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

问题:

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!

回答1:

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 

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 - left 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 idx=1. In this case left = A[0], right = A[1] and we know that target . Therefor we know that target - left and right - target >= 0 so target - left 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 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]) 

回答2:

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)) 

回答3:

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!!

回答4:

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.

回答5:

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]

回答6:

def finder(myList, target)     diff = ''     index = None     for i,num in enumerate(myList):         if abs(target - num) 

Hope this helps

EDIT:

If you'd like a one-liner, then you might like this better:

min(L, key=lambda x: abs(target-x)) 

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