From d73fa2131ea8796dbabd4313312a19d2feae06e8 Mon Sep 17 00:00:00 2001 From: Berna Ozer Date: Thu, 14 Dec 2023 17:06:49 +0100 Subject: [PATCH] Berna Ozer --- .DS_Store | Bin 0 -> 6148 bytes your-code/main.ipynb | 714 +++++++++++++++++++++++++------------------ 2 files changed, 410 insertions(+), 304 deletions(-) create mode 100644 .DS_Store diff --git a/.DS_Store b/.DS_Store new file mode 100644 index 0000000000000000000000000000000000000000..8abf80c9337c3f99b99b5ae0d6fca390f5efa55c GIT binary patch literal 6148 zcmeHK%}N6?5T3Nvb}2#+iXH=AE4I4SgO{b&7jQ)nDs|T_y0~s++ghX)_Np)BoA^A= zB&k?h!IOxcfytN5&xU+iG6?{P_AqJ!)BvD@N?35PSs>&mos)vK5DNW`9k>An5%j{* zTr?a0A_KH{4m{(o4IzUs``3RA6BVNOEqqMk{-Du#7lmSJWwl&(%GL6ff2T(NAQ%k$ z?I63L(Wz2NIPC}FS=<|SYwJfU9RzXO%XC5<^)Te}JWeAuYO7%y^>waq9F9|Qy0y*m zxV2xGdrfyzm*WH1txLD%P9{!eV|(}Tqpz}eZ5_%SMgZk*ehOUp~FAMUcB#Wbcm}y;2AF|a z28w1_qxygR{ri75iAT%;Gw`n%5T%aaY2%V?ZJk>j)mn*qjY>jsxxvpA?C4UAu~dq; bP_>|4QU}qqm>Wb33SR^?4cssTzskS|ewR#; literal 0 HcmV?d00001 diff --git a/your-code/main.ipynb b/your-code/main.ipynb index 38b51a7..4baa6a8 100644 --- a/your-code/main.ipynb +++ b/your-code/main.ipynb @@ -1,304 +1,410 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Before your start:\n", - "- Read the README.md file\n", - "- Comment as much as you can and use the resources in the README.md file\n", - "- Happy learning!" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Challenge 1 - Passing a Lambda Expression to a Function\n", - "\n", - "In the next excercise you will create a function that returns a lambda expression. Create a function called `modify_list`. The function takes two arguments, a list and a lambda expression. The function iterates through the list and applies the lambda expression to every element in the list." - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "Follow the steps as stated below:\n", - " 1. Define a list of any 10 numbers\n", - " 2. Define a simple lambda expression for eg that updates a number by 2\n", - " 3. Define an empty list\n", - " 4. Define the function -> use the lambda function to append the empty list\n", - " 5. Call the function with list and lambda expression\n", - " 6. print the updated list " - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": {}, - "outputs": [], - "source": [ - "# your code here\n" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "#### Now we will define a lambda expression that will transform the elements of the list. \n", - "\n", - "In the cell below, create a lambda expression that converts Celsius to Kelvin. Recall that 0°C + 273.15 = 273.15K" - ] - }, - { - "cell_type": "code", - "execution_count": 3, - "metadata": {}, - "outputs": [], - "source": [ - "# Your code here:\n" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "Finally, convert the list of temperatures below from Celsius to Kelvin." - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": {}, - "outputs": [], - "source": [ - "temps = [12, 23, 38, -55, 24]\n", - "\n", - "# Your code here:\n" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "#### In this part, we will define a function that returns a lambda expression\n", - "\n", - "In the cell below, write a lambda expression that takes two numbers and returns 1 if one is divisible by the other and zero otherwise. Call the lambda expression `mod`." - ] - }, - { - "cell_type": "code", - "execution_count": 5, - "metadata": {}, - "outputs": [], - "source": [ - "# Your code here:\n" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "#### Now create a function that returns mod. The function only takes one argument - the first number in the `mod` lambda function. \n", - "\n", - "Note: the lambda function above took two arguments, the lambda function in the return statement only takes one argument but also uses the argument passed to the function." - ] - }, - { - "cell_type": "code", - "execution_count": 6, - "metadata": {}, - "outputs": [], - "source": [ - "def divisor(a):\n", - " \"\"\"\n", - " input: a number\n", - " output: a function that returns 1 if the number is divisible by another number (to be passed later) and zero otherwise\n", - " \"\"\"\n", - " # Your code here:\n" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "Finally, pass the number 5 to `divisor`. Now the function will check whether a number is divisble by 5. Assign this function to `divisible5`" - ] - }, - { - "cell_type": "code", - "execution_count": 7, - "metadata": {}, - "outputs": [], - "source": [ - "# Your code here:\n" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "Test your function with the following test cases:" - ] - }, - { - "cell_type": "code", - "execution_count": null, - "metadata": {}, - "outputs": [], - "source": [ - "divisible5(10)" - ] - }, - { - "cell_type": "code", - "execution_count": null, - "metadata": {}, - "outputs": [], - "source": [ - "divisible5(8)" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Challenge 2 - Using Lambda Expressions in List Comprehensions\n", - "\n", - "In the following challenge, we will combine two lists using a lambda expression in a list comprehension. \n", - "\n", - "To do this, we will need to introduce the `zip` function. The `zip` function returns an iterator of tuples.