diff --git a/lab_lambda_functions.ipynb b/lab_lambda_functions.ipynb new file mode 100644 index 0000000..fc94d75 --- /dev/null +++ b/lab_lambda_functions.ipynb @@ -0,0 +1,420 @@ +{ + "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": "raw", + "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": 8, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "[8, 3, 96, 7, 11, 19, 23, 14, 68, 18]\n" + ] + } + ], + "source": [ + "l = [6,1,94,5,9,17,21,12,66,16] \n", + "f = lambda x: x + 2 \n", + "b = []\n", + "def modify_list(lst, fudduLambda):\n", + " for x in lst:\n", + " b.append(fudduLambda(x))\n", + "modify_list(l, f)\n", + "print(b)" + ] + }, + { + "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": 10, + "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": 11, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "[285.15, 296.15, 311.15, 218.14999999999998, 297.15]" + ] + }, + "execution_count": 11, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "temps = [12, 23, 38, -55, 24]\n", + "\n", + "# Your code here:\n", + "[celsius_to_kelvin(e) for e in temps]" + ] + }, + { + "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": 12, + "metadata": {}, + "outputs": [], + "source": [ + "# Your code here:\n", + "mod = lambda a, b : 1 if a % b == 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": 15, + "metadata": {}, + "outputs": [], + "source": [ + "def divisor(b):\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", + "\n", + " # Your code here:\n", + " return lambda x:mod(x,b)\n", + " \n", + " \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": 16, + "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": 17, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "1" + ] + }, + "execution_count": 17, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "divisible5(10)" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "0" + ] + }, + "execution_count": 18, + "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": 19, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "[('Green', 'eggs'),\n", + " ('cheese', 'cheese'),\n", + " ('English', 'cucumber'),\n", + " ('tomato', 'tomato')]" + ] + }, + "execution_count": 19, + "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": 3, + "metadata": {}, + "outputs": [], + "source": [ + "list1 = [1,2,3,4]\n", + "list2 = [2,3,4,5]\n", + "## Zip the lists together \n", + "## Print the zipped list " + ] + }, + { + "cell_type": "raw", + "metadata": {}, + "source": [ + "Complete the parts of the code marked as \"###\"" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "False\n", + "False\n", + "False\n", + "False\n" + ] + } + ], + "source": [ + "'''\n", + "\n", + "compare = lambda ###: print(\"True\") if ### else print(\"False\")\n", + "for ### in zip(list1,list2):\n", + " compare(###)\n", + " \n", + "''' \n", + "\n", + "compare = lambda a,b: print(\"True\") if a > b else print(\"False\")\n", + "\n", + "for n,m in zip(list1,list2):\n", + " compare(n,m)" + ] + }, + { + "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": 24, + "metadata": {}, + "outputs": [], + "source": [ + "list1 = ['Engineering', 'Computer Science', 'Political Science', 'Mathematics']\n", + "list2 = ['Lab', 'Homework', 'Essay', 'Module']\n", + "\n", + "# Your code here:\n", + "import operator\n", + "lambda_sort = lambda l1, l2: sorted(zip(l1,l2), key=operator.itemgetter(1))" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "[('Political Science', 'Essay'), ('Computer Science', 'Homework'), ('Engineering', 'Lab'), ('Mathematics', 'Module')]\n" + ] + } + ], + "source": [ + "print(lambda_sort(list1, list2))" + ] + }, + { + "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": 23, + "metadata": {}, + "outputs": [], + "source": [ + "d = {'Honda': 1997, 'Toyota': 1995, 'Audi': 2001, 'BMW': 2005}\n", + "\n", + "# Your code here:\n", + "lambda_sort_dict = lambda l: { k:v for k,v in sorted(l.items(), key=operator.itemgetter(1)) }" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "{'Toyota': 1995, 'Honda': 1997, 'Audi': 2001, 'BMW': 2005}\n" + ] + } + ], + "source": [ + "print(lambda_sort_dict(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": "3.8.8" + } + }, + "nbformat": 4, + "nbformat_minor": 2 +}