In Python, how would you check if a number is one of the integer types?

Question

In Python, how could you check if the type of a number is an integer without checking each integer type, i.e., 'int', 'numpy.int32', or 'numpy.int64'?

I thought to try if int(val) == val but this does not work when a float is set to an integer value (not type).

In [1]: vals = [3, np.ones(1, dtype=np.int32)[0], np.zeros(1, dtype=np.int64)[0], np.ones(1)[0]]
In [2]: for val in vals:
   ...:     print(type(val))
   ...:     if int(val) == val: 
   ...:         print('{} is an int'.format(val))
<class 'int'>
3 is an int
<class 'numpy.int32'>
1 is an int
<class 'numpy.int64'>
0 is an int
<class 'numpy.float64'>
1.0 is an int

I want to filter out the last value, which is a numpy.float64.

Answer 1

You can use isinstance with a tuple argument containing the types of interest.

To capture all python and numpy integer types use:

isinstance(value, (int, np.integer))

Here is an example showing the results for several data types:

vals = [3, np.int32(2), np.int64(1), np.float64(0)]
[(e, type(e), isinstance(e, (int, np.integer))) for e in vals]

Result:

[(3, <type 'int'>, True), 
 (2, <type 'numpy.int32'>, True), 
 (1, <type 'numpy.int64'>, True), 
 (0.0, <type 'numpy.float64'>, False)]

A second example which is true only for int and int64:

[(e, type(e), isinstance(e, (int, np.int64))) for e in vals]

Result:

[(3, <type 'int'>, True), 
(1, <type 'numpy.int32'>, False), 
(0, <type 'numpy.int64'>, True), 
(0.0, <type 'numpy.float64'>, False)]

Answer 2

In addition to the accepted answer, the standard library also provides the numbers module, which defines abstract classes for numerical types that can be used with isinstance:

isinstance(value, numbers.Integral)

This can be used to recognize both Python's int as well as NumPy's np.int64.

Answer 3

Use np.issubdtype:

for val in vals:
    if isinstance(val, int) or np.issubdtype(val, np.int):
        print('{} is an int'.format(val))