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

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!!