PY0101EN-2-3-Sets.ipynb

{"cells":[{"cell_type":"markdown","metadata":{},"source":["<center>\n","    <img src=\"https://cf-courses-data.s3.us.cloud-object-storage.appdomain.cloud/IBMDeveloperSkillsNetwork-PY0101EN-SkillsNetwork/IDSNlogo.png\" width=\"300\" alt=\"cognitiveclass.ai logo\"  />\n","</center>\n","\n","# Sets in Python\n","\n","Estimated time needed: **20** minutes\n","\n","## Objectives\n","\n","After completing this lab you will be able to:\n","\n","*   Work with sets in Python, including operations and logic operations.\n"]},{"cell_type":"markdown","metadata":{},"source":["<h2>Table of Contents</h2>\n","<div class=\"alert alert-block alert-info\" style=\"margin-thttps://op/?utm_medium=Exinfluencer&utm_source=Exinfluencer&utm_content=000026UJ&utm_term=10006555&utm_id=NA-SkillsNetwork-Channel-SkillsNetworkCoursesIBMDeveloperSkillsNetworkPY0101ENSkillsNetwork19487395-2021-01-01: 20px\">\n","    <ul>\n","        <li>\n","            <a href=\"https://#set\">Sets</a>\n","            <ul>\n","                <li><a href=\"https://content/?utm_medium=Exinfluencer&utm_source=Exinfluencer&utm_content=000026UJ&utm_term=10006555&utm_id=NA-SkillsNetwork-Channel-SkillsNetworkCoursesIBMDeveloperSkillsNetworkPY0101ENSkillsNetwork19487395-2021-01-01\">Set Content</a></li>\n","                <li><a href=\"op\">Set Operations</a></li>\n","                <li><a href=\"https://logic/?utm_medium=Exinfluencer&utm_source=Exinfluencer&utm_content=000026UJ&utm_term=10006555&utm_id=NA-SkillsNetwork-Channel-SkillsNetworkCoursesIBMDeveloperSkillsNetworkPY0101ENSkillsNetwork19487395-2021-01-01\">Sets Logic Operations</a></li>\n","            </ul>\n","        </li>\n","        <li>\n","            <a href=\"https://#quiz\">Quiz on Sets</a>\n","        </li>\n","    </ul>\n","\n","</div>\n","\n","<hr>\n"]},{"cell_type":"markdown","metadata":{},"source":["<h2 id=\"set\">Sets</h2>\n"]},{"cell_type":"markdown","metadata":{},"source":["<h3 id=\"content\">Set Content</h3>\n"]},{"cell_type":"markdown","metadata":{},"source":["A set is a unique collection of objects in Python. You can denote a set with a pair of curly brackets <b>{}</b>. Python will automatically remove duplicate items:\n"]},{"cell_type":"code","execution_count":null,"metadata":{},"outputs":[],"source":["# Create a set\n","\n","set1 = {\"pop\", \"rock\", \"soul\", \"hard rock\", \"rock\", \"R&B\", \"rock\", \"disco\"}\n","set1"]},{"cell_type":"markdown","metadata":{},"source":["The process of mapping is illustrated in the figure:\n"]},{"cell_type":"markdown","metadata":{},"source":["<img src=\"https://cf-courses-data.s3.us.cloud-object-storage.appdomain.cloud/IBMDeveloperSkillsNetwork-PY0101EN-SkillsNetwork/labs/Module%202/images/SetsUnique.png\" width=\"1100\" />\n"]},{"cell_type":"markdown","metadata":{},"source":["You can also  create a set from a list as follows:\n"]},{"cell_type":"code","execution_count":null,"metadata":{"scrolled":true},"outputs":[],"source":["# Convert list to set\n","\n","album_list = [ \"Michael Jackson\", \"Thriller\", 1982, \"00:42:19\", \\\n","              \"Pop, Rock, R&B\", 46.0, 65, \"30-Nov-82\", None, 10.0]\n","album_set = set(album_list)             \n","album_set"]},{"cell_type":"markdown","metadata":{},"source":["Now let us create a set of  genres:\n"]},{"cell_type":"code","execution_count":null,"metadata":{},"outputs":[],"source":["# Convert list to set\n","\n","music_genres = set([\"pop\", \"pop\", \"rock\", \"folk rock\", \"hard rock\", \"soul\", \\\n","                    \"progressive rock\", \"soft rock\", \"R&B\", \"disco\"])\n","music_genres"]},{"cell_type":"markdown","metadata":{},"source":["<h3 id=\"op\">Set