{ "cells": [ { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "# Introduction to Loops" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "slideshow": { "slide_type": "fragment" } }, "outputs": [], "source": [ "import numpy as np" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "## Problem 1\n", "\n", "**Write code to print the numbers 0 to 4 on each line below.**" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "### Solution" ] }, { "cell_type": "code", "execution_count": 1, "metadata": { "slideshow": { "slide_type": "fragment" } }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "0\n", "1\n", "2\n", "3\n" ] } ], "source": [ "print(0)\n", "print(1)\n", "print(2)\n", "print(3)" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "### Discussion\n", "* Okay, let's keep going to 10000...(!?)" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "## Loops\n", "\n", "* Loops in programing are used to simplify repetative work.\n", "* Above, each line is a print statement.\n", " * The only thing different on each line is the number we are printing.\n", "* We can write a loop statment to do the above problem as follows" ] }, { "cell_type": "code", "execution_count": 2, "metadata": { "slideshow": { "slide_type": "fragment" } }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "0\n", "1\n", "2\n", "3\n" ] } ], "source": [ "for i in range(4):\n", " print(i)" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "### Discussion\n", "* What are the parts of this: list them?\n", "* ```i``` can be any variable, not just \"i\"\n", "* Is there anything weird or unexpected here?" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "## Problem 2\n", "* We can put **any** code in the for loop\n", " * Variables, functions, expressions, math, etc.\n", "* We can put variables outside of the loop that are available inside of the loop\n", "\n" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "For the ideal gas law, pressure is given by\n", "$$P = nRT/V$$\n", "* n = 1 kmol\n", "* R = 8314 J/kmol*K\n", "* T = 300 K\n", "\n", "**For V = 2, 3, 4, 5, ... 20, use a loop to find and print the values of P for each V.**\n", " " ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "skip" } }, "source": [ "### Solution" ] }, { "cell_type": "code", "execution_count": 4, "metadata": { "slideshow": { "slide_type": "fragment" } }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "2 1247100.0\n", "3 831400.0\n", "4 623550.0\n", "5 498840.0\n", "6 415700.0\n", "7 356314.28571428574\n", "8 311775.0\n", "9 277133.3333333333\n", "10 249420.0\n", "11 226745.45454545456\n", "12 207850.0\n", "13 191861.53846153847\n", "14 178157.14285714287\n", "15 166280.0\n", "16 155887.5\n", "17 146717.64705882352\n", "18 138566.66666666666\n", "19 131273.68421052632\n", "20 124710.0\n" ] } ], "source": [ "n = 1 # kmol\n", "R = 8314 # J/kmol*K\n", "T = 300 # K\n", "\n", "for V in range(2,21):\n", " P = n*R*T/V\n", " print(V, P)" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "### Discussion\n", "* What if you wanted V to not step by integers?\n", " * Step by 2? Step by 5?\n", " * Step by 0.2?" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "## Problem 3 \n", "\n", "**Add all the numbers from 0 to 100**\n", "* Before you start the code, think about what you would do if solving this in your head?\n", " * How do you start?\n", " * What things do you keep track of?\n", " * What variables to use?\n", " * What is the loop syntax?\n", " * How do variables interact with the loop?\n", "* Think through the process of the problem to the solution\n", "* Talk to your neighbor\n", "* If you've solved it, help your neighbor." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "### Solution" ] }, { "cell_type": "code", "execution_count": 5, "metadata": { "slideshow": { "slide_type": "fragment" } }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "5050\n" ] } ], "source": [ "s = 0\n", "for i in range(101):\n", " s += i\n", "print(s)" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "## Nested loops\n", "\n", "* We can put **any** code in a ```for``` loop. Even another loop.\n", "* Look at the code below. \n", "* **Before running it, think of what you expect to see.**\n", " * Now run it and compare. Did you get what you thought you would?" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "slideshow": { "slide_type": "fragment" } }, "outputs": [], "source": [ "for i in range(3):\n", " print(\"i =\", i)\n", " for j in range(4):\n", " print(\" j =\", j)" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "* Note, each time through the top (outer) loop, **everything** inside is done.\n", "* This means the **whole** inner loop is computed for each outer loop.\n", "* **Think of an actual application for a nested loop.**" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "## Problem 4a\n", "\n", "* Given the following function\n", "$$f(x) = x^2 - 3,$$\n", "we can use Newton's method to find $x$ for $f(x)=0$. \n", "\n", "* Recall, Newton recurrence formula:\n", "$$x_{new} = x_{old} - \\frac{f(x_{old})}{f^\\prime(x_{old})}$$\n", "\n", "### We solved this previously in Excel\n", "* **Open Excel and solve this problem.**\n", "* Think about the steps involved, how we set it up.\n", "* **Use the simplest method possible.**\n" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "### Solution-Excel\n", "\n", "\n", "\n", "\n", "| Excel | Python|\n", "|:-----:|:-----:|\n", "|cell for guess | ?? |\n", "|Newton equation in terms of x cell
