牛的兄弟 ,你说的很清楚 牛的兄弟 ,你说的很清楚
问问为什么x不使用detach函数会报错?
2.5.4中d.backward()应该是d.sum().backward()吧
第5题,不使用d2l封装的函数
import torch
# x.grad.zero_()
# x = torch.rand(2, 4, requires_grad = True)
x = torch.linspace(0, 2 * torch.pi, 100, requires_grad = True) # 生成100
y = torch.sin(x)
y.sum().backward()
import numpy as np
import matplotlib.pyplot as plt
from matplotlib_inline import backend_inline
plt.figure(figsize=(8, 8))
plt.plot(x.detach().numpy(), y.detach().numpy(), "-", label="sin(x)")
plt.plot(x.detach().numpy(), x.grad.detach().numpy(), "r--", label="cos(x)")
# x1 = torch.linspace(0, 2 * torch.pi, 200)
# plt.plot(x1.numpy(), torch.zeros_like(x1).numpy(), "--", label="0")
plt.legend(loc='upper left')
# 添加标签和标题
plt.xlabel('X')
plt.ylabel('f(x)')
# 显示图表
plt.show()
I think it’s wrong in some sense.Because of the reshape method does not change the datatype,q still is a tensor.
第5题使用TensorFlow
import numpy as np
import matplotlib.pyplot as plt
import tensorflow as tf
def f(a):
c = tf.math.sin(a)
return c
a_values = tf.range(-10.0, 10.0, 0.1)
f_values = []
dx_values = []
with tf.GradientTape(persistent=True) as t:
for a in a_values:
a_tensor = tf.constant(a, dtype=tf.float32)
t.watch(a_tensor)
y = f(a_tensor)
f_values.append(y)
dy_dx = t.gradient(y, a_tensor)
dx_values.append(dy_dx)
plt.plot(a_values, f_values, label='f(x)')
plt.plot(a_values, dx_values, '--',label='f(x)')
plt.xlabel('x')
plt.legend()
plt.grid()
plt.show()
q4:
def f(a):
b = a*2
if a.norm() > 100:
b = a**2
else:
b = a**3
return b.sum()
a = torch.arange(100,104.,requires_grad=True)
print(a)
b = f(a)
b.backward()
a.grad
q5:
x = torch.arange(0,4.,0.1,requires_grad=True)
y = torch.sin(x)
y.sum().backward()
plot(x, [y, x.grad], ‘f(x)’, ‘x’)
(一个粗浅的理解)pytorch不支持对非标量进行backward。sum可以理解为对f(x)增加了一个复合函数,z=sum(y), 而 dz/dy 是一个单位向量或者单位矩阵,这样不影响最终的结果。
这里z=y.sum(),所以z相对于x的导数应该是
[3x1²,3x2²,3x3²],而不是9x**2
pytorch中这种一维的tensor全都看作列向量,所以结果就是一个列向量