使用 itertools 在 keras 中定义损失函数

Question

我想定义一个损失函数，它代表隐藏层输出点之间的距离。首先，我在没有 keras

的情况下写了这个

import numpy as np
import itertools
pts = np.array([
    [10,10,10],
    [10,11,20],
    [20,11,30],
    [20,10,10],
    [10,10,20],
    ])
diff = list(itertools.combinations(pts, 2))

ptdiff = lambda (p1,p2): (np.sqrt(np.sum((p1 - p2) ** 2)))
diffs = map(ptdiff, diff)
np.mean(diffs)

我得到了结果。我在keras中尝试了这个损失函数，z是隐藏层的输出，它是一个矩阵

定义损失函数

def vae_loss(z):
    z_diff = list(itertools.combinations(z,2))
    ptdiff = lambda (p1,p2): (np.sqrt(np.sum((p1 - p2) ** 2)))
    z_diffs = map(ptdiff, z_diff)
    loss = K.mean(z_diffs)
    return loss

但是它显示 TypeError: 'Tensor' object is not iterable.，我只是想知道如何解决这个问题。

Answer 1

基于 very helpful question, you can make use of Keras' broadcasting properties. I'm assuming here you run Keras on the TensorFlow backend. From the TF docs广播：

A special case arises, and is also supported, where each of the input arrays has a degenerate dimension at a different index. In this case, the result is an "outer operation".

您的 numpy 代码的可重现示例如下：

import numpy as np
import itertools

# Generate 100 random points in a 5-D space
n_dim = 5
matrix = np.random.rand(1000, 5)

# List all possible combinations
combinations = list(itertools.combinations(matrix.tolist(), 2))

def mse(tup):
    """MSE between first and second element of a tuple of lists"""
    return np.mean((np.array(tup[0]) - np.array(tup[1]))**2)

avg_mse = np.mean([mse(c) for c in combinations])
print('Average mse: {:.3f}'.format(avg_mse))

这个returns，就我而言，Average mse: 0.162

根据上面提到的问题，你可以构造你的损失函数如下：

import keras.backend as K

# Wrap our random matrix into a tensor
tensor = K.constant(value=matrix)

def loss_function(x):
    x_ = K.expand_dims(tensor, axis=0)
    x__ = K.expand_dims(tensor, axis=1)

    # Compute mse for all combinations, making use of broadcasting
    z = K.mean(K.square(x_ - x__), axis=-1)

    # Return average mse
    return(K.mean(z))

with K.get_session() as sess:
    print('Average mse: {:.3f}'.format(loss_function(tensor).eval()))

哪个returns对我来说Average mse: 0.162。

请注意，此实现并未完全复制您的 numpy 示例中的行为。不同之处在于，还考虑了行与自身的所有组合（itertools.combinations 不是这种情况）并且组合被考虑两次：mse((row1, row2)) 和 mse((row2, row1)) 都将被计算，这又不是您的 itertools 代码的情况。对于具有大量行的矩阵，这应该不会造成太大差异，如我的示例所示。

使用 itertools 在 keras 中定义损失函数

Define loss function in keras using itertools

python

tensorflow

keras

itertools

loss-function

定义损失函数