正如@Lescurel引用的帖子提到的，这个问题可以通过使用tf.einsum来解决：

volatility_matrix = tf.einsum("ij,ik->ijk", volas, volas)

解决了获得波动率矩阵的问题。从那里，它只是沿着第二个轴的标量乘法。具体来说，然后获得协方差矩阵，我做了

covariance_matrix = tf.linalg.matmul(volatility_matrix, correlation_matrix)

这些是当某些信息未被表示时，共价矩阵和波动性的应用矩阵乘积，有助于在从成批信息中阅读时确定目标关系。

[样品]：

import tensorflow as tf

Cell_A = 0.20
Cell_B = 0.20
Cell_C = 0.20

volatilities = tf.constant([[ Cell_A, Cell_B, Cell_C ]], dtype=tf.float32)
correlations = tf.constant([[ 33405, 33405, 33405 ], [ 33405, 40050, 33405 ], [ 33405, 33405, 37025 ]], dtype=tf.float32)

volatility_matrix = tf.einsum('ij,jk->ik', volatilities, correlations)
correlation_matrix = correlations

print( ' volatilities : ' )
print( volatilities )
print( ' correlations : ' )
print( correlations )
print( ' volatility_matrix : ' )
print( volatility_matrix )

print( ' Shortest: ' + str( tf.math.argmin( volatility_matrix[0] ).numpy() ) )
print( ' Hghest: ' + str( tf.math.argmax( volatility_matrix[0] ).numpy() ) )

[输出]：

volatilities :
tf.Tensor([[0.2 0.2 0.2]], shape=(1, 3), dtype=float32)
 correlations :
tf.Tensor(
[[33405. 33405. 33405.]
 [33405. 40050. 33405.]
 [33405. 33405. 37025.]], shape=(3, 3), dtype=float32)
 volatility_matrix :
tf.Tensor([[20043. 21372. 20767.]], shape=(1, 3), dtype=float32)
 Shortest: 0
 Hghest: 1