\n", - "\n", - "The way zip function works with list has been shown below:" - ] - }, - { - "cell_type": "code", - "execution_count": 1, - "metadata": {}, - "outputs": [ - { - "data": { - "text/plain": [ - "[('Green', 'eggs'),\n", - " ('cheese', 'cheese'),\n", - " ('English', 'cucumber'),\n", - " ('tomato', 'tomato')]" - ] - }, - "execution_count": 1, - "metadata": {}, - "output_type": "execute_result" - } - ], - "source": [ - "list1 = ['Green', 'cheese', 'English', 'tomato']\n", - "list2 = ['eggs', 'cheese', 'cucumber', 'tomato']\n", - "zipped = zip(list1,list2)\n", - "list(zipped)" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "In this exercise we will try to compare the elements on the same index in the two lists. \n", - "We want to zip the two lists and then use a lambda expression to compare if:\n", - "list1 element > list2 element " - ] - }, - { - "cell_type": "code", - "execution_count": 9, - "metadata": {}, - "outputs": [], - "source": [ - "list1 = [1,2,4,4]\n", - "list2 = [2,3,3,5]\n", - "## Zip the lists together \n", - "\n", - "## Print the zipped list \n", - "\n", - "## Use a lambda expression to compare if: list1 element > list2 element\n" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "Complete the parts of the code marked as \"###\"" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Challenge 3 - Using Lambda Expressions as Arguments\n", - "\n", - "#### In this challenge, we will zip together two lists and sort by the resulting tuple.\n", - "\n", - "In the cell below, take the two lists provided, zip them together and sort by the first letter of the second element of each tuple. Do this using a lambda function." - ] - }, - { - "cell_type": "code", - "execution_count": 10, - "metadata": {}, - "outputs": [], - "source": [ - "list1 = ['Engineering', 'Computer Science', 'Political Science', 'Mathematics']\n", - "list2 = ['Lab', 'Homework', 'Essay', 'Module']\n", - "\n", - "# Your code here:\n" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Bonus Challenge - Sort a Dictionary by Values\n", - "\n", - "Given the dictionary below, sort it by values rather than by keys. Use a lambda function to specify the values as a sorting key." - ] - }, - { - "cell_type": "code", - "execution_count": 11, - "metadata": {}, - "outputs": [], - "source": [ - "d = {'Honda': 1997, 'Toyota': 1995, 'Audi': 2001, 'BMW': 2005}\n", - "\n", - "# Your code here:\n" - ] - }, - { - "cell_type": "code", - "execution_count": null, - "metadata": {}, - "outputs": [], - "source": [] - } - ], - "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": "3.7.3" - } - }, - "nbformat": 4, - "nbformat_minor": 2 -} +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Before your start:\n", + "- Read the README.md file\n", + "- Comment as much as you can and use the resources in the README.md file\n", + "- Happy learning!" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Challenge 1 - Passing a Lambda Expression to a Function\n", + "\n", + "In the next excercise you will create a function that returns a lambda expression. Create a function called `modify_list`. The function takes two arguments, a list and a lambda expression. The function iterates through the list and applies the lambda expression to every element in the list." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Follow the steps as stated below:\n", + " 1. Define a list of any 10 numbers\n", + " 2. Define a simple lambda expression for eg that updates a number by 2\n", + " 3. Define an empty list\n", + " 4. Define the function -> use the lambda function to append the empty list\n", + " 5. Call the function with list and lambda expression\n", + " 6. print the updated list " + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "[70, 3, 84, 18, 47, 91, 54, 17, 92, 36]\n", + "[210, 9, 252, 54, 141, 273, 162, 51, 276, 108]\n" + ] + } + ], + "source": [ + "# your code here\n", + "import random\n", + "\n", + "list1 = random.sample(range(1, 101), 10)\n", + "print(list1)\n", + "multiply_by_3 = lambda x: x * 3\n", + "list2 = []\n", + "\n", + "def modify_list():\n", + " for item in list1:\n", + " result= multiply_by_3(item)\n", + " list2.append(result)\n", + "modify_list()\n", + "print(list2)\n", + "\n", + " \n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "#### Now we will define a lambda expression that will transform the elements of the list. \n", + "\n", + "In the cell below, create a lambda expression that converts Celsius to Kelvin. Recall that 0°C + 273.15 = 273.15K" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": {}, + "outputs": [], + "source": [ + "# Your code here:\n", + "\n", + "celsius_to_kelvin = lambda x : x + 273.15" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Finally, convert the list of temperatures below from Celsius to Kelvin." + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "[285.15, 296.15, 311.15, 218.14999999999998, 297.15]\n" + ] + } + ], + "source": [ + "temps = [12, 23, 38, -55, 24]\n", + "\n", + "# Your code here:\n", + "\n", + "temps_K = [celsius_to_kelvin(x) for x in temps]\n", + "print(temps_K)\n", + " \n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "#### In this part, we will define a function that returns a lambda expression\n", + "\n", + "In the cell below, write a lambda expression that takes two numbers and returns 1 if one is divisible by the other and zero otherwise. Call the lambda expression `mod`." + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": {}, + "outputs": [], + "source": [ + "# Your code here:\n", + "\n", + "mod = lambda x, y: 1 if (x % y == 0 or y % x == 0) else 0" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "#### Now create a function that returns mod. The function only takes one argument - the first number in the `mod` lambda function. \n", + "\n", + "Note: the lambda function above took two arguments, the lambda function in the return statement only takes one argument but also uses the argument passed to the function." + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": {}, + "outputs": [], + "source": [ + "def divisor(a):\n", + " \"\"\"\n", + " input: a number\n", + " output: a function that returns 1 if the number is divisible by another number (to be passed later) and zero otherwise\n", + " \"\"\"\n", + " # Your code here:\n", + " mod = lambda x: 1 if ( x % a == 0) else 0\n", + " return mod" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Finally, pass the number 5 to `divisor`. Now the function will check whether a number is divisble by 5. Assign this function to `divisible5`" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": {}, + "outputs": [], + "source": [ + "# Your code here:\n", + "\n", + "divisible5 = divisor(5)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Test your function with the following test cases:" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "1" + ] + }, + "execution_count": 9, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "divisible5(10)\n" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "0" + ] + }, + "execution_count": 10, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "divisible5(8)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Challenge 2 - Using Lambda Expressions in List Comprehensions\n", + "\n", + "In the following challenge, we will combine two lists using a lambda expression in a list comprehension. \n", + "\n", + "To do this, we will need to introduce the `zip` function. The `zip` function returns an iterator of tuples.\n", + "\n", + "The way zip function works with list has been shown below:" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "[('Green', 'eggs'),\n", + " ('cheese', 'cheese'),\n", + " ('English', 'cucumber'),\n", + " ('tomato', 'tomato')]" + ] + }, + "execution_count": 2, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "list3 = ['Green', 'cheese', 'English', 'tomato']\n", + "list4 = ['eggs', 'cheese', 'cucumber', 'tomato']\n", + "zipped = zip(list3,list4)\n", + "list(zipped)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "In this exercise we will try to compare the elements on the same index in the two lists. \n", + "We want to zip the two lists and then use a lambda expression to compare if:\n", + "list1 element > list2 element " + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "[(1, 2), (2, 3), (4, 3), (4, 5)]\n", + "[0, 0, 1, 0]\n" + ] + } + ], + "source": [ + "list5 = [1,2,4,4]\n", + "list6 = [2,3,3,5]\n", + "## Zip the lists together \n", + "zipped = list(zip(list5,list6))\n", + "\n", + "## Print the zipped list \n", + "print(zipped)\n", + "\n", + "## Use a lambda expression to compare if: list1 element > list2 element\n", + "comparison = lambda x1,x2 : 1 if x1>x2 else 0\n", + "\n", + "###compare and print result\n", + "\n", + "result = [comparison(x1,x2) for x1,x2 in zipped]\n", + "print(result)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Complete the parts of the code marked as \"###\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Challenge 3 - Using Lambda Expressions as Arguments\n", + "\n", + "#### In this challenge, we will zip together two lists and sort by the resulting tuple.\n", + "\n", + "In the cell below, take the two lists provided, zip them together and sort by the first letter of the second element of each tuple. Do this using a lambda function." + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "[('Engineering', 'Lab'), ('Computer Science', 'Homework'), ('Political Science', 'Essay'), ('Mathematics', 'Module')]\n", + "[('Political Science', 'Essay'), ('Computer Science', 'Homework'), ('Engineering', 'Lab'), ('Mathematics', 'Module')]\n" + ] + } + ], + "source": [ + "list7 = ['Engineering', 'Computer Science', 'Political Science', 'Mathematics']\n", + "list8 = ['Lab', 'Homework', 'Essay', 'Module']\n", + "\n", + "# Your code here:\n", + "\n", + "zipped = list(zip(list7,list8))\n", + "print(zipped)\n", + "\n", + "sorting = lambda zipped: zipped[1][0]\n", + "sorted_zipped_list = sorted(zipped, key = sorting)\n", + "print(sorted_zipped_list)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Bonus Challenge - Sort a Dictionary by Values\n", + "\n", + "Given the dictionary below, sort it by values rather than by keys. Use a lambda function to specify the values as a sorting key." + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "{'Toyota': 1995, 'Honda': 1997, 'Audi': 2001, 'BMW': 2005}\n" + ] + } + ], + "source": [ + "d = {'Honda': 1997, 'Toyota': 1995, 'Audi': 2001, 'BMW': 2005}\n", + "\n", + "# Your code here:\n", + "\n", + "sorting_by_values = lambda item:item[1]\n", + "sorted_d = dict(sorted(d.items(),key=sorting_by_values))\n", + "print(sorted_d)" + ] + } + ], + "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": "3.11.4" + } + }, + "nbformat": 4, + "nbformat_minor": 2 +}