How to compare two numpy.array based on numerical part and non-numerical part respectively?

Question

import numpy as np

a = np.array([[1., 2., 3.], 
              [4., 5., 'a']], dtype=object)
b = np.array([[1.00000001, 2., 3.], 
              [4., 5., 'a']], dtype=object)
print(a == b)

actual output:

[[False  True  True]
 [ True  True  True]]

expected output (since 1.00000001 is close enough to 1):

[[True  True  True]
 [True  True  True]]

I cannot use numpy.isclose() because there is non-numerical part in the array.

You could strip off the string and convert to float: a.ravel()[:-1].astype(float). isclose uses the difference, and also checks for floats like inf, which is why it needs the numeric dtype. Or you could leave the values object dtype, and do your own test of the differences a.ravel()[:-1] — hpaulj
– hpaulj, Commented Jan 6, 2022 at 4:15
The non-numerical part can be in any place in the array, not only in the last place. — 吴慈霆
– 吴慈霆, Commented Jan 6, 2022 at 4:53

Marco Spinaci · Accepted Answer · 2022-01-06 10:59:43Z

0

It's a bit hacky, but you could use masks to solve it. For example, using the is_numeric_3 function from this answer (which I'll report here, too, for completeness):

def is_float(val):
        try:
            float(val)
        except ValueError:
            return False
        else:
            return True

is_numeric_3 = np.vectorize(is_float, otypes = [bool]) # return numpy array

mask_a = is_numeric_3(a)
mask_b = is_numeric_3(b)
mask = mask_a & mask_b

result = a == b
result[mask] = np.isclose(a[mask], b[mask])

answered Jan 6, 2022 at 10:59

Marco Spinaci

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