如何在张量流中得到协方差矩阵？-Java 学习之路

如何在张量流中得到协方差矩阵？就像 numpy 中的 numpy.cov() 一样 .

例如，我想获得张量 A 的协方差矩阵，现在我必须使用numpy代替

A = sess.run(model.A, feed)
    cov = np.cov(np.transpose(A))

无论如何得到张量流而不是numpy的 cov ？

它与问题how to compute covariance in tensorflow不同，他们的问题是计算两个向量的协方差，而我的是使用tensorflow API有效地计算矩阵的协方差矩阵（2D张量）

3 回答

这是几个月的晚了，但无论如何发布完整性 .

import numpy as np
import tensorflow as tf

def tf_cov(x):
    mean_x = tf.reduce_mean(x, axis=0, keep_dims=True)
    mx = tf.matmul(tf.transpose(mean_x), mean_x)
    vx = tf.matmul(tf.transpose(x), x)/tf.cast(tf.shape(x)[0], tf.float32)
    cov_xx = vx - mx
    return cov_xx

data = np.array([[1., 4, 2], [5, 6, 24], [15, 1, 5], [7,3,8], [9,4,7]])

with tf.Session() as sess:
    print(sess.run(tf_cov(tf.constant(data, dtype=tf.float32))))


## validating with numpy solution
pc = np.cov(data.T, bias=True)
print(pc)

回复于 2024-04-24T22:45:19+08:00

相当于 np.cov(data) ：

import tensorflow as tf
import numpy as np

data = np.array([[1., 4, 2], [5, 6, 24], [15, 1, 5], [7,3,8], [9,4,7]])

def tf_conv(x):
    x = x - tf.expand_dims(tf.reduce_mean(x, axis=1), 1)
    fact = tf.cast(tf.shape(x)[1] - 1, tf.float32)
    return tf.matmul(x, tf.conj(tf.transpose(x))) / fact

with tf.Session() as sess:
    print(sess.run(tf_cov(tf.constant(data, dtype=tf.float32))))

回复于 2024-04-24T22:45:19+08:00

回答自2019年.Tensorflow概率现在支持毫不费力的相关性和协方差 .

https://www.tensorflow.org/probability/api_docs/python/tfp/stats/covariance

x = tf.random_normal(shape=(100, 2, 3))
y = tf.random_normal(shape=(100, 2, 3))

# cov[i, j] is the sample covariance between x[:, i, j] and y[:, i, j].
cov = tfp.stats.covariance(x, y, sample_axis=0, event_axis=None)

# cov_matrix[i, m, n] is the sample covariance of x[:, i, m] and y[:, i, n]
cov_matrix = tfp.stats.covariance(x, y, sample_axis=0, event_axis=-1)

回复于 2024-04-24T22:45:19+08:00

如何在张量流中得到协方差矩阵？

3 回答

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