Input: f, g, A, X, n epochs, n epoch f t Compute T as presented in Section 3.1. Compute TCF , ACF by Eqs. (3) and (4). // model training for epoch in range(n epochs) do Z = f(A, X). Get bA and bACF via g with Eqs. (6) and (7). Update Θf and Θg with L. (Eq. (11)) end for // decoder fine-tuning Freeze Θf and re-initialize Θg. Z = f(A, X). for epoch in range(n epochs f t) do Get bA via g with Eq. (6). Update Θg with LF . (Eq. (8)) end for // inference Z = f(A, X). Get bA and bACF via g with Eqs. (6) and (7). Output: bA for link prediction, bACF .
时间: 2023-06-18 13:06:37 浏览: 137
This appears to be a code snippet for a machine learning algorithm, possibly related to link prediction. The algorithm involves training a model (f) and a generator (g) using a dataset (A, X) for a specified number of epochs. The code then performs decoder fine-tuning and inference to output bA and bACF for link prediction.
相关问题
import numpy as npclass Perceptron: def __init__(self, input_size, lr=1, epochs=100): self.W = np.zeros(input_size+1) self.epochs = epochs self.lr = lr def activation_fn(self, x): return 1 if x >= 0 else 0 def predict(self, x): z = self.W.T.dot(x) a = self.activation_fn(z) return a def fit(self, X, d): for epoch in range(self.epochs): for i in range(d.shape[0]): x = np.insert(X[i], 0, 1) y = self.predict(x) e = d[i] - y self.W = self.W + self.lr * e * x X = np.array([[0,0], [0,1], [1,0], [1,1]])d = np.array([0, 0, 0, 1])perceptron = Perceptron(input_size=2)perceptron.fit(X, d)test_input = np.array([0, 1])print(perceptron.predict(np.insert(test_input, 0, 1)))帮我逐行解释这段代码
这段代码是实现一个感知机(Perceptron)模型的类,可以用于二分类问题。
首先,导入了numpy库并重命名为np。
接下来定义Perceptron类,__init__方法初始化模型参数,包括输入大小、学习率和迭代次数。其中self.W用于存储模型的权重参数,初始值为0。
activation_fn方法是定义激活函数,当输入的值大于等于0时返回1,否则返回0。
predict方法用于进行预测,输入一个样本向量x后,先将其与权重向量W进行内积得到z,然后通过激活函数得到预测值a,最后返回该预测值。
fit方法实现模型的训练,输入参数为训练样本矩阵X和标签向量d。在每个epoch(迭代次数)中,对于每个样本,预测其标签值y,并计算其与真实标签值d之间的误差e。然后根据感知机算法更新权重W,其中self.lr表示学习率。最终将训练得到的权重向量存储在self.W中。
接下来定义了一个X矩阵和对应的标签d用于训练。
声明了一个Perceptron对象perceptron,并用fit方法对其进行训练。
最后定义了一个test_input样本进行预测,并输出对应的预测值。
import numpy as np from sklearn import datasets from sklearn.linear_model import LinearRegression np.random.seed(10) class Newton(object): def init(self,epochs=50): self.W = None self.epochs = epochs def get_loss(self, X, y, W,b): """ 计算损失 0.5sum(y_pred-y)^2 input: X(2 dim np.array):特征 y(1 dim np.array):标签 W(2 dim np.array):线性回归模型权重矩阵 output:损失函数值 """ #print(np.dot(X,W)) loss = 0.5np.sum((y - np.dot(X,W)-b)2) return loss def first_derivative(self,X,y): """ 计算一阶导数g = (y_pred - y)*x input: X(2 dim np.array):特征 y(1 dim np.array):标签 W(2 dim np.array):线性回归模型权重矩阵 output:损失函数值 """ y_pred = np.dot(X,self.W) + self.b g = np.dot(X.T, np.array(y_pred - y)) g_b = np.mean(y_pred-y) return g,g_b def second_derivative(self,X,y): """ 计算二阶导数 Hij = sum(X.T[i]X.T[j]) input: X(2 dim np.array):特征 y(1 dim np.array):标签 output:损失函数值 """ H = np.zeros(shape=(X.shape[1],X.shape[1])) H = np.dot(X.T, X) H_b = 1 return H, H_b def fit(self, X, y): """ 线性回归 y = WX + b拟合,牛顿法求解 input: X(2 dim np.array):特征 y(1 dim np.array):标签 output:拟合的线性回归 """ self.W = np.random.normal(size=(X.shape[1])) self.b = 0 for epoch in range(self.epochs): g,g_b = self.first_derivative(X,y) # 一阶导数 H,H_b = self.second_derivative(X,y) # 二阶导数 self.W = self.W - np.dot(np.linalg.pinv(H),g) self.b = self.b - 1/H_bg_b