{ "cells": [ { "cell_type": "markdown", "id": "73bd968b-d970-4a05-94ef-4e7abf990827", "metadata": {}, "source": [ "Chapter 04\n", "\n", "# 矩阵逐项积\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": "cb498579-9952-45fc-93e8-50479104f8f2", "metadata": {}, "source": [ "\n", "该代码定义了两个 $2 \\times 2$ 矩阵 $A$ 和 $B$,并计算它们的 Hadamard 积(逐元素乘积)。矩阵 $A$ 和 $B$ 分别为:\n", "\n", "$$\n", "A = \\begin{bmatrix} 1 & 2 \\\\ 3 & 4 \\end{bmatrix}, \\quad B = \\begin{bmatrix} 5 & 6 \\\\ 7 & 8 \\end{bmatrix}\n", "$$\n", "\n", "Hadamard 积(逐元素乘积)的结果为每个对应元素相乘,计算公式为:\n", "\n", "$$\n", "A \\odot B = \\begin{bmatrix} 1 \\cdot 5 & 2 \\cdot 6 \\\\ 3 \\cdot 7 & 4 \\cdot 8 \\end{bmatrix} = \\begin{bmatrix} 5 & 12 \\\\ 21 & 32 \\end{bmatrix}\n", "$$\n", "\n", "代码使用 `np.multiply` 函数和 `*` 操作符两种方式来计算逐元素乘积,二者等效。该操作用于生成对应元素相乘的矩阵。" ] }, { "cell_type": "markdown", "id": "46ade452-7ec0-45e9-959c-1276be6d4724", "metadata": {}, "source": [ "## 导入所需库" ] }, { "cell_type": "code", "execution_count": 1, "id": "c90949ae-66bd-4373-ac80-42341574da7c", "metadata": {}, "outputs": [], "source": [ "import numpy as np # 导入NumPy库,用于数值计算" ] }, { "cell_type": "markdown", "id": "9b8e04f7-6950-42d1-9981-81ee65208cce", "metadata": {}, "source": [ "## 定义矩阵A和B" ] }, { "cell_type": "code", "execution_count": 2, "id": "bab49ebe-1142-4a6b-9662-ccc569aaf590", "metadata": {}, "outputs": [], "source": [ "A = np.array([[1, 2], # 定义矩阵A\n", " [3, 4]])" ] }, { "cell_type": "code", "execution_count": 3, "id": "c63f1c20-b2b9-4688-ab7d-800ed8510add", "metadata": {}, "outputs": [], "source": [ "B = np.array([[5, 6], # 定义矩阵B\n", " [7, 8]])" ] }, { "cell_type": "markdown", "id": "861a16bd-2946-438e-8aed-1b974e7d99d6", "metadata": {}, "source": [ "## 计算Hadamard积(逐元素乘积)" ] }, { "cell_type": "code", "execution_count": 4, "id": "f0947846-bc63-482d-a5ac-a6e891794832", "metadata": {}, "outputs": [ { "data": { "text/plain": [ "array([[ 5, 12],\n", " [21, 32]])" ] }, "execution_count": 4, "metadata": {}, "output_type": "execute_result" } ], "source": [ "A_times_B_piecewise = np.multiply(A, B) # 使用np.multiply计算A和B的逐元素乘积\n", "A_times_B_piecewise" ] }, { "cell_type": "code", "execution_count": 5, "id": "7e81c46d-d82b-45fa-b22b-5f99b812f2e4", "metadata": {}, "outputs": [ { "data": { "text/plain": [ "array([[ 5, 12],\n", " [21, 32]])" ] }, "execution_count": 5, "metadata": {}, "output_type": "execute_result" } ], "source": [ "A_times_B_piecewise_V2 = A * B # 使用*操作符计算A和B的逐元素乘积\n", "A_times_B_piecewise_V2" ] }, { "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 }