From 74142a21180418525565d5963410f7067293f42a Mon Sep 17 00:00:00 2001 From: Max Guzman Date: Wed, 3 Nov 2021 11:05:58 -0600 Subject: [PATCH] =?UTF-8?q?Solving=20lab-lambda-functions[Max=20Guzm=C3=A1?= =?UTF-8?q?n]?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../.ipynb_checkpoints/main-checkpoint.ipynb | 536 ++++++++++++++++++ your-code/main.ipynb | 322 +++++++++-- 2 files changed, 797 insertions(+), 61 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..d416bbd --- /dev/null +++ b/your-code/.ipynb_checkpoints/main-checkpoint.ipynb @@ -0,0 +1,536 @@ +{ + "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": 6, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "[range(0, 10)]" + ] + }, + "execution_count": 6, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "[range(10)]" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "[0, 2, 4, 6, 8, 10, 12, 14, 16, 18]\n" + ] + } + ], + "source": [ + "l = [i for i in range(10)]\n", + "f = lambda x: x*2 #define the lambda expression\n", + "b = []\n", + "\n", + "def modify_list(lst, fudduLambda):\n", + " for x in lst:\n", + " b.append(fudduLambda(x))\n", + " \n", + "\n", + "modify_list(l,f)\n", + "\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": 9, + "metadata": {}, + "outputs": [], + "source": [ + "# Your code here:\n", + "\n", + "f = lambda C: C + 273.15\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Finally, convert the list of temperatures below from Celsius to Kelvin." + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "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", + "kelvins = []\n", + "\n", + "for C in temps:\n", + " kelvins.append(f(C))\n", + "print(kelvins)" + ] + }, + { + "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": 34, + "metadata": {}, + "outputs": [], + "source": [ + "# Your code here:\n", + "mod = lambda x, y: 1 if x%y == 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": 64, + "metadata": {}, + "outputs": [], + "source": [ + "def divisor(b):\n", + " \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", + " return lambda x: 1 if x%b == 0 else 0\n", + "\n", + "# divisible5 = lambda x: i if x%5 == 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": 67, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "1" + ] + }, + "execution_count": 67, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "# Your code here:\n", + "divisible5 = divisor(5)\n", + "# divisible10 = divisor(10)\n", + "divisible5(20)\n", + "# divisible10(3)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Test your function with the following test cases:" + ] + }, + { + "cell_type": "code", + "execution_count": 68, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "1" + ] + }, + "execution_count": 68, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "divisible5(10)" + ] + }, + { + "cell_type": "code", + "execution_count": 69, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "0" + ] + }, + "execution_count": 69, + "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": 70, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "[('Green', 'eggs'),\n", + " ('cheese', 'cheese'),\n", + " ('English', 'cucumber'),\n", + " ('tomato', 'tomato')]" + ] + }, + "execution_count": 70, + "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": 71, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "[(1, 2), (2, 3), (3, 4), (4, 5), (7, 6)]\n" + ] + } + ], + "source": [ + "list1 = [1,2,3,4,7]\n", + "list2 = [2,3,4,5,6]\n", + "\n", + "## Zip the lists together \n", + "zipped = zip(list1,list2)\n", + "\n", + "\n", + "## Print the zipped list \n", + "print(list(zipped))" + ] + }, + { + "cell_type": "raw", + "metadata": {}, + "source": [ + "Complete the parts of the code marked as \"###\"" + ] + }, + { + "cell_type": "code", + "execution_count": 68, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "False\n", + "False\n", + "False\n", + "False\n", + "True\n" + ] + } + ], + "source": [ + "list1 = [1,2,3,4,7]\n", + "list2 = [2,3,4,5,6]\n", + "\n", + "compare = lambda x, y: print(\"True\") if x > y else print(\"False\") # Ver si el primer numero es mayor al segundo\n", + "\n", + "for x, y in zip(list1,list2): #sacar numeros de las listas\n", + " compare(x,y) # poner numeros en la func. lambda " + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": {}, + "outputs": [], + "source": [ + " " + ] + }, + { + "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": 91, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "[('Political Science', 'Essay'),\n", + " ('Computer Science', 'Homework'),\n", + " ('Engineering', 'Lab'),\n", + " ('Mathematics', 'Module')]" + ] + }, + "execution_count": 91, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "list1 = ['Engineering', 'Computer Science', 'Political Science', 'Mathematics']\n", + "list2 = ['Lab', 'Homework', 'Essay', 'Module']\n", + "# print(list(zip(list1,list2)))\n", + "# Your code here:\n", + "nueva_lista = []\n", + "acomodar = lambda l1, l2: sorted(list(zip(list2,list1)))\n", + "for tup in acomodar(list1,list2): #('Essay', 'Political Science') ('Homework', 'Computer Science')\n", + " nueva_lista.append((tup[1],tup[0]))\n", + "nueva_lista " + ] + }, + { + "cell_type": "code", + "execution_count": 90, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "[('Essay', 'Political Science'),\n", + " ('Homework', 'Computer Science'),\n", + " ('Lab', 'Engineering'),\n", + " ('Module', 'Mathematics')]" + ] + }, + "execution_count": 90, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "acomodar(list1,list2)" + ] + }, + { + "cell_type": "code", + "execution_count": 89, + "metadata": { + "collapsed": true + }, + "outputs": [ + { + "data": { + "text/plain": [ + "[('Engineering', 'Lab'),\n", + " ('Computer Science', 'Homework'),\n", + " ('Political Science', 'Essay'),\n", + " ('Mathematics', 'Module')]" + ] + }, + "execution_count": 89, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "list(zip(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": 135, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Estaba así [('Honda', 1997), ('Toyota', 1995), ('Audi', 2001), ('BMW', 2005)]\n", + "Organizado [('Toyota', 1995), ('Honda', 1997), ('Audi', 2001), ('BMW', 2005)]\n", + "En diccionario: {'Toyota': 1995, 'Honda': 1997, 'Audi': 2001, 'BMW': 2005}\n" + ] + } + ], + "source": [ + "d = {'Honda': 1997, 'Toyota': 1995, 'Audi': 2001, 'BMW': 2005}\n", + "d_nuevo = {} #{1997: 'Honda', 1995: 'Toyota', 2001: 'Audi', 2005: 'BMW'}\n", + "\n", + "# Your code here:\n", + "# diccionario = lambda marca, año: d_alreves[año]=marca\n", + "# sorted(d)\n", + "\n", + "año = list(d.values())\n", + "marca = list(d.keys())\n", + "año_marca = list(d.items())\n", + "print('Estaba así',año_marca)\n", + "\n", + "tuplas_organizadas = sorted(año_marca, key=lambda x: x[1])\n", + "print('Organizado: ',tuplas_organizadas)\n", + "\n", + "#pasar a diccionario marca: año\n", + "for m, a in tuplas_organizadas:\n", + " d_nuevo[m] = a\n", + "# print(d[m], \":\", m)\n", + "print('En diccionario:',d_nuevo)\n", + "# tuplas_organizadas = {key:value for key, value in sorted(año_marca, key=lambda x: x[1])}\n", + "# print(tuplas_organizadas)\n" + ] + } + ], + "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 +} diff --git a/your-code/main.ipynb b/your-code/main.ipynb index 66a9984..d416bbd 100644 --- a/your-code/main.ipynb +++ b/your-code/main.ipynb @@ -34,18 +34,50 @@ }, { "cell_type": "code", - "execution_count": null, + "execution_count": 6, "metadata": {}, - "outputs": [], + "outputs": [ + { + "data": { + "text/plain": [ + "[range(0, 10)]" + ] + }, + "execution_count": 