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‎.DS_Store‎

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‎Python_Basics/Anagram/.ipynb_checkpoints/AnagramsPython-checkpoint.ipynb‎

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{
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"cells": [
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"<h1 align=\"center\"> Anagrams using Python </h1>"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## What is an Anagram?"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once."
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Task: Write a program that takes in a word list and outputs a list of all the words that are anagrams of another word in the list."
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"While there are many different ways to solve this problem, this notebook gives a couple different approaches to solve this problem. \n",
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"For each of the approaches below, we first define a word list"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 1,
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"metadata": {},
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"outputs": [],
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"source": [
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"word_list = ['percussion',\n",
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" 'supersonic',\n",
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" 'car',\n",
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" 'tree',\n",
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" 'boy',\n",
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" 'girl',\n",
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" 'arc']"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Approach 1: For Loop"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 2,
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"metadata": {},
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"outputs": [],
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"source": [
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"temp = word_list[0]"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 3,
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"['c', 'e', 'i', 'n', 'o', 'p', 'r', 's', 's', 'u']"
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]
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},
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"execution_count": 3,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"sorted(temp)"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 4,
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"['percussion', 'supersonic', 'car', 'tree', 'boy', 'girl', 'arc']"
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]
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},
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"execution_count": 4,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"word_list"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 5,
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"metadata": {},
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"outputs": [],
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"source": [
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"# initialize a list\n",
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"anagram_list = []\n",
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"\n",
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"for word_1 in word_list: \n",
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" for word_2 in word_list: \n",
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" if word_1 != word_2 and (sorted(word_1)==sorted(word_2)):\n",
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" anagram_list.append(word_1)"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 6,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"['percussion', 'supersonic', 'car', 'arc']\n"
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]
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}
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],
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"source": [
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"print(anagram_list)"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 7,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"['c', 'e', 'i', 'n', 'o', 'p', 'r', 's', 's', 'u']\n"
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]
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}
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],
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"source": [
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"print(sorted('percussion'))"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 8,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"['c', 'e', 'i', 'n', 'o', 'p', 'r', 's', 's', 'u']\n"
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]
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}
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],
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"source": [
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"print(sorted('supersonic'))"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Approach 2: Dictionaries"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Sorting of lists in Python is O(nlogn) versus O(n) with a dictionary. If you have difficulty understanding the dictionary get method, I encourage you to see one of the following tutorials: [Python Dictionary and Dictionary Methods](https://hackernoon.com/python-basics-10-dictionaries-and-dictionary-methods-4e9efa70f5b9) or [Python Word Count](https://codeburst.io/python-basics-11-word-count-filter-out-punctuation-dictionary-manipulation-and-sorting-lists-3f6c55420855). "
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]
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},
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{
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"cell_type": "code",
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"execution_count": 9,
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"metadata": {},
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"outputs": [],
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"source": [
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"def freq(word):\n",
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" freq_dict = {}\n",
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" for char in word:\n",
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" freq_dict[char] = freq_dict.get(char, 0) + 1\n",
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" return freq_dict "
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]
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},
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{
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"cell_type": "code",
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"execution_count": 10,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"['percussion', 'supersonic', 'car', 'arc']\n"
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]
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}
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],
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"source": [
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"# initialize a list\n",
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"anagram_list = []\n",
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"for word_1 in word_list: \n",
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" for word_2 in word_list: \n",
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" if word_1 != word_2 and (freq(word_1) == freq(word_2)):\n",
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" anagram_list.append(word_1)\n",
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"print(anagram_list)"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 11,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"{'p': 1, 'e': 1, 'r': 1, 'c': 1, 'u': 1, 's': 2, 'i': 1, 'o': 1, 'n': 1}\n"
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]
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}
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],
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"source": [
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"print(freq('percussion'))"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 12,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"{'s': 2, 'u': 1, 'p': 1, 'e': 1, 'r': 1, 'o': 1, 'n': 1, 'i': 1, 'c': 1}\n"
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]
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}
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],
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"source": [
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"print(freq('supersonic'))"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 13,
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"[('c', 1),\n",
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" ('e', 1),\n",
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" ('i', 1),\n",
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" ('n', 1),\n",
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" ('o', 1),\n",
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" ('p', 1),\n",
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" ('r', 1),\n",
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" ('s', 2),\n",
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" ('u', 1)]"
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]
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},
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"execution_count": 13,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"sorted(list(freq('percussion').items()))"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 14,
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"[('c', 1),\n",
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" ('e', 1),\n",
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" ('i', 1),\n",
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" ('n', 1),\n",
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" ('o', 1),\n",
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" ('p', 1),\n",
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" ('r', 1),\n",
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" ('s', 2),\n",
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" ('u', 1)]"
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]
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},
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"execution_count": 14,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"sorted(list(freq('supersonic').items()))"
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]
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}
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],
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"metadata": {
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"anaconda-cloud": {},
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"kernelspec": {
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"display_name": "Python 3",
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"language": "python",
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"name": "python3"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.6.8"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 2
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}

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