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
.
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Answer
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)]