596 lines
15 KiB
Plaintext
596 lines
15 KiB
Plaintext
{
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"cells": [
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{
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"cell_type": "markdown",
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"id": "b4873f83",
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"metadata": {
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"origin_pos": 0
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},
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"source": [
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"# 自动微分\n",
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":label:`sec_autograd`\n",
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"\n",
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"正如 :numref:`sec_calculus`中所说，求导是几乎所有深度学习优化算法的关键步骤。\n",
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"虽然求导的计算很简单，只需要一些基本的微积分。\n",
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"但对于复杂的模型，手工进行更新是一件很痛苦的事情（而且经常容易出错）。\n",
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"\n",
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"深度学习框架通过自动计算导数，即*自动微分*（automatic differentiation）来加快求导。\n",
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"实际中，根据设计好的模型，系统会构建一个*计算图*（computational graph），\n",
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"来跟踪计算是哪些数据通过哪些操作组合起来产生输出。\n",
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"自动微分使系统能够随后反向传播梯度。\n",
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"这里，*反向传播*（backpropagate）意味着跟踪整个计算图，填充关于每个参数的偏导数。\n",
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"\n",
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"## 一个简单的例子\n",
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"\n",
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"作为一个演示例子，(**假设我们想对函数$y=2\\mathbf{x}^{\\top}\\mathbf{x}$关于列向量$\\mathbf{x}$求导**)。\n",
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"首先，我们创建变量`x`并为其分配一个初始值。\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 1,
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"id": "98cd8a9e",
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"metadata": {
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"execution": {
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"iopub.execute_input": "2023-08-18T07:07:31.627945Z",
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"iopub.status.busy": "2023-08-18T07:07:31.627424Z",
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"iopub.status.idle": "2023-08-18T07:07:32.686372Z",
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"shell.execute_reply": "2023-08-18T07:07:32.685559Z"
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},
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"origin_pos": 2,
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"tab": [
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"pytorch"
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]
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},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"tensor([0., 1., 2., 3.])"
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]
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},
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"execution_count": 1,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"import torch\n",
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"\n",
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"x = torch.arange(4.0)\n",
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"x"
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]
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},
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{
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"cell_type": "markdown",
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"id": "ec430520",
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"metadata": {
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"origin_pos": 5
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},
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"source": [
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"[**在我们计算$y$关于$\\mathbf{x}$的梯度之前，需要一个地方来存储梯度。**]\n",
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"重要的是，我们不会在每次对一个参数求导时都分配新的内存。\n",
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"因为我们经常会成千上万次地更新相同的参数，每次都分配新的内存可能很快就会将内存耗尽。\n",
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"注意，一个标量函数关于向量$\\mathbf{x}$的梯度是向量，并且与$\\mathbf{x}$具有相同的形状。\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 2,
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"id": "e27a5df4",
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"metadata": {
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"execution": {
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"iopub.execute_input": "2023-08-18T07:07:32.690633Z",
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"iopub.status.busy": "2023-08-18T07:07:32.689882Z",
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"iopub.status.idle": "2023-08-18T07:07:32.694159Z",
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"shell.execute_reply": "2023-08-18T07:07:32.693367Z"
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},
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"origin_pos": 7,
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"tab": [
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"pytorch"
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]
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},
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"outputs": [],
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"source": [
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"x.requires_grad_(True) # 等价于x=torch.arange(4.0,requires_grad=True)\n",
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"x.grad # 默认值是None"
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]
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},
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{
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"cell_type": "markdown",
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"id": "bd993524",
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"metadata": {
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"origin_pos": 10
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},
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"source": [
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"(**现在计算$y$。**)\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 3,
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"id": "4c3f80b7",
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"metadata": {
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"execution": {
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"iopub.execute_input": "2023-08-18T07:07:32.698006Z",
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"iopub.status.busy": "2023-08-18T07:07:32.697167Z",
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"iopub.status.idle": "2023-08-18T07:07:32.705385Z",
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"shell.execute_reply": "2023-08-18T07:07:32.704593Z"
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},
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"origin_pos": 12,
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"tab": [
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"pytorch"
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]
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},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"tensor(28., grad_fn=<MulBackward0>)"
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]
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},
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"execution_count": 3,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"y = 2 * torch.dot(x, x)\n",
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"y"
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]
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},
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{
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"cell_type": "markdown",
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"id": "35523dbc",
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"metadata": {
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"origin_pos": 15
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},
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"source": [
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"`x`是一个长度为4的向量，计算`x`和`x`的点积，得到了我们赋值给`y`的标量输出。\n",
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"接下来，[**通过调用反向传播函数来自动计算`y`关于`x`每个分量的梯度**]，并打印这些梯度。\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 4,
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"id": "a1c3a419",
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"metadata": {
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"execution": {
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"iopub.execute_input": "2023-08-18T07:07:32.708698Z",
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"iopub.status.busy": "2023-08-18T07:07:32.708196Z",
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"iopub.status.idle": "2023-08-18T07:07:32.713924Z",
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"shell.execute_reply": "2023-08-18T07:07:32.713091Z"
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},
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"origin_pos": 17,
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"tab": [
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"pytorch"
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]
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},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"tensor([ 0., 4., 8., 12.])"