print("itration:{} ".format(epoch), "loss:{:.4f}".format( self.get_loss(X, y , self.W,self.b))) def predict(): """ 需要自己实现的代码 """ pass def normalize(x): return (x - np.min(x))/(np.max(x) - np.min(x)) if name == "main": np.random.seed(2) X = np.random.rand(100,5) y = np.sum(X3 + X**2,axis=1) print(X.shape, y.shape) # 归一化 X_norm = normalize(X) X_train = X_norm[:int(len(X_norm)*0.8)] X_test = X_norm[int(len(X_norm)*0.8):] y_train = y[:int(len(X_norm)0.8)] y_test = y[int(len(X_norm)0.8):] # 牛顿法求解回归问题 newton=Newton() newton.fit(X_train, y_train) y_pred = newton.predict(X_test,y_test) print(0.5np.sum((y_test - y_pred)**2)) reg = LinearRegression().fit(X_train, y_train) y_pred = reg.predict(X_test) print(0.5np.sum((y_test - y_pred)**2)) ——修改代码中的问题,并补全缺失的代码,实现牛顿最优化算法
好的,我根据代码中的问题进行了修改和补全缺失的部分,如下所示:
```
import numpy as np
from sklearn import datasets
from sklearn.linear_model import LinearRegression
np.random.seed(10)
class Newton(object):
def __init__(self, epochs=50):
self.W = None
self.b = None
self.epochs = epochs
def get_loss(self, X, y, W, b):
"""
计算损失 0.5sum(y_pred-y)^2
input:
X(2 dim np.array):特征
y(1 dim np.array):标签
W(2 dim np.array):线性回归模型权重矩阵
b(1 dim np.array):线性回归模型偏置
output:损失函数值
"""
y_pred = np.dot(X, W) + b
loss = 0.5 * np.sum((y_pred - y) ** 2)
return loss
def first_derivative(self, X, y):
"""
计算一阶导数g = (y_pred - y)*x
input:
X(2 dim np.array):特征
y(1 dim np.array):标签
output:损失函数值
"""
y_pred = np.dot(X, self.W) + self.b
g = np.dot(X.T, np.array(y_pred - y))
g_b = np.mean(y_pred - y)
return g, g_b
def second_derivative(self, X, y):
"""
计算二阶导数 H_ij = sum(X.T[i]X.T[j])
input:
X(2 dim np.array):特征
y(1 dim np.array):标签
output:损失函数值
"""
H = np.dot(X.T, X)
H_b = 1
return H, H_b
def fit(self, X, y):
"""
线性回归 y = WX + b拟合,牛顿法求解
input:
X(2 dim np.array):特征
y(1 dim np.array):标签
output:拟合的线性回归
"""
self.W = np.random.normal(size=(X.shape[1]))
self.b = 0
for epoch in range(self.epochs):
g, g_b = self.first_derivative(X, y)
H, H_b = self.second_derivative(X, y)
H_inv = np.linalg.inv(H)
self.W = self.W - np.dot(H_inv, g)
self.b = self.b - H_inv @ np.array([g_b])
print("itration:{} ".format(epoch), "loss:{:.4f}".format(self.get_loss(X, y, self.W, self.b)))
def predict(self, X):
"""
线性回归预测
input:
X(2 dim np.array):特征
output:预测结果
"""
y_pred = np.dot(X, self.W) + self.b
return y_pred
def normalize(x):
return (x - np.min(x)) / (np.max(x) - np.min(x))
if __name__ == "__main__":
np.random.seed(2)
X = np.random.rand(100, 5)
y = np.sum(X ** 3 + X ** 2, axis=1)
print(X.shape, y.shape)
# 归一化
X_norm = normalize(X)
X_train = X_norm[:int(len(X_norm) * 0.8)]
X_test = X_norm[int(len(X_norm) * 0.8):]
y_train = y[:int(len(X_norm) * 0.8)]
y_test = y[int(len(X_norm) * 0.8):]
# 牛顿法求解回归问题
newton = Newton()
newton.fit(X_train, y_train)
y_pred = newton.predict(X_test)
print("Newton Method MSE: ", 0.5 * np.sum((y_test - y_pred) ** 2))
# 对比sklearn中的线性回归
reg = LinearRegression().fit(X_train, y_train)
y_pred = reg.predict(X_test)
print("Linear Regression MSE: ", 0.5 * np.sum((y_test - y_pred) ** 2))
```
在这个修改后的代码中,我主要做了以下几个修改和补充:
1. 在 `__init__` 函数中,添加了 `self.b = None`,用于存储线性回归模型的偏置;
2. 在 `get_loss` 函数中,修正了计算损失的公式,将 `b` 加入到了计算中;
3. 在 `fit` 函数中,对牛顿法的求解进行了修改,使用了矩阵求逆的方式来计算 Hessian 矩阵的逆;
4. 在 `predict` 函数中,添加了 `X` 作为输入参数,用于进行预测;
5. 在主函数中,修改了求解 MSE 的计算公式,以及添加了输出方法,用于输出牛顿法求解和 sklearn 中线性回归的 MSE 值。
希望这些修改能够帮助你。
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