download:Python数据分析入门到实战 { "cells": [ { "cell_type": "code", "execution_count": 1, "metadata": {}, "outputs": [], "source": [ "import numpy as np" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "1. 数组 a = np.random.rand(3,2,3) 能和 b = np.random.rand(3,2,2) 进行运算吗？能和 c = np.random.rand(3,1,1) 进行运算吗？请说明结果的原因。" ] }, { "cell_type": "code", "execution_count": 3, "metadata": {}, "outputs": [ { "ename": "ValueError", "evalue": "operands could not be broadcast together with shapes (3,2,3) (3,2,2) ", "output_type": "error", "traceback": [ "\u001b[1;31m---------------------------------------------------------------------------\u001b[0m", "\u001b[1;31mValueError\u001b[0m Traceback (most recent call last)", "\u001b[1;32m<ipython-input-3-8c0e41d1882e>\u001b[0m in \u001b[0;36m<module>\u001b[1;34m\u001b[0m\n\u001b[0;32m 1\u001b[0m \u001b[0ma\u001b[0m \u001b[1;33m=\u001b[0m \u001b[0mnp\u001b[0m\u001b[1;33m.\u001b[0m\u001b[0mrandom\u001b[0m\u001b[1;33m.\u001b[0m\u001b[0mrand\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;36m3\u001b[0m\u001b[1;33m,\u001b[0m\u001b[1;36m2\u001b[0m\u001b[1;33m,\u001b[0m\u001b[1;36m3\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0;32m 2\u001b[0m \u001b[0mb\u001b[0m \u001b[1;33m=\u001b[0m \u001b[0mnp\u001b[0m\u001b[1;33m.\u001b[0m\u001b[0mrandom\u001b[0m\u001b[1;33m.\u001b[0m\u001b[0mrand\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;36m3\u001b[0m\u001b[1;33m,\u001b[0m\u001b[1;36m2\u001b[0m\u001b[1;33m,\u001b[0m\u001b[1;36m2\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[1;32m----> 3\u001b[1;33m \u001b[0ma\u001b[0m \u001b[1;33m+\u001b[0m \u001b[0mb\u001b[0m\u001b[1;33m\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0m\u001b[0;32m 4\u001b[0m \u001b[1;31m# 答案：a和b不能参与运算，因为不满足广播的机制（对应形状的值不相等，并且没有等于1的值）\u001b[0m\u001b[1;33m\u001b[0m\u001b[1;33m\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n", "\u001b[1;31mValueError\u001b[0m: operands could not be broadcast together with shapes (3,2,3) (3,2,2) " ] } ], "source": [ "a = np.random.rand(3,2,3)\n", "b = np.random.rand(3,2,2)\n", "a + b\n", "# 答案：a和b不能参与运算，因为不满足广播的机制（对应形状的值不相等，并且没有等于1的值）" ] }, { "cell_type": "code", "execution_count": 4, "metadata": {}, "outputs": [ { "data": { "text/plain": [ "array([[[1.1966444 , 1.33968419, 1.21752683],\n", " [1.87658798, 1.33365452, 1.61317125]],\n", "\n", " [[0.84684334, 1.00225309, 0.83019466],\n", " [0.54902031, 1.14089275, 1.11052695]],\n", "\n", " [[1.04556413, 1.46687971, 1.68426195],\n", " [0.93094155, 1.52321132, 1.33566973]]])" ] }, "execution_count": 4, "metadata": {}, "output_type": "execute_result" } ], "source": [ "a = np.random.rand(3,2,3)\n", "c = np.random.rand(3,1,1)\n", "a + c\n", "# 答案：a和c可以参与运算，因为满足广播机制（a和b的形状值虽然不相等，但是不相等的地方都是等于1）" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "2. 