6, + "metadata": {}, + "output_type": "execute_result" + } + ], "source": [ - "l = [######]\n", - "f = lambda x: #define the lambda expression\n", + "[range(10)]" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "[0, 2, 4, 6, 8, 10, 12, 14, 16, 18]\n" + ] + } + ], + "source": [ + "l = [i for i in range(10)]\n", + "f = lambda x: x*2 #define the lambda expression\n", "b = []\n", + "\n", "def modify_list(lst, fudduLambda):\n", - " for x in ####:\n", - " b.append(#####(x))\n", - "#Call modify_list(##,##)\n", - "#print b" + " for x in lst:\n", + " b.append(fudduLambda(x))\n", + " \n", + "\n", + "modify_list(l,f)\n", + "\n", + "print(b)" ] }, { @@ -59,12 +91,13 @@ }, { "cell_type": "code", - "execution_count": 3, + "execution_count": 9, "metadata": {}, "outputs": [], "source": [ "# Your code here:\n", - "\n" + "\n", + "f = lambda C: C + 273.15\n" ] }, { @@ -76,13 +109,26 @@ }, { "cell_type": "code", - "execution_count": 4, + "execution_count": 13, "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", "\n", - "# Your code here:" + "# Your code here:\n", + "kelvins = []\n", + "\n", + "for C in temps:\n", + " kelvins.append(f(C))\n", + "print(kelvins)" ] }, { @@ -96,11 +142,12 @@ }, { "cell_type": "code", - "execution_count": 5, + "execution_count": 34, "metadata": {}, "outputs": [], "source": [ - "# Your code here:\n" + "# Your code here:\n", + "mod = lambda x, y: 1 if x%y == 0 else 0" ] }, { @@ -114,17 +161,21 @@ }, { "cell_type": "code", - "execution_count": 6, + "execution_count": 64, "metadata": {}, "outputs": [], "source": [ "def divisor(b):\n", + " \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", - " # Your code here:" + " return lambda x: 1 if x%b == 0 else 0\n", + "\n", + "# divisible5 = lambda x: i if x%5 == 0 else 0" ] }, { @@ -136,11 +187,26 @@ }, { "cell_type": "code", - "execution_count": 7, + "execution_count": 67, "metadata": {}, - "outputs": [], + "outputs": [ + { + "data": { + "text/plain": [ + "1" + ] + }, + "execution_count": 67, + "metadata": {}, + "output_type": "execute_result" + } + ], "source": [ - "# Your code here:\n" + "# Your code here:\n", + "divisible5 = divisor(5)\n", + "# divisible10 = divisor(10)\n", + "divisible5(20)\n", + "# divisible10(3)" ] }, { @@ -152,18 +218,40 @@ }, { "cell_type": "code", - "execution_count": null, + "execution_count": 68, "metadata": {}, - "outputs": [], + "outputs": [ + { + "data": { + "text/plain": [ + "1" + ] + }, + "execution_count": 68, + "metadata": {}, + "output_type": "execute_result" + } + ], "source": [ "divisible5(10)" ] }, { "cell_type": "code", - "execution_count": null, + "execution_count": 69, "metadata": {}, - "outputs": [], + "outputs": [ + { + "data": { + "text/plain": [ + "0" + ] + }, + "execution_count": 69, + "metadata": {}, + "output_type": "execute_result" + } + ], "source": [ "divisible5(8)" ] @@ -183,7 +271,7 @@ }, { "cell_type": "code", - "execution_count": 1, + "execution_count": 70, "metadata": {}, "outputs": [ { @@ -195,7 +283,7 @@ " ('tomato', 'tomato')]" ] }, - "execution_count": 1, + "execution_count": 70, "metadata": {}, "output_type": "execute_result" } @@ -218,14 +306,27 @@ }, { "cell_type": "code", - "execution_count": 3, + "execution_count": 71, "metadata": {}, - "outputs": [], + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "[(1, 2), (2, 3), (3, 4), (4, 5), (7, 6)]\n" + ] + } + ], "source": [ - "list1 = [1,2,3,4]\n", - "list2 = [2,3,4,5]\n", + "list1 = [1,2,3,4,7]\n", + "list2 = [2,3,4,5,6]\n", + "\n", "## Zip the lists together \n", - "## Print the zipped list " + "zipped = zip(list1,list2)\n", + "\n", + "\n", + "## Print the zipped list \n", + "print(list(zipped))" ] }, { @@ -237,28 +338,38 @@ }, { "cell_type": "code", - "execution_count": 4, + "execution_count": 68, "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" + "name": "stdout", + "output_type": "stream", + "text": [ + "False\n", + "False\n", + "False\n", + "False\n", + "True\n" + ] } ], "source": [ - "'''\n", + "list1 = [1,2,3,4,7]\n", + "list2 = [2,3,4,5,6]\n", "\n", - "compare = lambda ###: print(\"True\") if ### else print(\"False\")\n", - "for ### in zip(list1,list2):\n", - " compare(###)\n", - " \n", - "''' " + "compare = lambda x, y: print(\"True\") if x > y else print(\"False\") # Ver si el primer numero es mayor al segundo\n", + "\n", + "for x, y in zip(list1,list2): #sacar numeros de las listas\n", + " compare(x,y) # poner numeros en la func. lambda " + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": {}, + "outputs": [], + "source": [ + " " ] }, { @@ -274,14 +385,81 @@ }, { "cell_type": "code", - "execution_count": 12, + "execution_count": 91, "metadata": {}, - "outputs": [], + "outputs": [ + { + "data": { + "text/plain": [ + "[('Political Science', 'Essay'),\n", + " ('Computer Science', 'Homework'),\n", + " ('Engineering', 'Lab'),\n", + " ('Mathematics', 'Module')]" + ] + }, + "execution_count": 91, + "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" + "# print(list(zip(list1,list2)))\n", + "# Your code here:\n", + "nueva_lista = []\n", + "acomodar = lambda l1, l2: sorted(list(zip(list2,list1)))\n", + "for tup in acomodar(list1,list2): #('Essay', 'Political Science') ('Homework', 'Computer Science')\n", + " nueva_lista.append((tup[1],tup[0]))\n", + "nueva_lista " + ] + }, + { + "cell_type": "code", + "execution_count": 90, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "[('Essay', 'Political Science'),\n", + " ('Homework', 'Computer Science'),\n", + " ('Lab', 'Engineering'),\n", + " ('Module', 'Mathematics')]" + ] + }, + "execution_count": 90, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "acomodar(list1,list2)" + ] + }, + { + "cell_type": "code", + "execution_count": 89, + "metadata": { + "collapsed": true + }, + "outputs": [ + { + "data": { + "text/plain": [ + "[('Engineering', 'Lab'),\n", + " ('Computer Science', 'Homework'),\n", + " ('Political Science', 'Essay'),\n", + " ('Mathematics', 'Module')]" + ] + }, + "execution_count": 89, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "list(zip(list1,list2))" ] }, { @@ -295,21 +473,43 @@ }, { "cell_type": "code", - "execution_count": 13, + "execution_count": 135, "metadata": {}, - "outputs": [], + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Estaba así [('Honda', 1997), ('Toyota', 1995), ('Audi', 2001), ('BMW', 2005)]\n", + "Organizado [('Toyota', 1995), ('Honda', 1997), ('Audi', 2001), ('BMW', 2005)]\n", + "En diccionario: {'Toyota': 1995, 'Honda': 1997, 'Audi': 2001, 'BMW': 2005}\n" + ] + } + ], "source": [ "d = {'Honda': 1997, 'Toyota': 1995, 'Audi': 2001, 'BMW': 2005}\n", + "d_nuevo = {} #{1997: 'Honda', 1995: 'Toyota', 2001: 'Audi', 2005: 'BMW'}\n", "\n", - "# Your code here:" + "# Your code here:\n", + "# diccionario = lambda marca, año: d_alreves[año]=marca\n", + "# sorted(d)\n", + "\n", + "año = list(d.values())\n", + "marca = list(d.keys())\n", + "año_marca = list(d.items())\n", + "print('Estaba así',año_marca)\n", + "\n", + "tuplas_organizadas = sorted(año_marca, key=lambda x: x[1])\n", + "print('Organizado: ',tuplas_organizadas)\n", + "\n", + "#pasar a diccionario marca: año\n", + "for m, a in tuplas_organizadas:\n", + " d_nuevo[m] = a\n", + "# print(d[m], \":\", m)\n", + "print('En diccionario:',d_nuevo)\n", + "# tuplas_organizadas = {key:value for key, value in sorted(año_marca, key=lambda x: x[1])}\n", + "# print(tuplas_organizadas)\n" ] - }, - { - "cell_type": "code", - "execution_count": null, - "metadata": {}, - "outputs": [], - "source": [] } ], "metadata": { @@ -328,7 +528,7 @@ "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", - "version": "3.7.3" + "version": "3.8.8" } }, "nbformat": 4,