【问题标题】:TensorFlow: Implementing Spearman Distance as the Objective FunctionTensorFlow:将 Spearman 距离实现为目标函数
【发布时间】:2016-07-28 02:56:13
【问题描述】:

为了使我的问题可重现,我使用鸢尾花数据集(10 行任意行,所有列标准归一化)和最小神经网络模型(使用萼片长度预测花瓣宽度)生成了以下 .csv 文件,萼片宽度和花瓣长度)通过修改我在互联网上找到的 MNIST 示例。向下滚动查看我的问题!

iris.csv

"Sepal.Length","Sepal.Width","Petal.Length","Petal.Width","Species"
0.0551224773430978,-0.380319414627833,-0.335895230408602,-0.548226210538025,"versicolor"
1.48830688826362,-1.01418510567422,1.37931445678426,0.614677872421422,"virginica"
0.606347250774068,0.887411967464943,0.450242542888127,0.780807027129915,"virginica"
-0.606347250774067,-1.64805079672061,0.235841331989019,0.44854871771293,"virginica"
1.15757202420504,-1.01418510567422,0.950512034986045,0.44854871771293,"virginica"
-1.92928670700839,0.887411967464943,-2.33697319880027,-2.37564691233144,"setosa"
0.38585734140168,0.253546276418555,0.307308402288722,1.1130653365469,"virginica"
-0.826837160146455,0.253546276418555,-0.478829371008007,-0.548226210538025,"versicolor"
0.0551224773430978,1.52127765851133,-0.192961089809197,-0.21596790112104,"versicolor"
-0.385857341401679,0.253546276418555,0.021440121089911,0.282419563004437,"virginica"

nn.py

import pandas as pd
import numpy as np
import tensorflow as tf
import scipy.stats

# Import iris data
data = pd.read_csv("iris.csv")
input = data[["Sepal.Length", "Sepal.Width", "Petal.Length"]]
target = data[["Petal.Width"]]

# Parameters
learning_rate = 0.001
training_epochs = 6000

# Network Parameters
n_hidden_1 = 5 # 1st layer number of features
n_hidden_2 = 5 # 2nd layer number of features
n_input = 3 # data input
n_output = 1 # data output

# tf Graph input
x = tf.placeholder("float", [None, n_input])
y = tf.placeholder("float", [None, n_output])

# Create model
def multilayer_network(x, weights, biases):
  # Hidden layer with TanH activation
  layer_1 = tf.add(tf.matmul(x, weights['h1']), biases['b1'])
  layer_1 = tf.tanh(layer_1)
  # Hidden layer with TanH activation
  layer_2 = tf.add(tf.matmul(layer_1, weights['h2']), biases['b2'])
  layer_2 = tf.tanh(layer_2)
  # Output layer with linear activation
  out_layer = tf.matmul(layer_2, weights['out']) + biases['out']
  return out_layer

# Store layers weight & bias
weights = {
  'h1': tf.Variable(tf.random_normal([n_input, n_hidden_1])),
  'h2': tf.Variable(tf.random_normal([n_hidden_1, n_hidden_2])),
  'out': tf.Variable(tf.random_normal([n_hidden_2, n_output]))
}
biases = {
  'b1': tf.Variable(tf.random_normal([n_hidden_1])),
  'b2': tf.Variable(tf.random_normal([n_hidden_2])),
  'out': tf.Variable(tf.random_normal([n_output]))
}

# Construct model
pred = multilayer_network(x, weights, biases)

# Define loss and optimizer
cost = tf.reduce_mean(tf.square(pred-y))
optimizer = tf.train.AdamOptimizer(learning_rate=learning_rate).minimize(cost)

# Initializing the variables
init = tf.initialize_all_variables()

# Launch the graph
with tf.Session() as sess:
  sess.run(init)

  # Training cycle
  for epoch in range(training_epochs):
    # Run optimization op (backprop) and cost op (to get loss value)
    _, c = sess.run([optimizer, cost], feed_dict={x: input, y: target})

    # Display logs per epoch step
    if epoch % 1000 == 0:
      print "Epoch:", '%04d' % (epoch+1), "cost=", "{:.9f}".format(c)

print "Optimization Finished!"

这是一个示例培训课程结果:

$ python nn.py
Epoch: 0001 cost= 3.000185966
Epoch: 1001 cost= 0.031734336
Epoch: 2001 cost= 0.000614795
Epoch: 3001 cost= 0.000008422
Epoch: 4001 cost= 0.000000057
Epoch: 5001 cost= 0.000000000
Optimization Finished!

我的想法是用我最近了解到的 Spearman 距离代替均方误差作为我的目标函数。遵循定义:

我写了一个返回向量排名的函数:

import scipy.stats

def rank(vector):
  return scipy.stats.rankdata(vector, method="min")

使用TensorFlow的方法py_func,我定义了我的成本张量如下。

pred = tf.to_float(tf.py_func(rank, [pred], [tf.int64])[0])
y = tf.to_float(tf.py_func(rank, [y], [tf.int64])[0])

cost = tf.reduce_mean(tf.square(y-pred))

但是,这给了我错误

ValueError: No gradients provided for any variable: ((None, <tensorflow.python.ops.variables.Variable object at 0x7f67ffe4ee90>), (None, <tensorflow.python.ops.variables.Variable object at 0x7f66ed3c4990>), (None, <tensorflow.python.ops.variables.Variable object at 0x7f66ed357310>), (None, <tensorflow.python.ops.variables.Variable object at 0x7f66ed357190>), (None, <tensorflow.python.ops.variables.Variable object at 0x7f66ed380350>), (None, <tensorflow.python.ops.variables.Variable object at 0x7f66ed3801d0>))

我不明白根本问题是什么。您能提供给我的任何方向将不胜感激!

【问题讨论】:

  • 由于这看起来像是一个实现问题,我认为您在 StackOverflow 上会遇到更好的运气。我认为它很难找到一条贯穿你的损失函数的“路径”。
  • 看起来您正在使用 tf.to_floatypred 从 tensorflow 变量转换为浮点数,此时它们不再是 tensorflow 对象。成本也只是一个数值。但是,即使您解决了该问题,秩变换也无法微分,因此您在尝试使用任何基于梯度的方法进行训练时仍然会遇到问题。

标签: python ranking tensorflow


【解决方案1】:

您的错误来自tf.py_func 没有定义渐变。

无论如何,正如@user20160 在 cmets 中所说,rank 操作甚至不存在梯度,所以这不是您可以直接训练算法的损失。

【讨论】:

  • 我明白了。但是,我读过“损失函数的常见选择是 [...],以及平方秩距离之和。”在一本关于机器学习的书中。那怎么可能呢?
  • 你可以用它作为损失来看看你的模型有多坏/好,但你不能在它上面训练(因为没有梯度)。有一些方法可以通过强化学习进行不可微分损失的训练,但这非常复杂
  • 有道理!如果你知道的话,你能提供一些关于强化学习的自学资源吗?
  • 您可以阅读 RL 上 Karpathy 的blog post 中的“神经网络中的不可微分计算”部分。但是,我敢肯定有可区分的损失可以很好地解决您的问题
  • 谢谢,我去看看!
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