Question

我试图在Keras中构造一个损失函数，其中我要惩罚预测与一组给定值之间的最小距离。问题是我需要计算预测值和给定值之间的距离。

示例代码

def custom_loss(y_pred,y_test):


    #Given values
    centers=K.constant([[-2.5,-1],[-1.25,-2],[.5,-1],[1.5,.25]])
    num_centers=K.int_shape(centers)[0]


    #Begin constructing distance matrix
    height=K.int_shape(y_pred)[0]
    i=0
    current_center=K.reshape(K.repeat(K.reshape(centers[i,:],[1,-1]),height),[height,2])
    current_dist=K.sqrt(K.sum(K.square(y_pred-current_center),axis=1))


    #Values of distance matrix for first center
    Distance=K.reshape(current_dist,[height,1])


    for i in range(1,num_centers):
        current_center=K.reshape(K.repeat(K.reshape(centers[i,:],[1,-1]),height),[height,2])
        current_dist=K.sqrt(K.sum(K.square(y_pred-current_center),axis=-1))
        current_dist=K.reshape(current_dist,[height,1])


        #Iteratively concatenate distances of y_pred from remaining centers
        Distance=K.concatenate([Distance,current_dist],axis=-1)

    #Determine minimum distance from each predicted value to nearest center
    A=K.min(A,axis=1)


    #Return average minimum distance as loss
    return K.sum(A)/float(height)

但是，我无法消除函数对y_pred第一维的依赖，后者是可变的。我正在使用数组广播来计算y_pred与每个给定值之间的差，但是我明确使用批处理大小进行广播，因为我不知道如何在Keras中不使用批处理大小来进行此操作。但是，这会产生错误，因为在构造计算图时，批处理大小没有明确知道。

如何避免明确广播？由于当前方法非常笨拙，因此计算距离矩阵是否更有效？

Answer 1

可以使用隐式广播来实现您的损失功能，如下所示：

import keras.backend as K


def custom_loss(y_true, y_pred):
    centers = K.constant([[-2.5, -1], [-1.25, -2], [.5, -1], [1.5, .25]])

    # Expand dimensions to enable implicit broadcasting
    y_pred_r = y_pred[:, None, :]  # Shape: (batch_size, 1, 2)
    centers_r = centers[None, :, :]  # Shape: (1, nb_centers, 2)

    # Compute minimum distance to centers for each element
    distances = K.sqrt(K.sum(K.square(y_pred_r - centers_r), axis=-1))  # Shape=(batch_size, nb_centers)
    min_distances = K.min(distances, axis=-1)  # Shape=(batch_size,)

    # Output average of minimum distances
    return K.mean(min_distances)

注意：未测试。

Keras损失函数取决于批次大小

1 个答案: