From 808cb9911d1825ec5840077e92ac0cee8fc1dffa Mon Sep 17 00:00:00 2001 From: Salvador Carrasco Date: Mon, 8 Nov 2021 12:54:12 -0600 Subject: [PATCH] Laboratorio modificado por Salvador --- .../.ipynb_checkpoints/main-checkpoint.ipynb | 520 ++++++++++++++++++ your-code/main.ipynb | 258 +++++++-- 2 files changed, 741 insertions(+), 37 deletions(-) create mode 100644 your-code/.ipynb_checkpoints/main-checkpoint.ipynb diff --git a/your-code/.ipynb_checkpoints/main-checkpoint.ipynb b/your-code/.ipynb_checkpoints/main-checkpoint.ipynb new file mode 100644 index 0000000..df5b7ce --- /dev/null +++ b/your-code/.ipynb_checkpoints/main-checkpoint.ipynb @@ -0,0 +1,520 @@ +{ + "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": 9, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "[3, 4, 5, 6, 7, 8, 9, 10, 11, 12]" + ] + }, + "execution_count": 9, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "l = [1,2,3,4,5,6,7,8,9,10]\n", + "f = lambda x: x+2\n", + "b = []\n", + "def modify_list(lst, fudduLambda,lst_empty):\n", + " for i in lst: \n", + " b.append(fudduLambda(i))\n", + " return b\n", + "modify_list(l,f,b)\n", + " \n", + " #b.append(#####(x))\n", + "#Call modify_list(##,##)\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": [ + { + "data": { + "text/plain": [ + "283.15" + ] + }, + "execution_count": 10, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "# Your code here:\n", + "\n", + "celsius_to_kelvin= lambda x:x+273.15\n", + "celsius_to_kelvin(10)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Finally, convert the list of temperatures below from Celsius to Kelvin." + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "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", + "temps_kelvin=[]\n", + "\n", + "# Your code here:\n", + "\n", + "for i in temps:\n", + " temps_kelvin.append(celsius_to_kelvin(i))\n", + "print(temps_kelvin)\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": 24, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "'1'" + ] + }, + "execution_count": 24, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "# Your code here:\n", + "mod=lambda x,y:1 if x%y==0 else 0\n", + "mod(20,5)\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": 26, + "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:1 if x%b==0 else 0" + ] + }, + { + "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": 29, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "1" + ] + }, + "execution_count": 29, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "# Your code here:\n", + "divisible5=divisor(5)\n", + "divisible5(20)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Test your function with the following test cases:" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "1" + ] + }, + "execution_count": 30, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "divisible5(10)" + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "0" + ] + }, + "execution_count": 31, + "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": 32, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "[('Green', 'eggs'),\n", + " ('cheese', 'cheese'),\n", + " ('English', 'cucumber'),\n", + " ('tomato', 'tomato')]" + ] + }, + "execution_count": 32, + "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": 35, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "[(1, 2), (2, 3), (3, 4), (4, 5)]\n" + ] + } + ], + "source": [ + "list1 = [1,2,3,4]\n", + "list2 = [2,3,4,5]\n", + "zipped_lists=zip(list1,list2)\n", + "print(list(zipped_lists))\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": 4, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "'\\n\\ncompare = lambda ###: print(\"True\") if ### else print(\"False\")\\nfor ### in zip(list1,list2):\\n compare(###)\\n \\n'" + ] + }, + "execution_count": 4, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "'''\n", + "\n", + "compare = lambda ###: print(\"True\") if ### else print(\"False\")\n", + "for ### in zip(list1,list2):\n", + " compare(###)\n", + " \n", + "''' " + ] + }, + { + "cell_type": "code", + "execution_count": 37, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "False\n", + "False\n", + "False\n", + "False\n" + ] + } + ], + "source": [ + "list1 = [1,2,3,4]\n", + "list2 = [2,3,4,5]\n", + "\n", + "compare=lambda x,y:print(\"True\") if x > y else print(\"False\")\n", + "\n", + "for x, y in zip(list1,list2):\n", + " compare(x,y)" + ] + }, + { + "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": 41, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "[('Essay', 'Political Science'),\n", + " ('Homework', 'Computer Science'),\n", + " ('Lab', 'Engineering'),\n", + " ('Module', 'Mathematics')]" + ] + }, + "execution_count": 41, + "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", + "\n", + "acomodar=lambda l1, l2: sorted(list(zip(list2,list1)))\n", + "acomodar(list2,list1)" + ] + }, + { + "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": 53, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "[('Honda', 1997), ('Toyota', 1995), ('Audi', 2001), ('BMW', 2005)]\n", + "[('Toyota', 1995), ('Honda', 1997), ('Audi', 2001), ('BMW', 2005)]\n", + "{'Toyota': 1995, 'Honda': 1997, 'Audi': 2001, 'BMW': 2005}\n" + ] + } + ], + "source": [ + "d = {'Honda': 1997, 'Toyota': 1995, 'Audi': 2001, 'BMW': 2005}\n", + "d_nuevo={}\n", + "# Your code here:\n", + "#año=list(d.values())\n", + "#marca=list(d.keys())\n", + "#for key in diccionario:\n", + "# print key, \":\", diccionario[key]\n", + "\n", + "año_marca=list(d.items())\n", + "print(año_marca)\n", + "tuplas_organizadas=sorted(año_marca,key=lambda x: x[1])\n", + "print(tuplas_organizadas)\n", + "\n", + "for m,a in tuplas_organizadas:\n", + " d_nuevo[m]=a\n", + "print(d_nuevo)" + ] + }, + { + "cell_type": "code", + "execution_count": 54, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "{'Toyota': 1995, 'Honda': 1997, 'Audi': 2001, 'BMW': 2005}\n" + ] + } + ], + "source": [ + "tuplas_organizadas={key:value for key, value in sorted(año_marca,key=lambda x: x[1])}\n", + "print(tuplas_organizadas)" + ] + }, + { + "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": 4 +} diff --git a/your-code/main.ipynb b/your-code/main.ipynb index 66a9984..df5b7ce 100644 --- a/your-code/main.ipynb +++ b/your-code/main.ipynb @@ -34,16 +34,31 @@ }, { "cell_type": "code", - "execution_count": null, + "execution_count": 9, "metadata": {}, - "outputs": [], + "outputs": [ + { + "data": { + "text/plain": [ + "[3, 4, 5, 6, 7, 8, 9, 10, 11, 12]" + ] + }, + "execution_count": 9, + "metadata": {}, + "output_type": "execute_result" + } + ], "source": [ - "l = [######]\n", - "f = lambda x: #define the lambda expression\n", + "l = [1,2,3,4,5,6,7,8,9,10]\n", + "f = lambda x: x+2\n", "b = []\n", - "def modify_list(lst, fudduLambda):\n", - " for x in ####:\n", - " b.append(#####(x))\n", + "def modify_list(lst, fudduLambda,lst_empty):\n", + " for i in lst: \n", + " b.append(fudduLambda(i))\n", + " return b\n", + "modify_list(l,f,b)\n", + " \n", + " #b.append(#####(x))\n", "#Call modify_list(##,##)\n", "#print b" ] @@ -59,12 +74,25 @@ }, { "cell_type": "code", - "execution_count": 3, + "execution_count": 10, "metadata": {}, - "outputs": [], + "outputs": [ + { + "data": { + "text/plain": [ + "283.15" + ] + }, + "execution_count": 10, + "metadata": {}, + "output_type": "execute_result" + } + ], "source": [ "# Your code here:\n", - "\n" + "\n", + "celsius_to_kelvin= lambda x:x+273.15\n", + "celsius_to_kelvin(10)\n" ] }, { @@ -76,13 +104,26 @@ }, { "cell_type": "code", - "execution_count": 4, + "execution_count": 21, "metadata": {}, - "outputs": [], + "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", + "temps_kelvin=[]\n", + "\n", + "# Your code here:\n", "\n", - "# Your code here:" + "for i in temps:\n", + " temps_kelvin.append(celsius_to_kelvin(i))\n", + "print(temps_kelvin)\n" ] }, { @@ -96,11 +137,24 @@ }, { "cell_type": "code", - "execution_count": 5, + "execution_count": 24, "metadata": {}, - "outputs": [], + "outputs": [ + { + "data": { + "text/plain": [ + "'1'" + ] + }, + "execution_count": 24, + "metadata": {}, + "output_type": "execute_result" + } + ], "source": [ - "# Your code here:\n" + "# Your code here:\n", + "mod=lambda x,y:1 if x%y==0 else 0\n", + "mod(20,5)\n" ] }, { @@ -114,7 +168,7 @@ }, { "cell_type": "code", - "execution_count": 6, + "execution_count": 26, "metadata": {}, "outputs": [], "source": [ @@ -124,7 +178,8 @@ " 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:" + " # Your