python - 计算边界框重叠的百分比，用于图像检测器评估

Question

在大图像中测试对象检测算法时，我们根据为地面实况矩形给出的坐标检查检测到的边界框。

根据 Pascal VOC 挑战，有这个:

A predicted bounding box is considered correct if it overlaps more than 50% with a ground-truth bounding box, otherwise the bounding box is considered a false positive detection. Multiple detections are penalized. If a system predicts several bounding boxes that overlap with a single ground-truth bounding box, only one prediction is considered correct, the others are considered false positives.

这意味着我们需要计算重叠的百分比。这是否意味着 ground truth box 被检测到的边界框覆盖了 50%？或者边界框的 50% 被地面实况框吸收了？

我已经搜索过，但我还没有找到一个标准算法 - 这令人惊讶，因为我认为这在计算机视觉中很常见。 (我是新来的)。我错过了吗？有谁知道这类问题的标准算法是什么？

Answer 1

对于轴对齐的边界框，它相对简单。 “轴对齐”意味着边界框没有旋转；或者换句话说，框线与轴平行。下面介绍如何计算两个轴对齐的边界框的 IoU。

def get_iou(bb1, bb2):
    """
    Calculate the Intersection over Union (IoU) of two bounding boxes.

    Parameters
    ----------
    bb1 : dict
        Keys: {'x1', 'x2', 'y1', 'y2'}
        The (x1, y1) position is at the top left corner,
        the (x2, y2) position is at the bottom right corner
    bb2 : dict
        Keys: {'x1', 'x2', 'y1', 'y2'}
        The (x, y) position is at the top left corner,
        the (x2, y2) position is at the bottom right corner

    Returns
    -------
    float
        in [0, 1]
    """
    assert bb1['x1'] < bb1['x2']
    assert bb1['y1'] < bb1['y2']
    assert bb2['x1'] < bb2['x2']
    assert bb2['y1'] < bb2['y2']

    # determine the coordinates of the intersection rectangle
    x_left = max(bb1['x1'], bb2['x1'])
    y_top = max(bb1['y1'], bb2['y1'])
    x_right = min(bb1['x2'], bb2['x2'])
    y_bottom = min(bb1['y2'], bb2['y2'])

    if x_right < x_left or y_bottom < y_top:
        return 0.0

    # The intersection of two axis-aligned bounding boxes is always an
    # axis-aligned bounding box
    intersection_area = (x_right - x_left) * (y_bottom - y_top)

    # compute the area of both AABBs
    bb1_area = (bb1['x2'] - bb1['x1']) * (bb1['y2'] - bb1['y1'])
    bb2_area = (bb2['x2'] - bb2['x1']) * (bb2['y2'] - bb2['y1'])

    # compute the intersection over union by taking the intersection
    # area and dividing it by the sum of prediction + ground-truth
    # areas - the interesection area
    iou = intersection_area / float(bb1_area + bb2_area - intersection_area)
    assert iou >= 0.0
    assert iou <= 1.0
    return iou

解释

图片来自this answer