简单神经网络值误差 - 形状未对齐

Question

我是这方面的菜鸟，但我开始学习神经网络。 我想用Python和Numpy制作简单的 NN。我在 Youtube 上看过一个教程，我做了同样的事情，但我得到一个错误：

output = sigmoid(np.dot(input_layer, weights))
ValueError: shapes (13,3) and (13,1) not aligned: 3 (dim 1) != 13 (dim 0)

我知道我的输出数组应该看起来像 1D 数组，但由于某种原因我无法得到它。 我做错了什么

import numpy as np
import pandas as pd
df = pd.DataFrame({'input 1':[0.5, 0.3, 0, 0.1, 0.4, -0.4, 0.4, -0.1, -0.6, 0.2, 0.6, 0, 0.2],
'input 2':[0.3, 0.5, -0.4, -0.2, 0.9, 0, 0.35, -0.4, -0.9, 0.4, 0.3, -0.1, 0.1],
'input 3':[0, 0.4, 0, -0.1, 0.4, -0.2, 0.4, -0.3, -0.1, 0.1, 0.3, 0, 0.5],
'result':[1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1]})
print(df)
def sigmoid(x):
return 1 / (1 + np.exp(-x))
features = np.array(df.iloc[:,:-1])
results =  np.array(df.iloc[:,-1:]).T
np.random.seed(10)
weights = 2 * np.random.random((13,1)) - 1
print('These are my random weights:n')
print(weights)

for iteration in range(1):
input_layer = features
output = sigmoid(np.dot(input_layer, weights))
print('nOutput result:n', output)

Answer 1

我设法找到了结果：

import numpy as np
import pandas as pd

df = pd.DataFrame({'input 1':[0.5, 0.3, 0, 0.1, 0.4, -0.4, 0.4, 0.1, -0.6, 0.2, 0.6, 0, 0.2, 0.2, -0.1, -0.1, 0, 0.4, -0.2, -0.4],
'input 2':[0.3, 0.6, -0.4, -0.2, 0.9, 0, 0.35, -0.4, -0.9, 0.4, 0.3, -0.1, 0.1, 0.3, 0.1, 0.1, 0.3, 0.1, 0.3, 0.3],
'input 3':[0, 0.4, 0, -0.1, 0.4, -0.2, 0.7, -0.3, -0.1, 0.1, 0.3, 0, 0.5, 0.4, -0.31, 0.1, 0.3, 0.1, 0.1, 0.2],
'result':[1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0]})
print(df)
def sigmoid(x):
return 1 / (1 + np.exp(-x))
def sigmoid_derivate(x):
return x * (1 - x)

features = df.iloc[:,:-1].to_numpy()
results =  df.iloc[:,-1:].to_numpy()
np.random.seed(1)
weights = 2 * np.random.random((3,1)) - 1
print('These are my random weights:n')
print(weights)
for iteration in range(100000):
input_layer = features
outputs = sigmoid(np.dot(input_layer, weights))
error = results - outputs
adjustments = error * sigmoid_derivate(outputs)
weights += np.dot(input_layer.T, adjustments)

df['output prediction'] = outputs.round(0)
print(df)