没有隐藏层和线性激活函数的神经网络应该近似线性回归？

Question

据我所知，假设不使用隐藏层和线性激活函数，神经网络将产生与线性回归相同的方程形式。i、 e.y=总和（w_i*x_i+b_i），其中i为0，表示您拥有的特征数量

我试图通过使用线性回归的权重和偏差来证明这一点，并将其输入神经网络，看看结果是否相同。事实并非如此

我想知道我的理解是错误的，还是我的代码是错误的，或者两者都是错误的

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from sklearn.linear_model import LinearRegression
import tensorflow as tf
from tensorflow import keras
import numpy as np

linearModel = LinearRegression()
linearModel.fit(np.array(normTrainFeaturesDf), np.array(trainLabelsDf))

# Gets the weights of the linear model and the intercept in a form that can be passed into the neural network
linearWeights = np.array(linearModel.coef_)
intercept = np.array([linearModel.intercept_])

trialWeights = np.reshape(linearWeights, (len(linearWeights), 1))
trialWeights = trialWeights.astype('float32')
intercept = intercept.astype('float32')
newTrialWeights = [trialWeights, intercept]

# Create a neural network and set the weights of the model to the linear model
nnModel = keras.Sequential([keras.layers.Dense(1, activation='linear', input_shape=[len(normTrainFeaturesDf.keys())]),])
nnModel.set_weights(newTrialWeights)

# Print predictions of both models (the results are vastly different)
print(linearModel.predict(np.array(normTestFeaturesDf))
print(nnModel.predict(normTestFeaturesDf).flatten())

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Answer 1

是的，一个单层无激活函数的神经网络相当于线性回归

定义一些未包含的变量：

normTrainFeaturesDf = np.random.rand(100, 10)
normTestFeaturesDf = np.random.rand(10, 10)
trainLabelsDf = np.random.rand(100)

然后输出如预期的那样：

>>> linear_model_preds = linearModel.predict(np.array(normTestFeaturesDf))
>>> nn_model_preds = nnModel.predict(normTestFeaturesDf).flatten()

>>> print(linear_model_preds)
>>> print(nn_model_preds)
[0.46030349 0.69676376 0.43064266 0.4583325  0.50750268 0.51753189
 0.47254946 0.50654825 0.52998559 0.35908762]
[0.46030346 0.69676375 0.43064266 0.45833248 0.5075026  0.5175319
 0.47254944 0.50654817 0.52998555 0.3590876 ]

除了浮动精度引起的微小变化外，这些数字是相同的

>>> np.allclose(linear_model_preds, nn_model_preds)
True