Find elements of array one nearest to elements of array two
This answer explains how to find the nearest (sorted) array element to a single point, in a manner efficient for large arrays (slightly modified):
def arg_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 idx-1
else:
return idx
If, instead, we want to find the array elements nearest a set of points (i.e. a second array); are there efficient (by speed, for large arrays) ways of extending this besides using a for-loop?
Some test cases:
>>> xx = [0.2, 0.8, 1.3, 1.5, 2.0, 3.1, 3.8, 3.9, 4.5, 5.1, 5.5]
>>> yy = [1, 2, 3, 4, 5]
>>> of_x_nearest_y(xx, yy)
[0.5, 2.0, 3.1, 3.9, 5.1]
>>> xx = [0.2, 0.8, 1.3, 1.5, 2.0, 3.1, 3.8, 3.9, 4.5, 5.1, 5.5]
>>> yy = [-2, -1, 4.6, 5.8]
>>> of_x_nearest_y(xx, yy)
[0.2, 0.2, 4.5, 5.5]
Edit: assuming both arrays are sorted, you can do a little better than a completely naive for-loop by excluding values below those already matched, i.e.
def args_nearest(options, targets):
locs = np.zeros(targets.size, dtype=int)
prev = 0
for ii, tt in enumerate(targets):
locs[ii] = prev + arg_nearest(options[prev:], tt)
prev = locs[ii]
return locs
Solution 1:
You can make few changes to extend it for an array of elements in value
, like so -
idx = np.searchsorted(xx, yy, side="left").clip(max=xx.size-1)
mask = (idx > 0) & \
( (idx == len(xx)) | (np.fabs(yy - xx[idx-1]) < np.fabs(yy - xx[idx])) )
out = xx[idx-mask]
Explanation
Nomenclature : array
is the array in which we are looking to place elements from value
to maintain the sorted nature of array
.
Changes needed to extend the solution for a single element to many elements for searching :
1] Clip the indices array idx
obtained from np.searchsorted
at a max. of array.size-1
, because for elements in value
that are larger than the maximum of array
, we need to make idx
indexable by array
.
2] Introduce numpy
to replace math
to do those operations in a vectorized manner.
3] Replace the conditional statement by the trick of idx - mask
. In this case, internally Python would up-convert mask
to an int
array to match up with the datatype of idx
. Thus, all the True
elements become 1
and thus for True
elements we would effectively have idx-1
, which is the True
case of the IF conditional statement in the original code.