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]
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},
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"execution_count": 4,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"y.backward()\n",
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"x.grad"
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]
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},
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{
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"cell_type": "markdown",
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"id": "dca6a271",
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"metadata": {
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"origin_pos": 20
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},
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"source": [
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"函数$y=2\\mathbf{x}^{\\top}\\mathbf{x}$关于$\\mathbf{x}$的梯度应为$4\\mathbf{x}$。\n",
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"让我们快速验证这个梯度是否计算正确。\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 5,
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"id": "b8493d0a",
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"metadata": {
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"execution": {
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"iopub.execute_input": "2023-08-18T07:07:32.718858Z",
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"iopub.status.busy": "2023-08-18T07:07:32.718156Z",
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"iopub.status.idle": "2023-08-18T07:07:32.724091Z",
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"shell.execute_reply": "2023-08-18T07:07:32.723104Z"
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},
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"origin_pos": 22,
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"tab": [
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"pytorch"
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]
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},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"tensor([True, True, True, True])"
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]
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},
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"execution_count": 5,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"x.grad == 4 * x"
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]
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},
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{
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"cell_type": "markdown",
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"id": "2733c623",
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"metadata": {
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"origin_pos": 25
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},
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"source": [
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"[**现在计算`x`的另一个函数。**]\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 6,
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"id": "f2fcd392",
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"metadata": {
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"execution": {
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"iopub.execute_input": "2023-08-18T07:07:32.729368Z",
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"iopub.status.busy": "2023-08-18T07:07:32.728433Z",
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"iopub.status.idle": "2023-08-18T07:07:32.736493Z",
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"shell.execute_reply": "2023-08-18T07:07:32.735715Z"
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},
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"origin_pos": 27,
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"tab": [
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"pytorch"
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]
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},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"tensor([1., 1., 1., 1.])"
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]
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},
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"execution_count": 6,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"# 在默认情况下，PyTorch会累积梯度，我们需要清除之前的值\n",
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"x.grad.zero_()\n",
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"y = x.sum()\n",
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"y.backward()\n",
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"x.grad"
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]
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},
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{
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"cell_type": "markdown",
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"id": "58f4f459",
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"metadata": {
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"origin_pos": 30
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},
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"source": [
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"## 非标量变量的反向传播\n",
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"\n",
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"当`y`不是标量时，向量`y`关于向量`x`的导数的最自然解释是一个矩阵。\n",
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"对于高阶和高维的`y`和`x`，求导的结果可以是一个高阶张量。\n",
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"\n",
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"然而，虽然这些更奇特的对象确实出现在高级机器学习中（包括[**深度学习中**]），\n",
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"但当调用向量的反向计算时，我们通常会试图计算一批训练样本中每个组成部分的损失函数的导数。\n",
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"这里(**，我们的目的不是计算微分矩阵，而是单独计算批量中每个样本的偏导数之和。**)\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 7,
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"id": "f4e62a5d",
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"metadata": {
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"execution": {
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"iopub.execute_input": "2023-08-18T07:07:32.740109Z",
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"iopub.status.busy": "2023-08-18T07:07:32.739419Z",
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"iopub.status.idle": "2023-08-18T07:07:32.745803Z",
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"shell.execute_reply": "2023-08-18T07:07:32.744893Z"
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},
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"origin_pos": 32,
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"tab": [
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"pytorch"
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]
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},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"tensor([0., 2., 4., 6.])"