code here:\n", + " return lambda x:1 if x%b==0 else 0" ] }, { @@ -136,11 +191,24 @@ }, { "cell_type": "code", - "execution_count": 7, + "execution_count": 29, "metadata": {}, - "outputs": [], + "outputs": [ + { + "data": { + "text/plain": [ + "1" + ] + }, + "execution_count": 29, + "metadata": {}, + "output_type": "execute_result" + } + ], "source": [ - "# Your code here:\n" + "# Your code here:\n", + "divisible5=divisor(5)\n", + "divisible5(20)\n" ] }, { @@ -152,18 +220,40 @@ }, { "cell_type": "code", - "execution_count": null, + "execution_count": 30, "metadata": {}, - "outputs": [], + "outputs": [ + { + "data": { + "text/plain": [ + "1" + ] + }, + "execution_count": 30, + "metadata": {}, + "output_type": "execute_result" + } + ], "source": [ "divisible5(10)" ] }, { "cell_type": "code", - "execution_count": null, + "execution_count": 31, "metadata": {}, - "outputs": [], + "outputs": [ + { + "data": { + "text/plain": [ + "0" + ] + }, + "execution_count": 31, + "metadata": {}, + "output_type": "execute_result" + } + ], "source": [ "divisible5(8)" ] @@ -183,7 +273,7 @@ }, { "cell_type": "code", - "execution_count": 1, + "execution_count": 32, "metadata": {}, "outputs": [ { @@ -195,7 +285,7 @@ " ('tomato', 'tomato')]" ] }, - "execution_count": 1, + "execution_count": 32, "metadata": {}, "output_type": "execute_result" } @@ -218,12 +308,22 @@ }, { "cell_type": "code", - "execution_count": 3, + "execution_count": 35, "metadata": {}, - "outputs": [], + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "[(1, 2), (2, 3), (3, 4), (4, 5)]\n" + ] + } + ], "source": [ "list1 = [1,2,3,4]\n", "list2 = [2,3,4,5]\n", + "zipped_lists=zip(list1,list2)\n", + "print(list(zipped_lists))\n", "## Zip the lists together \n", "## Print the zipped list " ] @@ -261,6 +361,32 @@ "''' " ] }, + { + "cell_type": "code", + "execution_count": 37, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "False\n", + "False\n", + "False\n", + "False\n" + ] + } + ], + "source": [ + "list1 = [1,2,3,4]\n", + "list2 = [2,3,4,5]\n", + "\n", + "compare=lambda x,y:print(\"True\") if x > y else print(\"False\")\n", + "\n", + "for x, y in zip(list1,list2):\n", + " compare(x,y)" + ] + }, { "cell_type": "markdown", "metadata": {}, @@ -274,14 +400,31 @@ }, { "cell_type": "code", - "execution_count": 12, + "execution_count": 41, "metadata": {}, - "outputs": [], + "outputs": [ + { + "data": { + "text/plain": [ + "[('Essay', 'Political Science'),\n", + " ('Homework', 'Computer Science'),\n", + " ('Lab', 'Engineering'),\n", + " ('Module', 'Mathematics')]" + ] + }, + "execution_count": 41, + "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" + "# Your code here:\n", + "\n", + "acomodar=lambda l1, l2: sorted(list(zip(list2,list1)))\n", + "acomodar(list2,list1)" ] }, { @@ -295,13 +438,54 @@ }, { "cell_type": "code", - "execution_count": 13, + "execution_count": 53, "metadata": {}, - "outputs": [], + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "[('Honda', 1997), ('Toyota', 1995), ('Audi', 2001), ('BMW', 2005)]\n", + "[('Toyota', 1995), ('Honda', 1997), ('Audi', 2001), ('BMW', 2005)]\n", + "{'Toyota': 1995, 'Honda': 1997, 'Audi': 2001, 'BMW': 2005}\n" + ] + } + ], "source": [ "d = {'Honda': 1997, 'Toyota': 1995, 'Audi': 2001, 'BMW': 2005}\n", + "d_nuevo={}\n", + "# Your code here:\n", + "#año=list(d.values())\n", + "#marca=list(d.keys())\n", + "#for key in diccionario:\n", + "# print key, \":\", diccionario[key]\n", + "\n", + "año_marca=list(d.items())\n", + "print(año_marca)\n", + "tuplas_organizadas=sorted(año_marca,key=lambda x: x[1])\n", + "print(tuplas_organizadas)\n", "\n", - "# Your code here:" + "for m,a in tuplas_organizadas:\n", + " d_nuevo[m]=a\n", + "print(d_nuevo)" + ] + }, + { + "cell_type": "code", + "execution_count": 54, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "{'Toyota': 1995, 'Honda': 1997, 'Audi': 2001, 'BMW': 2005}\n" + ] + } + ], + "source": [ + "tuplas_organizadas={key:value for key, value in sorted(año_marca,key=lambda x: x[1])}\n", + "print(tuplas_organizadas)" ] }, { @@ -328,9 +512,9 @@ "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", - "version": "3.7.3" + "version": "3.8.8" } }, "nbformat": 4, - "nbformat_minor": 2 + "nbformat_minor": 4 }