diff --git a/your-code/lambdamain.ipynb b/your-code/lambdamain.ipynb new file mode 100644 index 0000000..cd307ac --- /dev/null +++ b/your-code/lambdamain.ipynb @@ -0,0 +1,375 @@ +{ + "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 - Passing a Lambda Expression to a Function\n", + "\n", + "In the next excercise you will create a function that takes a lambda expression as an argument. 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": "code", + "execution_count": 4, + "metadata": {}, + "outputs": [], + "source": [ + "def modify_list(lst, lmbda):\n", + " return list(map(lmbda,lst))\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": 6, + "metadata": {}, + "outputs": [], + "source": [ + "# your code here\n", + "c_to_k=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": 8, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "[285.15, 296.15, 311.15, 218.14999999999998, 297.15]" + ] + }, + "execution_count": 8, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "temps = [12, 23, 38, -55, 24]\n", + "\n", + "# your code here\n", + "temps = [12, 23, 38, -55, 24]\n", + "\n", + "# your code here\n", + "modify_list(lst = temps,lmbda = c_to_k)" + ] + }, + { + "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": 9, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "0" + ] + }, + "execution_count": 9, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "# your code here\n", + "mod = lambda x, y : 1 if x % y == 0 else 0\n", + "\n", + "mod(9,5)" + ] + }, + { + "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": 11, + "metadata": {}, + "outputs": [], + "source": [ + "\n", + " \"\"\"\n", + " Input: a number\n", + " Output: a function that returns 1 if the number is \n", + " divisible by another number (to be passed later) and zero otherwise.\n", + " \"\"\"\n", + " \n", + " # your code here\n", + " \n", + " def divisor(b):\n", + " return lambda a: mod(a, b)" + ] + }, + { + "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": 12, + "metadata": {}, + "outputs": [], + "source": [ + "# your code here\n", + "divisible5 = divisor(5)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Test your function with the following test cases:" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "1" + ] + }, + "execution_count": 13, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "divisible5(10)" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "0" + ] + }, + "execution_count": 14, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "divisible5(8)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Bonus Challenge - 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." + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "[(1,), (2,), (3,), (4,), (5,)]" + ] + }, + "execution_count": 10, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "# Here is an example of passing one list to the zip function. \n", + "# Since the zip function returns an iterator, we need to evaluate the iterator by using a list comprehension.\n", + "\n", + "l = [1,2,3,4,5]\n", + "[x for x in zip(l)]" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Using the `zip` function, let's iterate through two lists and add the elements by position." + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "['Green eggs', 'cheese cheese', 'English cucumber', 'tomato tomato']" + ] + }, + "execution_count": 15, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "list1 = ['Green', 'cheese', 'English', 'tomato']\n", + "list2 = ['eggs', 'cheese', 'cucumber', 'tomato']\n", + "\n", + "# your code here\n", + "[e[0]+ \" \"+ e[1] for e in zip(list1,list2)]" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Bonus Challenge - 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": 16, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "['Mathematics Module',\n", + " 'Engineering Lab',\n", + " 'Computer Science Homework',\n", + " 'Political Science Essay']" + ] + }, + "execution_count": 16, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "list1 = ['Engineering', 'Computer Science', 'Political Science', 'Mathematics']\n", + "list2 = ['Lab', 'Homework', 'Essay', 'Module']\n", + "\n", + "# your code here\n", + "sorted([e[0]+\" \"+e[1] for e in zip(list1,list2)],key= lambda e: e[1])" + ] + }, + { + "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": 17, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "{'Toyota': 1995, 'Honda': 1997, 'Audi': 2001, 'BMW': 2005}" + ] + }, + "execution_count": 17, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "d = {'Honda': 1997, 'Toyota': 1995, 'Audi': 2001, 'BMW': 2005}\n", + "\n", + "# your code here\n", + "dict(sorted(d.items(),key=lambda e: e[1]))" + ] + }, + { + "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": "3.8.8" + } + }, + "nbformat": 4, + "nbformat_minor": 2 +}