diff --git a/your-code/.ipynb_checkpoints/main-checkpoint.ipynb b/your-code/.ipynb_checkpoints/main-checkpoint.ipynb new file mode 100644 index 0000000..dced226 --- /dev/null +++ b/your-code/.ipynb_checkpoints/main-checkpoint.ipynb @@ -0,0 +1,514 @@ +{ + "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": 1, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "[5, 3, 10, 7, 0, 1, 10, 2, 5, 4]" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "# your code here\n", + "\n", + "#1. Define a list of any 10 numbers\n", + "\n", + "import random\n", + "\n", + "list_numbers = [random.randint(0, 10) for number in range(10)]\n", + "list_numbers" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "7\n" + ] + } + ], + "source": [ + "#2. Define a simple lambda expression for eg that updates a number by 2\n", + "\n", + "add_two = lambda x: x+2\n", + "\n", + "print(add_two(5))" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "[7, 5, 12, 9, 2, 3, 12, 4, 7, 6]\n" + ] + } + ], + "source": [ + "#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\n", + "\n", + "random_list = []\n", + "\n", + "def update_list(lst, lambda_func):\n", + " return [lambda_func(x) for x in lst]\n", + "\n", + "random_list = update_list(list_numbers, add_two)\n", + "\n", + "print(random_list)" + ] + }, + { + "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", + "convert_temperature = 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": 7, + "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", + "kelvin = [convert_temperature(celsius) for celsius in temps]\n", + "print(kelvin)" + ] + }, + { + "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": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "1\n", + "0\n" + ] + } + ], + "source": [ + "# Your code here:\n", + "\n", + "mod = lambda number1, number2: 1 if (number1 % number2 == 0) or (number2 % number1 == 0) else 0\n", + "\n", + "# Use cases : \n", + "\n", + "result1 = mod(10, 2) \n", + "result2 = mod(15, 7) \n", + "\n", + "print(result1) \n", + "print(result2) " + ] + }, + { + "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": 22, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Enter a number: 3\n", + "Enter a number: 4\n", + "0\n" + ] + } + ], + "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", + " \n", + "# Create function to return a lambda function\n", + "def divisor(number1):\n", + " return lambda number2: 1 if (number1 % number2 == 0) or (number2 % number1 == 0) else 0\n", + "\n", + "# Create the lambda function using user_input_number_1\n", + "user_input_number_1 = float(input(\"Enter a number: \"))\n", + "\n", + "mod_function = divisor(user_input_number_1)\n", + "\n", + "# Input the second number\n", + "user_input_number_2 = float(input(\"Enter a number: \"))\n", + "\n", + "result = mod_function(user_input_number_2) # check if user_input_number_1 is divisible by user_input_number_2 using the lambda function\n", + "\n", + "print(result)" + ] + }, + { + "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": 15, + "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": 16, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "1" + ] + }, + "execution_count": 16, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "divisible5(10)" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "0" + ] + }, + "execution_count": 17, + "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": 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": 25, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "[False, False, True, False]" + ] + }, + "execution_count": 25, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "list1 = [1,2,4,4]\n", + "list2 = [2,3,3,5]\n", + "\n", + "## Zip the lists together \n", + "zipped = zip(list1,list2)\n", + "\n", + "## Print the zipped list \n", + "list(zipped)\n", + "\n", + "## Use a lambda expression to compare if: list1 element > list2 element\n", + "compare = lambda element: element[0] > element[1]\n", + "\n", + "compared_lists = [compare(element) for element in zip(list1, list2)]\n", + "compared_lists" + ] + }, + { + "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": 29, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "[('Political Science', 'Essay'), ('Computer Science', 'Homework'), ('Engineering', 'Lab'), ('Mathematics', 'Module')]\n" + ] + } + ], + "source": [ + "list1 = ['Engineering', 'Computer Science', 'Political Science', 'Mathematics']\n", + "list2 = ['Lab', 'Homework', 'Essay', 'Module']\n", + "\n", + "# Your code here:\n", + "zipped = zip(list1,list2)\n", + "\n", + "zipped_list = list(zipped)\n", + "\n", + "second_element = lambda element : element[1][0]\n", + "\n", + "sorted_list = sorted(zipped_list, key=second_element)\n", + "\n", + "print(sorted_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": 35, + "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", + "value = lambda item : item[1]\n", + "\n", + "sorted_dic = sorted(d.items(), key=value)\n", + "\n", + "print(dict(sorted_dic))" + ] + }, + { + "cell_type": "code", + "execution_count": 33, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "{'Toyota': 1995, 'Honda': 1997, 'Audi': 2001, 'BMW': 2005}\n" + ] + } + ], + "source": [ + "#Other way to write :\n", + "sorted_dict = dict(sorted(d.items(), key=lambda item: item[1]))\n", + "\n", + "print(sorted_dict)" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 3 (ipykernel)", + "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": "3.10.9" + } + }, + "nbformat": 