Effificient distance-like matrix computation (manual metric function)

Question

I want to compute a \"distance\" matrix, similarly to scipy.spatial.distance.cdist, but using the intersection over union (IoU) between \"bounding boxes\" (4-dimensional vec

Answer 1

If you want a performant solution you could use cython or numba. With both it is quite straight forward to outperform your cdist approach by 3 orders of magnitude.

Template function

For other distance functions like Minkowski distance I have written an answer a month ago.

import numpy as np
import numba as nb
from scipy.spatial.distance import cdist

def gen_cust_dist_func(kernel,parallel=True):

    kernel_nb=nb.njit(kernel,fastmath=True)

    def cust_dot_T(A,B):
        assert B.shape[1]==A.shape[1]

        out=np.empty((A.shape[0],B.shape[0]),dtype=np.float64)
        for i in nb.prange(A.shape[0]):
            for j in range(B.shape[0]):
                out[i,j]=kernel_nb(A[i,:],B[j,:])
        return out

    if parallel==True:
        return nb.njit(cust_dot_T,fastmath=True,parallel=True)
    else:
        return nb.njit(cust_dot_T,fastmath=True,parallel=False)

Example with your custom function

def compute_iou(bbox_a, bbox_b):
    xA = max(bbox_a[0], bbox_b[0])
    yA = max(bbox_a[1], bbox_b[1])
    xB = min(bbox_a[2], bbox_b[2])
    yB = min(bbox_a[3], bbox_b[3])

    interArea = max(0, xB - xA + 1) * max(0, yB - yA + 1)
    boxAArea = (bbox_a[2] - bbox_a[0] + 1) * (bbox_a[3] - bbox_a[1] + 1)
    boxBArea = (bbox_b[2] - bbox_b[0] + 1) * (bbox_b[3] - bbox_b[1] + 1)

    iou = interArea / float(boxAArea + boxBArea - interArea)

    return iou

#generarte custom distance function
cust_dist=gen_cust_dist_func(compute_iou,parallel=True)


A_bboxes = np.array([[0, 0, 10, 10], [5, 5, 15, 15]]*100)
B_bboxes = np.array([[1, 1, 11, 11], [4, 4, 13, 13], [9, 9, 13, 13]]*1000)

%timeit cust_dist(A_bboxes,B_bboxes)
#1.74 ms ± 13.3 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit cdist(A_bboxes, B_bboxes, lambda u, v: compute_iou(u, v))
#3.33 s ± 11.8 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)