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]
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},
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"execution_count": 7,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"# 对非标量调用backward需要传入一个gradient参数，该参数指定微分函数关于self的梯度。\n",
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"# 本例只想求偏导数的和，所以传递一个1的梯度是合适的\n",
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"x.grad.zero_()\n",
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"y = x * x\n",
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"# 等价于y.backward(torch.ones(len(x)))\n",
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"y.sum().backward()\n",
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"x.grad"
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]
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},
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{
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"cell_type": "markdown",
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"id": "80f510c4",
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"metadata": {
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"origin_pos": 35
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},
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"source": [
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"## 分离计算\n",
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"\n",
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"有时，我们希望[**将某些计算移动到记录的计算图之外**]。\n",
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"例如，假设`y`是作为`x`的函数计算的，而`z`则是作为`y`和`x`的函数计算的。\n",
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"想象一下，我们想计算`z`关于`x`的梯度，但由于某种原因，希望将`y`视为一个常数，\n",
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"并且只考虑到`x`在`y`被计算后发挥的作用。\n",
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"\n",
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"这里可以分离`y`来返回一个新变量`u`，该变量与`y`具有相同的值，\n",
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"但丢弃计算图中如何计算`y`的任何信息。\n",
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"换句话说，梯度不会向后流经`u`到`x`。\n",
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"因此，下面的反向传播函数计算`z=u*x`关于`x`的偏导数，同时将`u`作为常数处理，\n",
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"而不是`z=x*x*x`关于`x`的偏导数。\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 8,
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"id": "8dab493d",
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"metadata": {
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||
"execution": {
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||
"iopub.execute_input": "2023-08-18T07:07:32.749398Z",
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||
"iopub.status.busy": "2023-08-18T07:07:32.748759Z",
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||
"iopub.status.idle": "2023-08-18T07:07:32.755280Z",
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||
"shell.execute_reply": "2023-08-18T07:07:32.754543Z"
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},
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"origin_pos": 37,
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"tab": [
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"pytorch"
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]
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},
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||
"outputs": [
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{
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"data": {
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"text/plain": [
|
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"tensor([True, True, True, True])"
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]
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},
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"execution_count": 8,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"x.grad.zero_()\n",
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"y = x * x\n",
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"u = y.detach()\n",
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"z = u * x\n",
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"\n",
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"z.sum().backward()\n",
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"x.grad == u"
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]
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},
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{
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||
"cell_type": "markdown",
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||
"id": "f8fe6f9c",
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||
"metadata": {
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||
"origin_pos": 40
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||
},
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"source": [
|
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"由于记录了`y`的计算结果，我们可以随后在`y`上调用反向传播，\n",
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"得到`y=x*x`关于的`x`的导数，即`2*x`。\n"
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]
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},
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{
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||
"cell_type": "code",
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||
"execution_count": 9,
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||
"id": "271a9b3a",
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||
"metadata": {
|
||
"execution": {
|
||
"iopub.execute_input": "2023-08-18T07:07:32.759344Z",
|
||
"iopub.status.busy": "2023-08-18T07:07:32.758633Z",
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||
"iopub.status.idle": "2023-08-18T07:07:32.764663Z",
