{ "cells": [ { "cell_type": "markdown", "id": "73bd968b-d970-4a05-94ef-4e7abf990827", "metadata": {}, "source": [ "Chapter 04\n", "\n", "# 矩阵乘幂比较\n", "Book_4《矩阵力量》 | 鸢尾花书:从加减乘除到机器学习 (第二版)" ] }, { "cell_type": "markdown", "id": "f7cb8538-a1e9-4d16-9e49-6772327912c0", "metadata": {}, "source": [ "\n", "该代码展示了 `np.matrix` 和 `np.array` 在平方操作中的不同行为。\n", "\n", "1. **矩阵平方(`np.matrix` 类型)**:当 $A$ 是 `np.matrix` 类型时,`A**2` 计算的是矩阵乘法,即 $A @ A$。对于矩阵:\n", "\n", " $$\n", " A = \\begin{bmatrix} 1 & 3 \\\\ 2 & 4 \\end{bmatrix}\n", " $$\n", "\n", " 其平方 $A^2$ 计算为:\n", "\n", " $$\n", " A^2 = \\begin{bmatrix} 1 & 3 \\\\ 2 & 4 \\end{bmatrix} @ \\begin{bmatrix} 1 & 3 \\\\ 2 & 4 \\end{bmatrix} = \\begin{bmatrix} 7 & 15 \\\\ 10 & 22 \\end{bmatrix}\n", " $$\n", "\n", "2. **逐元素平方(`np.array` 类型)**:当 $B$ 是 `np.array` 类型时,`B**2` 计算的是逐元素平方,即每个元素单独平方。对于数组:\n", "\n", " $$\n", " B = \\begin{bmatrix} 1 & 3 \\\\ 2 & 4 \\end{bmatrix}\n", " $$\n", "\n", " 逐元素平方的结果为:\n", "\n", " $$\n", " B\\odot B = \\begin{bmatrix} 1^2 & 3^2 \\\\ 2^2 & 4^2 \\end{bmatrix} = \\begin{bmatrix} 1 & 9 \\\\ 4 & 16 \\end{bmatrix}\n", " $$\n", "\n", "此代码展示了 `np.matrix` 和 `np.array` 在平方操作上的关键差异:一个执行矩阵乘法,另一个进行逐元素运算。" ] }, { "cell_type": "markdown", "id": "9734eeec-2b37-40f9-add5-c4ccd409546a", "metadata": {}, "source": [ "## 导入所需库" ] }, { "cell_type": "code", "execution_count": 1, "id": "4b42a3b3-504c-45eb-8c6b-2830e6a11be8", "metadata": {}, "outputs": [], "source": [ "import numpy as np # 导入NumPy库,用于数值计算" ] }, { "cell_type": "markdown", "id": "79f50411-80c3-487e-b20d-e42867cea09a", "metadata": {}, "source": [ "## 定义矩阵A并计算其矩阵平方" ] }, { "cell_type": "code", "execution_count": 2, "id": "5879a711-cd82-4352-a41f-c9b950865d71", "metadata": {}, "outputs": [], "source": [ "A = np.matrix([[1, 3], # 定义为np.matrix类型的矩阵A\n", " [2, 4]])" ] }, { "cell_type": "code", "execution_count": 3, "id": "02a0e4c0-e50d-4955-8e72-351468e22d57", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "[[ 7 15]\n", " [10 22]]\n" ] } ], "source": [ "print(A**2) # 打印矩阵A的矩阵平方结果" ] }, { "cell_type": "markdown", "id": "8dee9c28-b297-4f4d-9f0b-b0c8880d792e", "metadata": {}, "source": [ "## 定义数组B并计算其逐元素平方" ] }, { "cell_type": "code", "execution_count": 4, "id": "7ab6ed46-fca3-46de-aa69-8d1a16b67d24", "metadata": {}, "outputs": [], "source": [ "B = np.array([[1, 3], # 定义为np.array类型的数组B\n", " [2, 4]])" ] }, { "cell_type": "code", "execution_count": 5, "id": "8e1194bb-59fa-40a2-b0a2-e2c85b7cc667", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "[[ 1 9]\n", " [ 4 16]]\n" ] } ], "source": [ "print(B**2) # 打印数组B的逐元素平方结果" ] }, { "cell_type": "code", "execution_count": null, "id": "85a80909-2aac-49ed-bb7a-f8cc6b80ee7d", "metadata": {}, "outputs": [], "source": [] }, { "cell_type": "code", "execution_count": null, "id": "ecd322f4-f919-4be2-adc3-69d28ef25e69", "metadata": {}, "outputs": [], "source": [] } ], "metadata": { "kernelspec": { "display_name": "Python 3 (ipykernel)", "language": "python", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.12.7" } }, "nbformat": 4, "nbformat_minor": 5 }