Operations</h3> \n"]},{"cell_type":"markdown","metadata":{},"source":["Let us go over set operations, as these can be used to change the set. Consider the set <b>A</b>:\n"]},{"cell_type":"code","execution_count":null,"metadata":{},"outputs":[],"source":["# Sample set\n","\n","A = set([\"Thriller\", \"Back in Black\", \"AC/DC\"])\n","A"]},{"cell_type":"markdown","metadata":{},"source":["We can add an element to a set using the <code>add()</code> method:\n"]},{"cell_type":"code","execution_count":null,"metadata":{},"outputs":[],"source":["# Add element to set\n","\n","A.add(\"NSYNC\")\n","A"]},{"cell_type":"markdown","metadata":{},"source":["If we add the same element twice, nothing will happen as there can be no duplicates in a set:\n"]},{"cell_type":"code","execution_count":null,"metadata":{},"outputs":[],"source":["# Try to add duplicate element to the set\n","\n","A.add(\"NSYNC\")\n","A"]},{"cell_type":"markdown","metadata":{},"source":["We can remove an item from a set using the <code>remove</code> method:\n"]},{"cell_type":"code","execution_count":null,"metadata":{},"outputs":[],"source":["# Remove the element from set\n","\n","A.remove(\"NSYNC\")\n","A"]},{"cell_type":"markdown","metadata":{},"source":["We can verify if an element is in the set using the <code>in</code> command:\n"]},{"cell_type":"code","execution_count":null,"metadata":{},"outputs":[],"source":["# Verify if the element is in the set\n","\n","\"AC/DC\" in A"]},{"cell_type":"markdown","metadata":{},"source":["<h3 id=\"logic\">Sets Logic Operations</h3>\n"]},{"cell_type":"markdown","metadata":{},"source":["Remember that with sets you can check the difference between sets, as well as the symmetric difference, intersection, and union:\n"]},{"cell_type":"markdown","metadata":{},"source":["Consider the following two sets:\n"]},{"cell_type":"code","execution_count":null,"metadata":{},"outputs":[],"source":["# Sample Sets\n","\n","album_set1 = set([\"Thriller\", 'AC/DC', 'Back in Black'])\n","album_set2 = set([ \"AC/DC\", \"Back in Black\", \"The Dark Side of the Moon\"])"]},{"cell_type":"markdown","metadata":{},"source":["<img src=\"https://cf-courses-data.s3.us.cloud-object-storage.appdomain.cloud/IBMDeveloperSkillsNetwork-PY0101EN-SkillsNetwork/labs/Module%202/images/SetsSamples.png\" width=\"650\" />\n"]},{"cell_type":"code","execution_count":null,"metadata":{"scrolled":true},"outputs":[],"source":["# Print two sets\n","\n","album_set1, album_set2"]},{"cell_type":"markdown","metadata":{},"source":["As both sets contain <b>AC/DC</b> and <b>Back in Black</b> we represent these common elements with the intersection of two circles.\n"]},{"cell_type":"markdown","metadata":{},"source":["<img src = \"https://cf-courses-data.s3.us.cloud-object-storage.appdomain.cloud/IBMDeveloperSkillsNetwork-PY0101EN-SkillsNetwork/labs/Module%202/images/SetsLogic.png\" width = \"650\" />\n"]},{"cell_type":"markdown","metadata":{},"source":["You can find the intersect of two sets as follow using <code>&</code>:\n"]},{"cell_type":"code","execution_count":null,"metadata":{},"outputs":[],"source":["# Find the intersections\n","\n","intersection = album_set1 & album_set2\n","intersection"]},{"cell_type":"markdown","metadata":{},"source":["You can find all the elements that are only contained in <code>album_set1</code> using the <code>difference</code> method:\n"]},{"cell_type":"code","execution_count":null,"metadata":{},"outputs":[],"source":["# Find the difference in set1 but not set2\n","\n","album_set1.difference(album_set2)  "]},{"cell_type":"markdown","metadata":{},"source":["You only need to consider elements in <code>album_set1</code>; all the elements in <code>album_set2</code>, including the intersection, are not included.