Filter integers in numpy float array

Is there any built in function to discard integer and keep only float number in numpy.

import numpy as np

input = np.array([0.0, 0.01, 1.0, 2.0, 2.001, 2.002])

desired_ouput = some_function(input)
# Expected ouput
# desired_output = np.array([0.01, 2.001, 2.002])


Mask with whether each element is equal to it as an integer.

arr = np.array([0.0, 0.01, 1.0, 2.0, 2.001, 2.002])
out = arr[arr != arr.astype(int)]
#np.array([0.01, 2.001, 2.002])


I don't think so. My approach would be

import numpy as np
a = np.array([0.0, 0.01, 1.0, 2.0, 2.001, 2.002])
mask = np.isclose(a, a.astype(int))

print(a[~mask])
#[ 0.01 2.001 2.002]


I know of no in-built function. But you can create one yourself:

import numpy as np

A = np.array([0.0, 0.01, 1.0, 2.0, 2.001, 2.002])

def remove_ints(arr):
return arr[~(arr == arr.astype(int))]

res = remove_ints(A)

array([ 0.01 , 2.001, 2.002])


Aside, you should not use a built-in class such as input as a variable name.

I've always used np.equal with np.mod:

>>> A[~np.equal(np.mod(A, 1), 0)]
array([0.01 , 2.001, 2.002])


If you do not have to much data (short list), maybe do not need numpy:

>>> i = [0.0, 0.01, 1.0, 2.0, 2.001, 2.002]
>>> a=[j for j in i if not j.is_integer()]
>>> a
['0.01', '2.001', '2.002']


Otherwise see Joe Iddon answer

I don't know any builtin for this but you can filter those floats using:

filter(lambda x: int(str(x).split('.')[1]) != 0, input)


The lambda expression here checks if the decimal places are zero which I interpret as the number being an int.