sep 24 '20 at 10:24

2

One More Way:

>>> x = np.array([1.,2.0,9/3])
>>> [not (i%1) for i in x]
[True, True, True]

answered Sep 24 '20 at 10:24

Shantanu Tripathi

score 1 · Answer 3 · edited Jun 17 '22 at 18:00

1

Here is a oneliner using is_integer():

>>> x = np.array([1., 2.0, 9/3])
>>> all([i.is_integer() for i in x])
True

edited Jun 17 '22 at 18:00

Michael Currie

answered Sep 24 '20 at 10:13

Grayrigel

I have added one line solution using `is_integer()`, which is not cumbersome. – Grayrigel Sep 24 '20 at 10:59
`all(i.is_integer() for i in x)` is 'more correct'. remove the [] and it will be faster (i.e. stop at the first non integer encounter). There is no point in creating a temporary list. – Andreas H. Mar 08 '23 at 12:56

3 Answers3