\n"]},{"cell_type":"markdown","metadata":{},"source":["<img src=\"https://cf-courses-data.s3.us.cloud-object-storage.appdomain.cloud/IBMDeveloperSkillsNetwork-PY0101EN-SkillsNetwork/labs/Module%202/images/SetsLeft.png\" width=\"650\" />\n"]},{"cell_type":"markdown","metadata":{},"source":["The elements in <code>album_set2</code> but not in <code>album_set1</code> is given by:\n"]},{"cell_type":"code","execution_count":null,"metadata":{},"outputs":[],"source":["album_set2.difference(album_set1)  "]},{"cell_type":"markdown","metadata":{},"source":["<img src = \"https://cf-courses-data.s3.us.cloud-object-storage.appdomain.cloud/IBMDeveloperSkillsNetwork-PY0101EN-SkillsNetwork/labs/Module%202/images/SetsRight.png\" width=\"650\" />\n"]},{"cell_type":"markdown","metadata":{},"source":["You can also find the intersection of <code>album_list1</code> and <code>album_list2</code>, using the <code>intersection</code> method:\n"]},{"cell_type":"code","execution_count":null,"metadata":{},"outputs":[],"source":["# Use intersection method to find the intersection of album_list1 and album_list2\n","\n","album_set1.intersection(album_set2)   "]},{"cell_type":"markdown","metadata":{},"source":["This corresponds to the intersection of the two circles:\n"]},{"cell_type":"markdown","metadata":{},"source":["<img src=\"https://cf-courses-data.s3.us.cloud-object-storage.appdomain.cloud/IBMDeveloperSkillsNetwork-PY0101EN-SkillsNetwork/labs/Module%202/images/SetsIntersect.png\" width=\"650\" />\n"]},{"cell_type":"markdown","metadata":{},"source":["The union corresponds to all the elements in both sets, which is represented by coloring both circles:\n"]},{"cell_type":"markdown","metadata":{},"source":["<img src = \"https://cf-courses-data.s3.us.cloud-object-storage.appdomain.cloud/IBMDeveloperSkillsNetwork-PY0101EN-SkillsNetwork/labs/Module%202/images/SetsUnion.png\" width=\"650\" />\n"]},{"cell_type":"markdown","metadata":{},"source":["The union is given by:\n"]},{"cell_type":"code","execution_count":null,"metadata":{"collapsed":true},"outputs":[],"source":["# Find the union of two sets\n","\n","album_set1.union(album_set2)"]},{"cell_type":"markdown","metadata":{},"source":["And you can check if a set is a superset or subset of another set, respectively, like this:\n"]},{"cell_type":"code","execution_count":null,"metadata":{},"outputs":[],"source":["# Check if superset\n","\n","set(album_set1).issuperset(album_set2)   "]},{"cell_type":"code","execution_count":null,"metadata":{},"outputs":[],"source":["# Check if subset\n","\n","set(album_set2).issubset(album_set1)     "]},{"cell_type":"markdown","metadata":{},"source":["Here is an example where <code>issubset()</code> and <code>issuperset()</code> return true:\n"]},{"cell_type":"code","execution_count":null,"metadata":{},"outputs":[],"source":["# Check if subset\n","\n","set({\"Back in Black\", \"AC/DC\"}).issubset(album_set1) "]},{"cell_type":"code","execution_count":null,"metadata":{},"outputs":[],"source":["# Check if superset\n","\n","album_set1.issuperset({\"Back in Black\", \"AC/DC\"})   "]},{"cell_type":"markdown","metadata":{},"source":["<hr>\n"]},{"cell_type":"markdown","metadata":{},"source":["<h2 id=\"quiz\">Quiz on Sets</h2>\n"]},{"cell_type":"markdown","metadata":{},"source":["Convert the list <code>\\['rap','house','electronic music', 'rap']</code> to a set:\n"]},{"cell_type":"code","execution_count":null,"metadata":{},"outputs":[],"source":["# Write your code below and press Shift+Enter to execute\n"]},{"cell_type":"markdown","metadata":{},"source":["<details><summary>Click here for the solution</summary>\n","\n","```python\n","set(['rap','house','electronic music','rap'])\n","\n","```\n","\n","</details>\n"]},{"cell_type":"markdown","metadata":{},"source":["<hr>\n"]},{"cell_type":"markdown","metadata":{},"source":["Consider the list <code>A = \\[1, 2, 2, 1]</code> and set <code>B = set(\\[1, 2, 2, 1])</code>, does <code>sum(A) == sum(B)</code>?