A2 - (A2^2-3)/(2*A2)| ?? |\n", "| repeat: fill down | ?? |\n" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "## Problem 4b\n", "### Now solve in Python using loops\n", "\n", "* **Try to write the simplest solution you can think of with as few lines of code as possible.**\n", "* **We want 3 decimal places of accuracy.**\n", "* Consider: \n", " * What variables are needed?\n", " * How to setup and use the loop.\n", " * What to print and where?" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "### Solution-Python" ] }, { "cell_type": "code", "execution_count": 7, "metadata": { "slideshow": { "slide_type": "fragment" } }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "2.0\n", "1.75\n", "1.7321428571428572\n", "1.7320508100147276\n", "1.7320508075688772\n" ] } ], "source": [ "x = 1\n", "for i in range(5):\n", " x = x - (x**2-3)/(2*x)\n", " print(x)" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "### Discussion\n", "- Note that `i` isn't used, it's just convenient for making the loop\n", "- x is reassigned in place (x is replaced with the updated value on each loop)" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "### Explicitly write out the four steps\n", "* We don't have to code it from top to bottom...\n", "* Compare to Excel\n", "\n", "| Excel | Python|\n", "|:-----:|:-----:|\n", "|cell for guess | variable for x |\n", "|Newton equation in terms of x cell
A2 - (A2^2-3)/(2*A2)| Newton equation in terms of x
x = x - (x\\*\\*2-3)/(2*x)|\n", "| repeat: fill down | repeat: wrap in loop |" ] }, { "cell_type": "code", "execution_count": 8, "metadata": { "slideshow": { "slide_type": "slide" } }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "2.0\n", "1.75\n", "1.7321428571428572\n", "1.7320508100147276\n", "1.7320508075688772\n", "1.7320508075688774\n" ] } ], "source": [ "# step 1: initial guess\n", "x = 1\n", "\n", "# step 2: add Newton equation\n", "x=1\n", "x = x - (x**2 - 3)/(2*x)\n", "\n", "# step 3: wrap in loop\n", "x=1\n", "for i in range(6):\n", " x = x - (x**2 - 3)/(2*x)\n", "\n", "# step 4: add a print\n", "x=1\n", "for i in range(6):\n", " x = x - (x**2 - 3)/(2*x)\n", " print(x)" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "## Problem 4c\n", "\n", "**Redo Solution 1 using functions for $f(x)$ and $f^\\prime(x)$.**\n", "* Recall: we discussed functions two classes back." ] }, { "cell_type": "code", "execution_count": 9, "metadata": { "slideshow": { "slide_type": "fragment" } }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "2.0\n", "1.75\n", "1.7321428571428572\n", "1.7320508100147276\n", "1.7320508075688772\n" ] } ], "source": [ "def f(x):\n", " return x**2-3\n", "def fp(x):\n", " return 2*x\n", "\n", "x = 1\n", "for i in range(5):\n", " x = x - f(x)/fp(x)\n", " print(x)" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "## Problem 4d\n", "\n", "* Now, to get to three decimal places we only needed 3 iterations.\n", "* But we didn't know this at the beginning, so we guessed the number of iterations.\n", " * If 6 wasn't enough, we would do more.\n", "* This guess and check is not very efficient.\n", "\n", "**Modify the code above to do just the right number of iterations.**\n", "* What Python features do you need to use?\n", "* How do you decide when you are done?\n", "* **Talk with your neighbor.**" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "skip" } }, "source": [ "### Solution" ] }, { "cell_type": "code", "execution_count": 15, "metadata": { "slideshow": { "slide_type": "slide" } }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "2.0\n", "1.75\n", "1.7321428571428572\n", "1.7320508100147276\n" ] } ], "source": [ "def f(x):\n", " return x**2-3\n", "def fp(x):\n", " return 2*x\n", "\n", "eps = 0.001\n", "xold = 1\n", "for i in range(5):\n", " xnew = xold - f(xold)/fp(xold)\n", " print(xnew)\n", " rerr = np.abs((xold - xnew)/xnew)\n", " if rerr < eps: \n", " break\n", " xold = xnew" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "### Discussion\n", "- Keep track of old and new x: `xold` and `xnew`.\n", " - Need to reset `xold=xnew` at the end to get set for next interation.\n", "- New keyword `break`\n", "- Define the error and compare it to the desired accuracy.\n", " - Use absolute value: `np.abs()`" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [] } ], "metadata": { "celltoolbar": "Slideshow", "kernelspec": { "display_name": "base", "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": "3.14.6" }, "varInspector": { "cols": { "lenName": 16, "lenType": 16, "lenVar": 40 }, "kernels_config": { "python": { "delete_cmd_postfix": "", "delete_cmd_prefix": "del ", "library": "var_list.py", "varRefreshCmd": "print(var_dic_list())" }, "r": { "delete_cmd_postfix": ") ", "delete_cmd_prefix": "rm(", "library": "var_list.r", "varRefreshCmd": "cat(var_dic_list()) " } }, "types_to_exclude": [ "module", "function", "builtin_function_or_method", "instance", "_Feature" ], "window_display": false } }, "nbformat": 4, "nbformat_minor": 4 }