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||
"shell.execute_reply": "2023-08-18T07:07:32.763922Z"
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||
},
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||
"origin_pos": 42,
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"tab": [
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"pytorch"
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||
]
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||
},
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||
"outputs": [
|
||
{
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||
"data": {
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"text/plain": [
|
||
"tensor([True, True, True, True])"
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]
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},
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"execution_count": 9,
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||
"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"x.grad.zero_()\n",
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"y.sum().backward()\n",
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"x.grad == 2 * x"
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]
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},
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{
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"cell_type": "markdown",
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||
"id": "fd79d12f",
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"metadata": {
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||
"origin_pos": 45
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},
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"source": [
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"## Python控制流的梯度计算\n",
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"\n",
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"使用自动微分的一个好处是：\n",
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"[**即使构建函数的计算图需要通过Python控制流（例如，条件、循环或任意函数调用），我们仍然可以计算得到的变量的梯度**]。\n",
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"在下面的代码中，`while`循环的迭代次数和`if`语句的结果都取决于输入`a`的值。\n"
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]
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},
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{
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||
"cell_type": "code",
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"execution_count": 10,
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||
"id": "6323b2ff",
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||
"metadata": {
|
||
"execution": {
|
||
"iopub.execute_input": "2023-08-18T07:07:32.769249Z",
|
||
"iopub.status.busy": "2023-08-18T07:07:32.768616Z",
|
||
"iopub.status.idle": "2023-08-18T07:07:32.773175Z",
|
||
"shell.execute_reply": "2023-08-18T07:07:32.772293Z"
|
||
},
|
||
"origin_pos": 47,
|
||
"tab": [
|
||
"pytorch"
|
||
]
|
||
},
|
||
"outputs": [],
|
||
"source": [
|
||
"def f(a):\n",
|
||
" b = a * 2\n",
|
||
" while b.norm() < 1000:\n",
|
||
" b = b * 2\n",
|
||
" if b.sum() > 0:\n",
|
||
" c = b\n",
|
||
" else:\n",
|
||
" c = 100 * b\n",
|
||
" return c"
|
||
]
|
||
},
|
||
{
|
||
"cell_type": "markdown",
|
||
"id": "51aaf333",
|
||
"metadata": {
|
||
"origin_pos": 50
|
||
},
|
||
"source": [
|
||
"让我们计算梯度。\n"
|
||
]
|
||
},
|
||
{
|
||
"cell_type": "code",
|
||
"execution_count": 11,
|
||
"id": "7719d6b6",
|
||
"metadata": {
|
||
"execution": {
|
||
"iopub.execute_input": "2023-08-18T07:07:32.777740Z",
|
||
"iopub.status.busy": "2023-08-18T07:07:32.777207Z",
|
||
"iopub.status.idle": "2023-08-18T07:07:32.782254Z",
|
||
"shell.execute_reply": "2023-08-18T07:07:32.781458Z"
|
||
},
|
||
"origin_pos": 52,
|
||
"tab": [
|
||
"pytorch"
|
||
]
|
||
},
|
||
"outputs": [],
|
||
"source": [
|
||
"a = torch.randn(size=(), requires_grad=True)\n",
|
||
"d = f(a)\n",
|
||
"d.backward()"
|
||
]
|
||
},
|
||
{
|
||
"cell_type": "markdown",
|
||
"id": "816a1ac2",
|
||
"metadata": {
|
||
"origin_pos": 55
|
||
},
|
||
"source": [
|
||
"我们现在可以分析上面定义的`f`函数。\n",
|
||
"请注意，它在其输入`a`中是分段线性的。\n",
|
||
"换言之，对于任何`a`，存在某个常量标量`k`，使得`f(a)=k*a`，其中`k`的值取决于输入`a`，因此可以用`d/a`验证梯度是否正确。\n"
|
||
]
|
||
},
|
||
{
|
||
"cell_type": "code",
|
||
"execution_count": 12,
|
||
"id": "2595bdc0",
|
||
"metadata": {
|
||
"execution": {
|
||
"iopub.execute_input": "2023-08-18T07:07:32.785728Z",
|
||
"iopub.status.busy": "2023-08-18T07:07:32.785179Z",
|
||
"iopub.status.idle": "2023-08-18T07:07:32.790672Z",
|
||
"shell.execute_reply": "2023-08-18T07:07:32.789892Z"
|
||
},
|
||
"origin_pos": 57,
|
||
"tab": [
|
||
"pytorch"
|
||
]
|
||
},
|
||
"outputs": [
|
||
{
|
||
"data": {
|
||
"text/plain": [
|
||
"tensor(True)"
|
||
]
|
||
},
|
||
"execution_count": 12,
|
||
"metadata": {},
|
||
"output_type": "execute_result"
|
||
}
|
||
],
|
||
"source": [
|
||
"a.grad == d / a"
|
||
]
|
||
},
|
||
{
|
||
"cell_type": "markdown",
|
||
"id": "67fb5517",
|
||
"metadata": {
|
||
"origin_pos": 60
|
||
},
|
||
"source": [
|
||
"## 小结\n",
|
||
"\n",
|
||
"* 深度学习框架可以自动计算导数：我们首先将梯度附加到想要对其计算偏导数的变量上，然后记录目标值的计算，执行它的反向传播函数，并访问得到的梯度。\n",
|
||
"\n",
|
||
"## 练习\n",
|
||
"\n",
|
||
"1. 为什么计算二阶导数比一阶导数的开销要更大？\n",
|
||
"1. 在运行反向传播函数之后，立即再次运行它，看看会发生什么。\n",
|
||
"1. 在控制流的例子中，我们计算`d`关于`a`的导数，如果将变量`a`更改为随机向量或矩阵，会发生什么？\n",
|
||
"1. 重新设计一个求控制流梯度的例子，运行并分析结果。\n",
|
||
"1. 使$f(x)=\\sin(x)$，绘制$f(x)$和$\\frac{df(x)}{dx}$的图像，其中后者不使用$f'(x)=\\cos(x)$。\n"
|
||
]
|
||
},
|
||
{
|
||
"cell_type": "markdown",
|
||
"id": "530f74f8",
|
||
"metadata": {
|
||
"origin_pos": 62,
|
||
"tab": [
|
||
"pytorch"
|
||
]
|
||
},
|
||
"source": [
|
||
"[Discussions](https://discuss.d2l.ai/t/1759)\n"
|
||
]
|
||
}
|
||
],
|
||
"metadata": {
|
||
"language_info": {
|
||
"name": "python"
|
||
},
|
||
"required_libs": []
|
||
},
|
||
"nbformat": 4,
|
||
"nbformat_minor": 5
|
||
} |