numpy.dot-形状错误 - 神经网络

发布于 2025-02-03 00:20:58 字数 451 浏览 1 评论 0 原文

我正在尝试将这些 a1 和 w2 矩阵（ z2 = w2.dot（a1））倍增：

A1 : [[0.42940542]
 [0.55013895]]
W2 : [[-0.4734037  -0.39642393 -0.05440914 -0.24011293 -0.03670913 -0.37523234]
 [-0.45501004  0.23881832  0.21831658  0.32237388  0.25674681  0.27956714]]

但是我得到了此错误形状（2,6）和（2,1）未对齐：6（DIM 1）！= 2（DIM 0），为什么？用（2,6）矩阵将（2,1）乘不正常？

因为我有一个带有2个节点的隐藏图层，并带有 6节点的输出图层

原文

I am trying to multiply these A1 and W2 matrices (Z2 = W2.dot(A1)):

A1 : [[0.42940542]
 [0.55013895]]
W2 : [[-0.4734037  -0.39642393 -0.05440914 -0.24011293 -0.03670913 -0.37523234]
 [-0.45501004  0.23881832  0.21831658  0.32237388  0.25674681  0.27956714]]

But I am getting this error shapes (2,6) and (2,1) not aligned: 6 (dim 1) != 2 (dim 0), why? Isn't it normal to multiply a (2,1) with a (2,6) matrix?

Because I have a hidden layer with 2 nodes and output layer with 6 nodes

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

嘴硬脾气大 2025-02-10 00:20:58

从数学上讲，这是不可能的，因为您将A（2，6）矩阵乘以（2，1）。您需要做的就是转置W2。

ps：请注意，在线性代数np.dot（W2.T，a1）中与np.dot（a1.t，w2）不同，

import numpy as np

A1 = np.asarray([[0.42940542], [0.55013895]])
W2 = np.asarray([[
    -0.4734037, -0.39642393, -0.05440914, -0.24011293, -0.03670913, -0.37523234
], [-0.45501004, 0.23881832, 0.21831658, 0.32237388, 0.25674681, 0.27956714]])
print(W2.shape, A1.shape)  # (2, 6), (2, 1)
Z2 = W2.T @ A1
print(Z2)

结果是：[[-0.45360086] [[-0.45360086] [[ -0.03884332] [0.09674087] [0.07424463] [0.12548332] [-0.00732603]]

Mathematically this is impossible because your multiplying a (2, 6) matrix by (2, 1). All you need to do is to transpose W2.

P.S: Note that in linear algebra np.dot(W2.T, A1) is not the same as np.dot(A1.T, W2)

import numpy as np

A1 = np.asarray([[0.42940542], [0.55013895]])
W2 = np.asarray([[
    -0.4734037, -0.39642393, -0.05440914, -0.24011293, -0.03670913, -0.37523234
], [-0.45501004, 0.23881832, 0.21831658, 0.32237388, 0.25674681, 0.27956714]])
print(W2.shape, A1.shape)  # (2, 6), (2, 1)
Z2 = W2.T @ A1
print(Z2)

The result would be: [[-0.45360086] [-0.03884332] [ 0.09674087] [ 0.07424463] [ 0.12548332] [-0.00732603]]

回复收藏 0 原文

~没有更多了~