Find index of first element in array close to float within tolerance, with numpy

Question

I need to find the index of the first element in an array, that is close to a float within a given tolerance.

I can do this with a for block:

import numpy as np

# Define array of floats
a = np.random.uniform(0., 10., 10.)

# Float to search, and tolerance.
val, tol = 6.87, 0.1

for i, _ in enumerate(a):
    if np.isclose(_, val, tol):
        print('Index is: {}'.format(i))
        break

but I was wondering if there might be a one-liner solution using some numpy function.

Notice that I'm interested in the index of the first element that is close to val, no matter that there might be closer elements further down the a array. The solutions I've found are interested in the index of the nearest value, no matter where it is located within the array.

ForceBru · Accepted Answer · 2016-12-07 17:00:54Z

8

Here's a one-liner:

Index = next(i for i, _ in enumerate(a) if np.isclose(_, val, tol))

What is this?

The code in parentheses is a generator expression and next returns (you guessed it!) the next (in this case, the first) value the generator will produce. If there'll be no next value, a StopIteration exception will be raised.

Advantages

It isn't memory-consuming as it doesn't need to compute all the possible values and store them.
It's fast as it doesn't continue looping through the array if the needed value is already found.
It will raise an exception if no value could be found.

It could be easily turned into a one-liner function:

FirstIndex = lambda a, val, tol: next(i for i, _ in enumerate(a) if np.isclose(_, val, tol))

i = FirstIndex(a, val, tol) # call it

edited Dec 7, 2016 at 17:00

answered Dec 7, 2016 at 16:41

ForceBru

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5 Comments

jmd_dk Over a year ago

This is indeed a much more efficient solution than those of José Sánchez and Divakar.

Divakar Over a year ago

Looks promising! I tried it on the sample posted in my solution and I got 0 as the index. I was expecting 1.

ForceBru Over a year ago

@Divakar, NumPy 1.8.0 on Pyton 3.5 doesn't show any result when running the OP's code, but your solution returns 0 given the exact same array as in your answer...

ForceBru Over a year ago

@Divakar, this is because numpy.isclose(a, val, rtol, atol) uses a different formula, namely absolute(a - val) <= (atol + rtol * absolute(val)).

Divakar Over a year ago

Hmm I guess it will work then, thanks for the clarifications!

José Sánchez · Accepted Answer · 2016-12-07 16:43:46Z

5

You could use numpy.where:

np.where(np.isclose(a, val, tol))

And the just get the lowest index (i.e. the first result) that is returned by where.

Hope this helps.

answered Dec 7, 2016 at 16:43

José Sánchez

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Comments

Divakar · Accepted Answer · 2016-12-07 16:58:55Z

4

Here's a vectorized one-liner -

(np.abs(a - val) <= tol).argmax()

Sample step-by-step run -

In [57]: a
Out[57]: array([5, 3, 9, 6, 8, 3, 5, 1])

In [58]: val = 2

In [59]: tol = 2

In [60]: (np.abs(a - val) < tol) # Many might satisfy
Out[60]: array([False,True,False,False,False,True,False,True], dtype=bool)

In [61]: (np.abs(a - val) <= tol).argmax() # Choose the first one with argmax
Out[61]: 1

edited Dec 7, 2016 at 16:58

answered Dec 7, 2016 at 16:42

Divakar

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1 Comment

Bijou Smith Over a year ago

Fails if there is no closest value, since it will return 0, which is wrong! You should return 'None' if there is no element closest to 'val'.

potato · Accepted Answer · 2019-03-31 20:59:49Z

1

Come on folks! None of the above are robust. Even Divakar's answer fails if there is no closest value, you'd get Index=0 which is an error. It should probably return 'None' in that case. So at a minimum a one-liner is out of the question. The original question is not well-posed. You need to consider, "...and what if there is no such close value within tolerance?" and then give specifications for what to do in that case.

i_array = np.where(np.isclose(a, val, tol))

is fair enough, but then an empty result still has to be dealt with. So, I'd say a one-liner is going to be clunky, and at minimum a two-liner is preferable:

i_array = np.where(np.isclose(a, val, tol))
my_i = None if (len(i_array[0])==0) else int(i_array[0])

edited Mar 31, 2019 at 20:59

potato

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answered Mar 13, 2019 at 3:39

Bijou Smith

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