想要将数组 a = np.arange(15).reshape(3,5) 和 b = np.arange(100,124).reshape(6,4) 叠加在一起，其中 a 在 b 的上面，并且在 b 的第1列后面（下标从0开始）新增一列，用0来填充。" ] }, { "cell_type": "code", "execution_count": 27, "metadata": {}, "outputs": [ { "data": { "text/plain": [ "array([[ 0., 1., 2., 3., 4.],\n", " [ 5., 6., 7., 8., 9.],\n", " [ 10., 11., 12., 13., 14.],\n", " [100., 101., 0., 102., 103.],\n", " [104., 105., 0., 106., 107.],\n", " [108., 109., 0., 110., 111.],\n", " [112., 113., 0., 114., 115.],\n", " [116., 117., 0., 118., 119.],\n", " [120., 121., 0., 122., 123.]])" ] }, "execution_count": 27, "metadata": {}, "output_type": "execute_result" } ], "source": [ "# 第二题答案：\n", "# 1. 准备好数据\n", "a = np.arange(15).reshape(3,5)\n", "b = np.arange(100,124).reshape(6,4)\n", "\n", "# 2. 因为b只有4列，无法直接和a堆叠，并且题目要求要在第1列后面添加一列\n", "# 所以先将b数组在下标为1的地方切割，然后添加完0数组后再进行拼接\n", "bsplits = np.hsplit(b,[2])\n", "# 3. 创建一个全0的6行1列的数组\n", "bzero = np.zeros((6,1))\n", "# 4. 将b的前半部分，0，b的后半部分组合在一起形成一个新的数组\n", "c = np.hstack([bsplits[0],bzero,bsplits[1]])\n", "# 5. 将a和新生成的数组进行堆叠得到结果\n", "result = np.vstack([a,c])\n", "result" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "3. 将数组 a = np.random.rand(4,5) 扁平化成一维数组，可以使用 flatten 和 ravel ，对两者的返回值进行操作，哪个会影响到数组 a ？对会影响到 a 数组的那个函数，请说明原因。" ] }, { "cell_type": "code", "execution_count": 31, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "[[0.88760841 0.18955808 0.50391074 0.66762016 0.44185892]\n", " [0.74659343 0.4997936 0.12576716 0.16264012 0.82619357]\n", " [0.35162609 0.27658575 0.32955319 0.17778362 0.49154326]\n", " [0.63413851 0.0808167 0.3285915 0.98401184 0.41864415]]\n", "==============================\n", "[[20. 0.18955808 0.50391074 0.66762016 0.44185892]\n", " [ 0.74659343 0.4997936 0.12576716 0.16264012 0.82619357]\n", " [ 0.35162609 0.27658575 0.32955319 0.17778362 0.49154326]\n", " [ 0.63413851 0.0808167 0.3285915 0.98401184 0.41864415]]\n" ] } ], "source": [ "# 第三题答案：\n", "a = np.random.rand(4,5)\n", "a1 = a.flatten()\n", "a1[0] = 10\n", "print(a)\n", "print(\"=\"*30)\n", "a2 = a.ravel()\n", "a2[0] = 20\n", "print(a)\n", "# 结果：ravel会影响原来的数组。原因是因为ravel返回的是一个浅拷贝（视图），虽然在栈中的内存不一样，但是指向的堆\n", "# 区的内存地址还是一样，所以修改了a2，会影响到原来堆区的值。但是flatten返回的确实一个深拷贝，也即栈区和堆区都\n", "# 进行了拷贝。" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "4. 使用 numpy 自带的 csv 方法读取出 stock.csv 文件中 preClosePrice 、 openPrice 、 highestPrice 、 lowestPrice 的数据（提示：使用skiprows和usecols参数）。" ] }, { "cell_type": "code", "execution_count": 35, "metadata": {}, "outputs": [ { "data": { "text/plain": [ "array([[13.38 , 13.4 , 13.48 , 12.96 ],\n", " [31.22 , 30.5 , 32.03 , 30.5 ],\n", " [25.56 , 25.41 , 26.4 , 25.18 ],\n", " ...,\n", " [ 0.507, 0.507, 0.508, 0.507],\n", " [ 1.675, 1.67 , 1.69 , 1.67 ],\n", " [ 0.976, 0.976, 0.988, 0.976]])" ] }, "execution_count": 35, "metadata": {}, "output_type": "execute_result" } ], "source": [ "# 第四题答案：\n", "stocks = np.loadtxt(\"stock.csv\",dtype=np.float,delimiter=\",\",skiprows=1,usecols=[6,7,8,9])\n", "stocks\n", "# help(np.loadtxt)" ] } ], "metadata": { "kernelspec": { "display_name": "Python 3", "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": "3.7.1" } }, "nbformat": 4, "nbformat_minor": 2 }
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