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python - 当输出的矩阵维度的大小与标签的大小不匹配时，如何计算神经网络中的误差？

问题描述

我尝试在不使用 KERAS 等库的情况下使用神经网络编写图像分类代码。

但是当我想计算误差时我有困难，因为输出前馈的矩阵大小与矩阵标签的大小不匹配热多类

以下是一个热门多类标签的代码，这里我使用 20 个类，每个类有 15 个变体（300 个输入，大小 = 64 x 64 像素）：

labels_train = np.array([0] * 15 + [1] * 15 + [2] * 15 + [3] * 15 + [4] * 15 + [5] * 15 + [6] * 15 + [7] 
                        * 15 + [8] * 15 + [9] * 15 + [10] * 15 + [11] * 15 + [12] * 15 + [13] * 15 + [14] 
                        * 15 + [15] * 15 + [16] * 15 + [17] * 15 + [18] * 15 + [19] * 15)

y_train = np.zeros((300,20))

for i in range(300):
    y_train[i, labels_train[i]] = 1

以下是前馈和计算误差值的代码：

for epoch in range(1):

    hasil_ih = np.dot(W_ih, x_train) + B_ih
    output_ih = sigmoid(hasil_ih)

    hasil_hh = np.dot(W_hh, output_ih) + B_hh
    output_hh = sigmoid(hasil_hh)

    hasil_ho = np.dot(W_ho, output_hh) + B_ho
    output_ho = sigmoid(hasil_ho)

    print(output_ho.shape)
    print(y_train.shape)
    error_o = labels_train - output_ho

以下是程序运行时的结果：

Traceback (most recent call last):
(300, 4096)
  File "D:/Pengenalan_Induk_Aksara/NN2.py", line 71, in <module>
(300, 20)
    error_o = labels_train - output_ho
ValueError: operands could not be broadcast together with shapes (300,) (300,4096) 

Process finished with exit code 1

请帮助我，我仍然是初学者，我还没有找到解决方案。以下是完整代码：

import numpy as np
from PIL import Image
import matplotlib.pyplot as plt
import os


dir = "data_train/LBP_R1/"
kategori = ["ka","ga","nga","pa","ba","ma","ta","da","na","ca","ja","nya","ya","a","la","ra","sa","wa","ha","gha"]
features = []; label = []

for y in kategori:
    path = os.path.join(dir, y)
    class_num = kategori.index(y)
    label.append(class_num)

    for img in os.listdir(path):
        im = Image.open(os.path.join(path,img))
        imgs = list(im.getdata())
        features.append(imgs)


x_train = np.vstack([features])
x_train = (x_train/255)


labels_train = np.array([0] * 15 + [1] * 15 + [2] * 15 + [3] * 15 + [4] * 15 + 
                        [5] * 15 + [6] * 15 + [7] * 15 + [8] * 15 + [9] * 15 + 
                        [10] * 15 + [11] * 15 + [12] * 15 + [13] * 15 + [14] * 
                        15 + [15] * 15 + [16] * 15 + [17] * 15 + [18] * 15 + [19] 
                        * 15)

y_train = np.zeros((300,20))

for i in range(300):
    y_train[i, labels_train[i]] = 1

#Training Phase
def sigmoid(x):
    return 1/(1+np.exp(-x)) 

def dsigmoid(y):
    return y*(1-y) 

attributes = x_train.shape[0] 
hidden_nodes = 4 
output_nodes = 300 

np.random.seed(42)
W_ih = np.random.rand(hidden_nodes,attributes) #bobot input layer 1
B_ih = np.random.rand(hidden_nodes,1) #Bias input layer 1

W_hh = np.random.rand(hidden_nodes, hidden_nodes) #weight hidden layer 2
B_hh = np.random.rand(hidden_nodes, 1) #bias hidden layer 2

W_ho = np.random.rand(output_nodes, hidden_nodes) #weight hidden output
B_ho = np.random.rand(output_nodes, 1) #bias hidden output

lr = 0.25; #Learning rate

for epoch in range(1):

    hasil_ih = np.dot(W_ih, x_train) + B_ih
    output_ih = sigmoid(hasil_ih)

    hasil_hh = np.dot(W_hh, output_ih) + B_hh
    output_hh = sigmoid(hasil_hh)

    hasil_ho = np.dot(W_ho, output_hh) + B_ho
    output_ho = sigmoid(hasil_ho)

    print(output_ho.shape)
    print(y_train.shape)

    error_o = labels_train - output_ho
    print(output_ho)
    print(y_train)
    print(error_o)

标签： pythonnumpyneural-networkoutputone-hot-encoding

解决方案

只需获取具有最大概率的索引即可获得该类。但这会给你一些失败的案例。

output_ho_idx = np.argmax(output_ho)
error_o = labels_train - output_ho_idx

您还可以将标签转换为 one-hot-encoding 并使用 sigmoid 输出估计错误。

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