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Problem: I want to compare each element of a Numpy array with a float, returning an array with the smaller value. For example, using the inputs:

import numpy as np
input_a = 3
input_b = np.array([1,2,3,4,5])

the output should be

output = np.array([1,2,3,3,3])

My current solution is working by making a new np.array with only the constant, then using np.minimum(). 

c = np.copy(input_b)
c.fill(input_a)
output = np.minimum(input_b, c)

However, I am afraid that this is not the most efficient solution. Is there a more elegant / efficient way to achieve this?

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5 Answers 5

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3

Your best bet is to use logical indexing.

import numpy as np
input_a = 3
input_b = np.array([1,2,3,4,5])

input_b[input_b > input_a] = input_a

print(input_b)
# [1 2 3 3 3]

input_b > input_a will return a mask array of either True or False values, where in this case the element will be True if the corresponding element in input_b is greater than input_a. You can then use this to index input_b and modify only those values.

Note that using logical indexing is quicker than using numpy.where for this particular array, though I can't tell you why exactly.

setup = 'from __main__ import np, input_a, input_b'
print(timeit.timeit('input_b[input_b > input_a] = input_a', setup=setup)) 
# 2.2448947575996456
print(timeit.timeit('np.where(input_b < input_a, input_b, input_a)', setup=setup)) 
# 5.35540746395358
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2

I think np.minimum is fine for this operation:

>>> np.minimum(input_b, 3)
array([1, 2, 3, 3, 3])

If you want to modify input_b directly, use the out keyword argument to fill input_b with the pair-wise minimum values.

>>> np.minimum(input_b, 3, out=input_b)
>>> input_b
array([1, 2, 3, 3, 3])

This is quicker than using boolean indexing and then assigning values:

>>> %timeit input_b[input_b > input_a] = input_a
100000 loops, best of 3: 4.16 µs per loop

>>> %timeit np.minimum(input_b, 3, out=input_b)
100000 loops, best of 3: 2.53 µs per loop
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1

There's a builtin to do this: clip

output = input_b.clip(max=input_a)

or if you want set input_b itself

np.clip(input_b, None, out=input_b)

Here it's doing the same as minimum, but it can also do maximum in the same call. Some versions accept the max keyword, others don't.

clip has a modest edge over minimum in my timings. But I'd recommend which ever one seems clearest in intent.

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A one-liner for this would be to use numpy.where:

>>> np.where(input_b < input_a, input_b, input_a)
array([ 1.,  2.,  3.,  3.,  3.])

Here we pass numpy.where three arguments, where the first is a boolean array of where input_b < input_a. Whenever the value in this first argument is True, we take the value at the corresponding index from the second argument (input_b). Otherwise we take the value of input_a.

Edit: In fact, as @Kasra's answer shows, you can pass input_a directly without converting it to an np.array.

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You can use numpy.where : 

>>> np.where(input_b < input_a, input_b, input_a)
array([1, 2, 3, 3, 3])
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