深度学习(七十二)ssd物体检测

深藏阁楼爱情的钟 2022-05-30 05:18 346阅读 0赞

def ssd_anchor_one_layer(img_shape,
                             feat_shape,
                             sizes,
                             ratios,
                             step,
                             offset=0.5,
                             dtype=np.float32):
        # 计算每个default center归一化后的坐标，因为是center，所以要加offset
        y, x = np.mgrid[0:feat_shape[0], 0:feat_shape[1]]
        y = (y.astype(dtype) + offset) * step / img_shape[0]
        x = (x.astype(dtype) + offset) * step / img_shape[1]
    
    
        y = np.expand_dims(y, axis=-1)
        x = np.expand_dims(x, axis=-1)
    
        num_anchors = len(sizes) + len(ratios)
        h = np.zeros((num_anchors, ), dtype=dtype)
        w = np.zeros((num_anchors, ), dtype=dtype)
    
    
        #第一种默认的default box的情况，也就是正方形，正方形的边长为sk
        h[0] = sizes[0] / img_shape[0]
        w[0] = sizes[0] / img_shape[1]
        di = 1
    
        #第二种默认的default box情况，也是正方形，只不过正方形的边长为sqrt(sk * sk+1)
        if len(sizes) > 1:
            h[1] = math.sqrt(sizes[0] * sizes[1]) / img_shape[0]
            w[1] = math.sqrt(sizes[0] * sizes[1]) / img_shape[1]
            di += 1
    
        #第三种~第六种default box，矩形边长比例为2、0.5,3,1/3的情况，对于conv4、和最后两层没有3、1/3两种default box
        for i, r in enumerate(ratios):
            h[i+di] = sizes[0] / img_shape[0] / math.sqrt(r)
            w[i+di] = sizes[0] / img_shape[1] * math.sqrt(r)
        return y, x, h, w