Individually check the truth value for array elements

Question

My code is:-

import numpy as np
def f(x):
    if x<=0.5:
        return 0
    else:
        return 1
        
X=np.array([0.1,0.2,0.6,0.5,0.01,1])
print(f(X))

This code gives an error

ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

but I don't know how a.any() or a.all() will be useful for me. I want the output to be:-

[0,0,1,0,0,1]

I can use the for loop, but is there any shortcut syntax so that I get the desired output in just one go?

The easiest would be np.where(X<=0.5, 0, 1). See the docs here. — Péter Leéh
– Péter Leéh, Commented Aug 15, 2020 at 8:26

bigbounty · Accepted Answer · 2020-08-15 08:26:12Z

0

You can use vectorize - https://numpy.org/doc/stable/reference/generated/numpy.vectorize.html

In [1]: import numpy as np

In [2]: X=np.array([0.1,0.2,0.6,0.5,0.01,1])

In [3]: f = lambda x: 0 if x<=0.5 else 1

In [4]: func = np.vectorize(f)

In [5]: func(X)
Out[5]: array([0, 0, 1, 0, 0, 1])

OR

where - https://numpy.org/doc/stable/reference/generated/numpy.where.html

In [7]: np.where(X<=0.5, 0,1)
Out[7]: array([0, 0, 1, 0, 0, 1])

answered Aug 15, 2020 at 8:26

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