Is there a numpy-thonic way, e.g. function, to find the *nearest value* in an array?

Example:

```
np.find_nearest( array, value )
```

0 votes

Is there a numpy-thonic way, e.g. function, to find the *nearest value* in an array?

Example:

```
np.find_nearest( array, value )
```

0 votes

I think the most pythonic way would be:

```
num = 65 # Input number
array = n.random.random((10))*100 # Given array
nearest_idx = n.where(abs(array-num)==abs(array-num).min())[0] # If you want the index of the element of array (array) nearest to the the given number (num)
nearest_val = array[abs(array-num)==abs(array-num).min()] # If you directly want the element of array (array) nearest to the given number (num)
```

This is the basic code. You can use it as a function if you want

0 votes

```
import numpy as np
def find_nearest(array,value):
idx = (np.abs(array-value)).argmin()
return array[idx]
array = np.random.random(10)
print(array)
# [ 0.21069679 0.61290182 0.63425412 0.84635244 0.91599191 0.00213826
# 0.17104965 0.56874386 0.57319379 0.28719469]
value = 0.5
print(find_nearest(array, value))
# 0.568743859261
```

0 votes

With slight modification, the answer above works with arrays of arbitrary dimension (1d, 2d, 3d, ...):

```
def find_nearest(a, a0):
"Element in nd array `a` closest to the scalar value `a0`"
idx = np.abs(a - a0).argmin()
return a.flat[idx]
```

Or, written as a single line:

```
a.flat[np.abs(a - a0).argmin()]
```

0 votes

Here's a version that will handle a non-scalar "values" array:

```
import numpy as np
def find_nearest(array, values):
indices = np.abs(np.subtract.outer(array, values)).argmin(0)
return array[indices]
```

Or a version that returns a numeric type (e.g. int, float) if the input is scalar:

```
def find_nearest(array, values):
values = np.atleast_1d(values)
indices = np.abs(np.subtract.outer(array, values)).argmin(0)
out = array[indices]
return out if len(out) > 1 else out[0]
```

0 votes

Here's an extension to find the nearest vector in an array of vectors.

```
import numpy as np
def find_nearest_vector(array, value):
idx = np.array([np.linalg.norm(x+y) for (x,y) in array-value]).argmin()
return array[idx]
A = np.random.random((10,2))*100
""" A = array([[ 34.19762933, 43.14534123],
[ 48.79558706, 47.79243283],
[ 38.42774411, 84.87155478],
[ 63.64371943, 50.7722317 ],
[ 73.56362857, 27.87895698],
[ 96.67790593, 77.76150486],
[ 68.86202147, 21.38735169],
[ 5.21796467, 59.17051276],
[ 82.92389467, 99.90387851],
[ 6.76626539, 30.50661753]])"""
pt = [6, 30]
print find_nearest_vector(A,pt)
# array([ 6.76626539, 30.50661753])
```

0 votes

If you don't want to use numpy this will do it:

```
def find_nearest(array, value):
n = [abs(i-value) for i in array]
idx = n.index(min(n))
return array[idx]
```

0 votes

*IF* your array is sorted and is very large, this is a much faster solution:

```
def find_nearest(array,value):
idx = np.searchsorted(array, value, side="left")
if idx > 0 and (idx == len(array) or math.fabs(value - array[idx-1]) < math.fabs(value - array[idx])):
return array[idx-1]
else:
return array[idx]
```

This scales to very large arrays. You can easily modify the above to sort in the method if you can't assume that the array is already sorted. It’s overkill for small arrays, but once they get large this is much faster.

0 votes

For large arrays, the (excellent) answer given by @Demitri is far faster than the answer currently marked as best. I've adapted his exact algorithm in the following two ways:

The function below works whether or not the input array is sorted.

The function below returns the

*index*of the input array corresponding to the closest value, which is somewhat more general.

Note that the function below also handles a specific edge case that would lead to a bug in the original function written by @Demitri. Otherwise, my algorithm is identical to his.

```
def find_idx_nearest_val(array, value):
idx_sorted = np.argsort(array)
sorted_array = np.array(array[idx_sorted])
idx = np.searchsorted(sorted_array, value, side="left")
if idx >= len(array):
idx_nearest = idx_sorted[len(array)-1]
elif idx == 0:
idx_nearest = idx_sorted[0]
else:
if abs(value - sorted_array[idx-1]) < abs(value - sorted_array[idx]):
idx_nearest = idx_sorted[idx-1]
else:
idx_nearest = idx_sorted[idx]
return idx_nearest
```

0 votes

Here is a version with scipy for @Ari Onasafari, answer "** to find the nearest vector in an array of vectors**"

```
In [1]: from scipy import spatial
In [2]: import numpy as np
In [3]: A = np.random.random((10,2))*100
In [4]: A
Out[4]:
array([[ 68.83402637, 38.07632221],
[ 76.84704074, 24.9395109 ],
[ 16.26715795, 98.52763827],
[ 70.99411985, 67.31740151],
[ 71.72452181, 24.13516764],
[ 17.22707611, 20.65425362],
[ 43.85122458, 21.50624882],
[ 76.71987125, 44.95031274],
[ 63.77341073, 78.87417774],
[ 8.45828909, 30.18426696]])
In [5]: pt = [6, 30] # <-- the point to find
In [6]: A[spatial.KDTree(A).query(pt)[1]] # <-- the nearest point
Out[6]: array([ 8.45828909, 30.18426696])
#how it works!
In [7]: distance,index = spatial.KDTree(A).query(pt)
In [8]: distance # <-- The distances to the nearest neighbors
Out[8]: 2.4651855048258393
In [9]: index # <-- The locations of the neighbors
Out[9]: 9
#then
In [10]: A[index]
Out[10]: array([ 8.45828909, 30.18426696])
```

0 votes

Here is a fast vectorized version of @Dimitri's solution if you have many `values`

to search for (`values`

can be multi-dimensional array):

```
def get_closest(array, values):
# get insert positions
idxs = np.searchsorted(array, values, side="left")
# find indexes where previous index is closer
prev_idx_is_less = ((idxs == len(array))|(np.fabs(values - array[np.maximum(idxs-1, 0)]) < np.fabs(values - array[np.minimum(idxs, len(array)-1)])))
idxs[prev_idx_is_less] -= 1
return array[idxs]
```

**Benchmarks**

> 100 times faster than using a `for`

loop with @Demitri's solution`

```
>>> %timeit ar=get_closest(np.linspace(1, 1000, 100), np.random.randint(0, 1050, (1000, 1000)))
139 ms ± 4.04 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
>>> %timeit ar=[find_nearest(np.linspace(1, 1000, 100), value) for value in np.random.randint(0, 1050, 1000*1000)]
took 21.4 seconds
```

...