4, + "nbformat_minor": 2 +} diff --git a/your-code/main.ipynb b/your-code/main.ipynb index 38b51a7..dced226 100644 --- a/your-code/main.ipynb +++ b/your-code/main.ipynb @@ -1,304 +1,514 @@ -{ - "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": 1, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "[5, 3, 10, 7, 0, 1, 10, 2, 5, 4]" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "# your code here\n", + "\n", + "#1. Define a list of any 10 numbers\n", + "\n", + "import random\n", + "\n", + "list_numbers = [random.randint(0, 10) for number in range(10)]\n", + "list_numbers" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "7\n" + ] + } + ], + "source": [ + "#2. Define a simple lambda expression for eg that updates a number by 2\n", + "\n", + "add_two = lambda x: x+2\n", + "\n", + "print(add_two(5))" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "[7, 5, 12, 9, 2, 3, 12, 4, 7, 6]\n" + ] + } + ], + "source": [ + "#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\n", + "\n", + "random_list = []\n", + "\n", + "def update_list(lst, lambda_func):\n", + " return [lambda_func(x) for x in lst]\n", + "\n", + "random_list = update_list(list_numbers, add_two)\n", + "\n", + "print(random_list)" + ] + }, + { + "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", + "convert_temperature = 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": 7, + "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", + "kelvin = [convert_temperature(celsius) for celsius in temps]\n", + "print(kelvin)" + ] + }, + { + "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": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "1\n", + "0\n" + ] + } + ], + "source": [ + "# Your code here:\n", + "\n", + "mod = lambda number1, number2: 1 if (number1 % number2 == 0) or (number2 % number1 == 0) else 0\n", + "\n", + "# Use cases : \n", + "\n", + "result1 = mod(10, 2) \n", + "result2 = mod(15, 7) \n", + "\n", + "print(result1) \n", + "print(result2) " + ] + }, + { + "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": 22, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Enter a number: 3\n", + "Enter a number: 4\n", + "0\n" + ] + } + ], + "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", + " \n", + "# Create function to return a lambda function\n", + "def divisor(number1):\n", + " return lambda number2: 1 if (number1 % number2 == 0) or (number2 % number1 == 0) else 0\n", + "\n", + "# Create the lambda function using user_input_number_1\n", + "user_input_number_1 = float(input(\"Enter a number: \"))\n", + "\n", + "mod_function = divisor(user_input_number_1)\n", + "\n", + "# Input the second number\n", + "user_input_number_2 = float(input(\"Enter a number: \"))\n", + "\n", + "result = mod_function(user_input_number_2) # check if user_input_number_1 is divisible by user_input_number_2 using the lambda function\n", + "\n", + "print(result)" + ] + }, + { + "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": 15, + "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": 16, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "1" + ] + }, + "execution_count": 16, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "divisible5(10)" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "0" + ] + }, + "execution_count": 17, + "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": 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": 25, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "[False, False, True, False]" + ] + }, + "execution_count": 25, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "list1 = [1,2,4,4]\n", + "list2 = [2,3,3,5]\n", + "\n", + "## Zip the lists together \n", + "zipped = zip(list1,list2)\n", + "\n", + "## Print the zipped list \n", + "list(zipped)\n", + "\n", + "## Use a lambda expression to compare if: list1 element > list2 element\n", + "compare = lambda element: element[0] > element[1]\n", + "\n", + "compared_lists = [compare(element) for element in zip(list1, list2)]\n", + "compared_lists" + ] + }, + { + "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": 29, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "[('Political Science', 'Essay'), ('Computer Science', 'Homework'), ('Engineering', 'Lab'), ('Mathematics', 'Module')]\n" + ] + } + ], + "source": [ + "list1 = ['Engineering', 'Computer Science', 'Political Science', 'Mathematics']\n", + "list2 = ['Lab', 'Homework', 'Essay', 'Module']\n", + "\n", + "# Your code here:\n", + "zipped = zip(list1,list2)\n", + "\n", + "zipped_list = list(zipped)\n", + "\n", + "second_element = lambda element : element[1][0]\n", + "\n", + "sorted_list = sorted(zipped_list, key=second_element)\n", + "\n", + "print(sorted_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": 35, + "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", + "value = lambda item : item[1]\n", + "\n", + "sorted_dic = sorted(d.items(), key=value)\n", + "\n", + "print(dict(sorted_dic))" + ] + }, + { + "cell_type": "code", + "execution_count": 33, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "{'Toyota': 1995, 'Honda': 1997, 'Audi': 2001, 'BMW': 2005}\n" + ] + } + ], + "source": [ + "#Other way to write :\n", + "sorted_dict = dict(sorted(d.items(), key=lambda item: item[1]))\n", + "\n", + "print(sorted_dict)" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 3 (ipykernel)", + "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": "3.10.9" + } + }, + "nbformat": 4, + "nbformat_minor": 2 +}