Tensorflow 2.X Debug中的Tensor.numpy问题 @tf.function

news2025/5/17 23:48:59

我在调试YOLOv3模型过程中想查看get_pred函数下面的get_anchors_and_decode函数里grid_shape的数值

#---------------------------------------------------#
#   将预测值的每个特征层调成真实值
#---------------------------------------------------#
def get_anchors_and_decode(feats, anchors, num_classes, input_shape, calc_loss=False):
    num_anchors = len(anchors)
    #------------------------------------------#
    #   grid_shape指的是特征层的高和宽
    #------------------------------------------#
    grid_shape = K.shape(feats)[1:3]
    # print(grid_shape)
    #--------------------------------------------------------------------#
    #   获得各个特征点的坐标信息。生成的shape为(13, 13, num_anchors, 2)
    #--------------------------------------------------------------------#

    # with tf.compat.v1.Session() as sess:
    #     sess.run(tf.compat.v1.global_variables_initializer())
    #     grid_shape_value = sess.run(grid_shape)
    print("Grid Shape: ", grid_shape)


    grid_x  = K.tile(K.reshape(K.arange(0, stop=grid_shape[1]), [1, -1, 1, 1]), [grid_shape[0], 1, num_anchors, 1])
    grid_y  = K.tile(K.reshape(K.arange(0, stop=grid_shape[0]), [-1, 1, 1, 1]), [1, grid_shape[1], num_anchors, 1])
    grid    = K.cast(K.concatenate([grid_x, grid_y]), K.dtype(feats))
    #---------------------------------------------------------------#
    #   将先验框进行拓展，生成的shape为(13, 13, num_anchors, 2)
    #---------------------------------------------------------------#
    anchors_tensor = K.reshape(K.constant(anchors), [1, 1, num_anchors, 2])
    anchors_tensor = K.tile(anchors_tensor, [grid_shape[0], grid_shape[1], 1, 1])

    #---------------------------------------------------#
    #   将预测结果调整成(batch_size,13,13,3,85)
    #   85可拆分成4 + 1 + 80
    #   4代表的是中心宽高的调整参数
    #   1代表的是框的置信度
    #   80代表的是种类的置信度
    #---------------------------------------------------#
    feats           = K.reshape(feats, [-1, grid_shape[0], grid_shape[1], num_anchors, num_classes + 5])
    #------------------------------------------#
    #   对先验框进行解码，并进行归一化
    #------------------------------------------#
    box_xy          = (K.sigmoid(feats[..., :2]) + grid) / K.cast(grid_shape[::-1], K.dtype(feats))
    box_wh          = K.exp(feats[..., 2:4]) * anchors_tensor / K.cast(input_shape[::-1], K.dtype(feats))
    #------------------------------------------#
    #   获得预测框的置信度
    #------------------------------------------#
    box_confidence  = K.sigmoid(feats[..., 4:5])
    box_class_probs = K.sigmoid(feats[..., 5:])
    
    #---------------------------------------------------------------------#
    #   在计算loss的时候返回grid, feats, box_xy, box_wh
    #   在预测的时候返回box_xy, box_wh, box_confidence, box_class_probs
    #---------------------------------------------------------------------#
    if calc_loss == True:
        return grid, feats, box_xy, box_wh
    return box_xy, box_wh, box_confidence, box_class_probs

@tf.function    
def get_pred(self, image_data, input_image_shape):
        out_boxes, out_scores, out_classes = self.yolo_model([image_data, input_image_shape], training=False)
        return out_boxes, out_scores, out_classes

直接打印的话只有这个变量的属性，看不到具体数值：

Tensor("yolo_eval/strided_slice:0", shape=(2,), dtype=int32)

如果试图用grid_shape.numpy()查看数值呢，又会报错：

AttributeError: 'Tensor' object has no attribute 'numpy'

根本原因在于get_pred函数调用不是直接调用的，而是通过@tf.function这个特殊的tf装饰器调用的

@tf.function是一个加速程序运行，提升效率的装饰器

把@tf.function去掉不影响程序大致运行逻辑，会取消一些并行加速计算，但是这么做对调试就方便很多

去掉@tf.function之后的get_pred函数：

def get_pred(self, image_data, input_image_shape):
        out_boxes, out_scores, out_classes = self.yolo_model([image_data, input_image_shape], training=False)
        return out_boxes, out_scores, out_classes

此时，调试运行至get_pred下面调用的get_anchors_and_decode函数，在python控制台输入grid_shape.numpy()，回车：