\n"]},{"cell_type":"code","execution_count":null,"metadata":{},"outputs":[],"source":["# Write your code below and press Shift+Enter to execute\n"]},{"cell_type":"markdown","metadata":{},"source":["<details><summary>Click here for the solution</summary>\n","\n","```python\n","A = [1, 2, 2, 1]  \n","B = set([1, 2, 2, 1])\n","print(\"the sum of A is:\", sum(A))\n","print(\"the sum of B is:\", sum(B))\n","\n","```\n","\n","</details>\n"]},{"cell_type":"markdown","metadata":{},"source":["<hr>\n"]},{"cell_type":"markdown","metadata":{},"source":["Create a new set <code>album_set3</code> that is the union of <code>album_set1</code> and <code>album_set2</code>:\n"]},{"cell_type":"code","execution_count":null,"metadata":{},"outputs":[],"source":["# Write your code below and press Shift+Enter to execute\n","\n","album_set1 = set([\"Thriller\", 'AC/DC', 'Back in Black'])\n","album_set2 = set([ \"AC/DC\", \"Back in Black\", \"The Dark Side of the Moon\"])"]},{"cell_type":"markdown","metadata":{},"source":["<details><summary>Click here for the solution</summary>\n","\n","```python\n","album_set3 = album_set1.union(album_set2)\n","album_set3\n","\n","```\n","\n","</details>\n"]},{"cell_type":"markdown","metadata":{},"source":["<hr>\n"]},{"cell_type":"markdown","metadata":{},"source":["Find out if <code>album_set1</code> is a subset of <code>album_set3</code>:\n"]},{"cell_type":"code","execution_count":null,"metadata":{},"outputs":[],"source":["# Write your code below and press Shift+Enter to execute\n"]},{"cell_type":"markdown","metadata":{},"source":["<details><summary>Click here for the solution</summary>\n","\n","```python\n","album_set1.issubset(album_set3)\n","\n","```\n","\n","</details>\n"]},{"cell_type":"markdown","metadata":{},"source":["<hr>\n","<h2>The last exercise!</h2>\n","<p>Congratulations, you have completed your first lesson and hands-on lab in Python. \n","<hr>\n"]},{"cell_type":"markdown","metadata":{},"source":["## Author\n","\n","<a href=\"https://www.linkedin.com/in/joseph-s-50398b136/?utm_medium=Exinfluencer&utm_source=Exinfluencer&utm_content=000026UJ&utm_term=10006555&utm_id=NA-SkillsNetwork-Channel-SkillsNetworkCoursesIBMDeveloperSkillsNetworkPY0101ENSkillsNetwork19487395-2021-01-01\" target=\"_blank\">Joseph Santarcangelo</a>\n","\n","## Other contributors\n","\n","<a href=\"https://www.linkedin.com/in/jiahui-mavis-zhou-a4537814a?utm_medium=Exinfluencer&utm_source=Exinfluencer&utm_content=000026UJ&utm_term=10006555&utm_id=NA-SkillsNetwork-Channel-SkillsNetworkCoursesIBMDeveloperSkillsNetworkPY0101ENSkillsNetwork19487395-2021-01-01\">Mavis Zhou</a>\n","\n","## Change Log\n","\n","| Date (YYYY-MM-DD) | Version | Changed By | Change Description                 |\n","| ----------------- | ------- | ---------- | ---------------------------------- |\n","| 2022-01-10        | 2.1     | Malika     | Removed the readme for GitShare    |\n","| 2020-08-26        | 2.0     | Lavanya    | Moved lab to course repo in GitLab |\n","|                   |         |            |                                    |\n","|                   |         |            |                                    |\n","\n","## <h3 align=\"center\"> © IBM Corporation 2020. All rights reserved. <h3/>\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":"3.7.6"}},"nbformat":4,"nbformat_minor":2}