[Python] Numpy Batched IoU 코드

def compute_iou(bboxes1, bboxes2, bbox_format='cxcywh', normalized=True):
    if bbox_format == 'cxcywh':
        # if bboxes are composed of center point xy and wh/2.
        cx1, cy1, w1, h1 = np.split(bboxes1, 4, axis=1)
        cx2, cy2, w2, h2 = np.split(bboxes2, 4, axis=1)

        x11 = cx1 - w1 / 2
        y11 = cy1 - h1 / 2
        x12 = cx1 + w1 / 2 
        y12 = cy1 + h1 / 2

        x21 = cx2 - w2 /2
        y21 = cy2 - h2 /2
        x22 = cx2 + w2 /2
        y22 = cy2 + h2 /2
    elif bbox_format == 'tlxtlywh':
        # if bboxes are composed of top left point xy and wh.
        tlx1, tly1, w1, h1 = np.split(bboxes1, 4, axis=1)
        tlx2, tly2, w2, h2 = np.split(bboxes2, 4, axis=1)
        
        x11 = tlx1
        y11 = tly1
        x12 = tlx1 + w1 
        y12 = tly1 + h1

        x21 = tlx2
        y21 = tly2
        x22 = tlx2 + w2
        y22 = tly2 + h2
    elif bbox_format == 'tlxtlybrxbry':
        # if bboxes are composed of top left point xy and bottom right xy.
        x11, y11, x12, y12 = np.split(bboxes1, 4, axis=1)
        x21, y21, x22, y22 = np.split(bboxes2, 4, axis=1)

    # determine the (x, y)-coordinates of the intersection rectangle
    inter_x1 = np.maximum(x11, x21.T)
    inter_y1 = np.maximum(y11, y21.T)
    inter_x2 = np.minimum(x12, x22.T)
    inter_y2 = np.minimum(y12, y22.T)

    # # compute the area of intersection rectangle
    offset = 0. if normalized else 1.
    
    inter_area = np.maximum(inter_x2 - inter_x1 + offset , 1e-6) * np.maximum(inter_y2 - inter_y1 + offset , 1e-6)

    # # compute the area of both the prediction and ground-truth rectangles
    bboxes1_area = (x12 - x11 + offset ) * (y12 - y11 + offset)
    bboxes2_area = (x22 - x21 + offset ) * (y22 - y21 + offset)

    iou = inter_area / ((bboxes1_area + bboxes2_area.T - inter_area) + 1e-5)
    return iou