Finding the nearest value and return the index of array in Python

I found this post: Python: finding an element in an array

and it's about returning the index of an array through matching the values.

On the other hand, what I am thinking of doing is similar but different. I would like to find the nearest value for the target value. For example I am looking for 4.2 but I know in the array there is no 4.2 but I want to return the index of the value 4.1 instead of 4.4.

What would be the fastest way of doing it?

I am thinking of doing it the old way like how I used to do it with Matlab, which is using the array A where I want to get the index from to minus the target value and take the absolute of it, then select the min. Something like this:-

[~,idx] = min(abs(A - target))

That is Matlab code but I am newbie in Python so I am thinking, is there a fast way of doing it in Python?

Thank you so much for your help!

This is similar to using bisect_left, but it'll allow you to pass in an array of targets

def find_closest(A, target):
    #A must be sorted
    idx = A.searchsorted(target)
    idx = np.clip(idx, 1, len(A)-1)
    left = A[idx-1]
    right = A[idx]
    idx -= target - left < right - target
    return idx

Some explanation:

First the general case: idx = A.searchsorted(target) returns an index for each target such that target is between A[index - 1] and A[index]. I call these left and right so we know that left < target <= right. target - left < right - target is True (or 1) when target is closer to left and False (or 0) when target is closer to right.

Now the special case: when target is less than all the elements of A, idx = 0. idx = np.clip(idx, 1, len(A)-1) replaces all values of idx < 1 with 1, so idx=1. In this case left = A[0], right = A[1] and we know that target <= left <= right. Therefor we know that target - left <= 0 and right - target >= 0 so target - left < right - target is True unless target == left == right and idx - True = 0.

There is another special case if target is greater than all the elements of A, In that case idx = A.searchsorted(target) and np.clip(idx, 1, len(A)-1) replaces len(A) with len(A) - 1 so idx=len(A) -1 and target - left < right - target ends up False so idx returns len(A) -1. I'll let you work though the logic on your own.

For example:

In [163]: A = np.arange(0, 20.)

In [164]: target = np.array([-2, 100., 2., 2.4, 2.5, 2.6])

In [165]: find_closest(A, target)
Out[165]: array([ 0, 19,  2,  2,  3,  3])

The corresponding Numpy code is almost the same, except you use numpy.argmin to find the minimum index.

idx = numpy.argmin(numpy.abs(A - target))

Well, more than 2 years have gone by and I have found a very simple implementation from this URL in fact: Find nearest value in numpy array

The implementation is:

def getnearpos(array,value):
    idx = (np.abs(array-value)).argmin()
    return idx   

Cheers!!

Tested and timed two solutions:

idx = np.searchsorted(sw, sCut)

and

idx = np.argmin(np.abs(sw - sCut))

for computation in a time expensive method. timing was 113s for computation with the second solution, and 132s for computation with the first one.

Possible solution:

>>> a = [1.0, 3.2, -2.5, -3.1]
>>> i = -1.5
>>> diff = [(abs(i - x),idx) for (idx,x) in enumerate(a)]
>>> diff
[(2.5, 0), (4.7, 1), (1.0, 2), (1.6, 3)]
>>> diff.sort()
>>> diff
[(1.0, 2), (1.6, 3), (2.5, 0), (4.7, 1)]

You'll have the index of nearest value in diff[0][1]