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Warm-up: numpy¶
一个三次多项式,从 \(-\pi\) 到 \(\pi\) 的 \(y=\sin(x)\),通过最小化欧几里得距离的平方来训练预测。
该实现使用 numpy 手动计算前向传递、损失和反向传递。
numpy数组是一个通用的n维数组;它不了解深度学习、梯度或计算图,只是用于执行通用数值计算的一个库。
import numpy as np
import math
# Create random input and output data
x = np.linspace(-math.pi, math.pi, 2000)
y = np.sin(x)
# Randomly initialize weights
a = np.random.randn()
b = np.random.randn()
c = np.random.randn()
d = np.random.randn()
learning_rate = 1e-6
for t in range(2000):
# Forward pass: compute predicted y
# y = a + b x + c x^2 + d x^3
y_pred = a + b * x + c * x ** 2 + d * x ** 3
# Compute and print loss
loss = np.square(y_pred - y).sum()
if t % 100 == 99:
print(t, loss)
# Backprop to compute gradients of a, b, c, d with respect to loss
grad_y_pred = 2.0 * (y_pred - y)
grad_a = grad_y_pred.sum()
grad_b = (grad_y_pred * x).sum()
grad_c = (grad_y_pred * x ** 2).sum()
grad_d = (grad_y_pred * x ** 3).sum()
# Update weights
a -= learning_rate * grad_a
b -= learning_rate * grad_b
c -= learning_rate * grad_c
d -= learning_rate * grad_d
print(f'Result: y = {a} + {b} x + {c} x^2 + {d} x^3')
Total running time of the script: ( 0 minutes 0.000 seconds)