用于鸡尾酒配方生成的条件变分自动编码器答案

【问题标题】：Conditional Variational Autoencoder for cocktail recipe generation用于鸡尾酒配方生成的条件变分自动编码器
【发布时间】：2018-10-18 12:13:05
【问题描述】：

我想使用条件变分自动编码器来生成鸡尾酒配方。我从this repo 修改了代码，所以它可以读取我自己的数据。输入是一个包含所有可能成分的数组，因此大多数条目的值为 0。如果存在一种成分，它会得到一个值，该值是按 250 毫升标准化的量。最后一个索引是“剩余的”，以确保鸡尾酒总是将 op 添加到 1。

例子：

0,0.0,0.0,0.24,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.6,0.0,0.0,0.0,0.0,0.0,0.06,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.088,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0120000000000000

带有 softmax 激活函数的输出看起来有点像这样：

[5.8228267e-10 6.7397465e-10 1.9761790e-08 2.3713847e-01 3.1315527e-11
 4.9592632e-11 4.2637563e-05 7.6098106e-10 2.9357905e-05 1.3291576e-08
 2.6885323e-09 4.2986945e-10 3.0274603e-09 8.6994453e-11 3.2391853e-10
 3.3694150e-10 4.9642315e-11 2.2861177e-10 2.5966980e-11 3.3872125e-10
 4.8175470e-12 1.1207919e-09 7.8108942e-10 1.0438563e-09 4.7190268e-12
 2.2692757e-09 3.3177341e-10 4.7493649e-09 1.6603904e-08 2.7854623e-11
 1.1586791e-07 2.3917833e-08 1.0172608e-09 2.2049740e-06 4.0200213e-10
 4.8334226e-05 1.9393491e-09 4.0731374e-10 4.5671125e-10 8.5878060e-10
 1.3625046e-10 1.7755342e-09 2.4927729e-09 3.8919952e-09 1.6791472e-10
 1.5160178e-09 9.0631114e-10 1.2043951e-08 2.1420650e-01 1.4531254e-10
 3.9913628e-10 4.6368896e-06 6.8399265e-11 2.4654754e-09 6.5392605e-12
 5.8443012e-10 2.7861690e-11 4.7215394e-08 5.1503157e-09 5.4484850e-10
 1.9266211e-10 7.2835156e-09 6.4243433e-10 4.2432866e-09 4.2630177e-08
 1.1281617e-12 1.8015703e-08 3.5657147e-10 3.4241193e-11 4.8394988e-10
 9.6064046e-11 2.9857121e-02 3.8048144e-11 1.1893182e-10 5.1867032e-01]

如何确保值仅分布在几种成分中，而其余成分为 0，类似于输入？这是改变激活函数的问题吗？

谢谢:)

【问题讨论】：

更具体地说，我只将代码中的第 84 行更改为 prob = tf.nn.softmax(logits) 并将第 98 行更改为 recon_loss = tf.reduce_sum(tf.nn.softmax_cross_entropy_with_logits_v2(logits=logits, labels=X)) 。其他的激活函数和损失计算还是一样的。

标签： python function autoencoder activation softmax

【解决方案1】：

我不确定你想在这里使用概率。您似乎正在对某些特定值进行回归。因此，不使用 softmax 并使用简单的均方误差损失是有意义的。

请注意，如果您的损失中某些值总是有偏差，您可以对损失使用额外的权重，或使用一些抽象（例如 Keras 的 class_weight）。

对于这个任务，您可以考虑使用 Keras，尤其是对于这个任务，它会很有意义。有一个例子签入master：https://github.com/keras-team/keras/blob/master/examples/variational_autoencoder.py

对于这项任务，使用 GAN 可能实际上是有意义的：https://github.com/keras-team/keras/blob/master/examples/mnist_acgan.py。您将让它区分随机鸡尾酒和“真正的”鸡尾酒。它将学习区分两者，并在此过程中训练权重，以便能够创建一个生成器，为您生成鸡尾酒！

【讨论】：