如何计算张量流的协方差？

本文介绍了如何计算张量流的协方差？的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我有一个问题，我不知道如何计算两个张量的协方差。我已经尝试过 contrib.metrics.streaming_covariance 。但是总是返回 0 。必须存在一些错误。

I have a problem that I don't know how to compute the covariance of two tensor. I have tried the contrib.metrics.streaming_covariance. But is always returns 0. There must be some errors.

推荐答案

您可以使用两个随机变量 X的协方差的定义。 和 Y 的期望值为 x0 和 y0 ：

You could use the definition of the covariance of two random variables X and Y with the expected values x0 and y0:

cov_xx = 1 /（N-1）* Sum_i（（x_i-x0）^ 2）

cov_yy = 1 /（N-1）* Sum_i（（y_i-y0）^ 2）

cov_xy = 1 /（N-1）* Sum_i（（x_i-x0）*（y_i-y0））

关键点是估计 x0 和 y0

The crucial point is to estimate x0 and y0 here, since you normally do not know the probability distribution. In many cases, the mean of the x_i or y_i is estimated to be x_0 or y_0, respectively, i.e. the distribution is estimated to be uniform.

然后您可以如下计算协方差矩阵的元素：

Then you can compute the elements of the covariance matrix as follows:

import tensorflow as tf x = tf.constant([1, 4, 2, 5, 6, 24, 15], dtype=tf.float64) y = tf.constant([8, 5, 4, 6, 2, 1, 1], dtype=tf.float64) cov_xx = 1 / (tf.shape(x)[0] - 1) * tf.reduce_sum((x - tf.reduce_mean(x))**2) cov_yy = 1 / (tf.shape(x)[0] - 1) * tf.reduce_sum((y - tf.reduce_mean(y))**2) cov_xy = 1 / (tf.shape(x)[0] - 1) * tf.reduce_sum((x - tf.reduce_mean(x)) * (y - tf.reduce_mean(y))) with tf.Session() as sess: sess.run([cov_xx, cov_yy, cov_xy]) print(cov_xx.eval(), cov_yy.eval(), cov_xy.eval())

当然，如果您需要矩阵形式，您可以修改最后一部分，如下所示：

Of course, if you need the covariance in a matrix form, you can modify the last part as follows:

with tf.Session() as sess: sess.run([cov_xx, cov_yy, cov_xy]) print(cov_xx.eval(), cov_yy.eval(), cov_xy.eval()) cov = tf.constant([[cov_xx.eval(), cov_xy.eval()], [cov_xy.eval(), cov_yy.eval()]]) print(cov.eval())

要验证TensorFlow方式的元素，可以使用numpy进行检查：

To verify the elements of the TensorFlow way, you can check with numpy:

import numpy as np x = np.array([1,4,2,5,6, 24, 15], dtype=float) y = np.array([8,5,4,6,2,1,1], dtype=float) pc = np.cov(x,y) print(pc)