Count all the letters in a sentence. Both A and a should count for the same letter.The best way to get an indexed letter value is subtract ord("a") or ord("A") from yourcurrent letter.String methods: isalpha(), lower(), upper(), islower(), isupper()String functions: ord(), len()def getLetterFrequency(sentence):freq = ]for i in range(26):%3Dfreq.append(0)# TODO: count each letter herereturn freqprint(getLetterFrequency("Hello World"))# remove thewhen you're ready to rest fullyprint("Testing")assert( getLetterFrequency("Hello World!") == [0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 3, 0, 0, 2, 0,0, 1, 0, 0, 0, 0, 1, 0, 0, 0])assert( getLetterFrequency("Wow, such tricky...") == [0, 0, 2, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0,1, 0, 0, 1, 1, 1, 1, 0, 2, 0, 1, 0])assert( getLetterFrequency("The quick brown fox jumps over the lazy dog.") == [1, 1, 1,%3D%3D1, 3, 1, 1, 2, 1, 1, 1, 1, 1, 1, 4, 1, 1, 2, 1, 2, 2, 1, 1, 1, 1, 1])print("Done!")# remove thewhen you're ready to rest fullyExpected OutputExample, the string "Hello World" returns list: [0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 3, 0, 0, 2, 0,0.100.001.0.0.01 Complete the function getLetterFrequency() to returna list containing the number of times each letter of thealphabet occurs in the string passed to it. The Othindex corresponds to the letter "a" or "A", the 1stcorresponds to the letter "b" or "B", ..., the 25thcorresponds to "z" or "Z".You should ignore any non-alphabet characters thatoccur in the string. But make sure that both lowercaseand uppercase letters contribute to the same count.You must use the ord() function to convert letters intoindexed based values. Note that you can offset anumber by ord("a") or ord("A") to get the index thatyou need.You should use at least some of the (or all of) followingstring methods:• isalpha()lower()• upper()• isupper()• islower()and the following string functions:ord()len()

According to the Question below the Complete Program: Program Output:

Count all the letters in a sentence. Both A and a should count for the same letter. The best way to get an indexed letter value is subtract ord("a") or ord("A") from your current letter. String methods: isalpha(), lower(), upper(), islower(), isupper() String functions: ord(), len() def getletterFrequency(sentence): freq = 0 for i in range(26): freq.append(0) # TODO: count each letter here return freq print(getLetterFrequency("Hello World")) # remove the when you're ready to rest fully print("Testing") assert( getLetterFrequency("Hello World!") == [0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 3, 0, 0, 2, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0]) assert( getLetterFrequency("Wow, such tricky..") == [0, 0, 2, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1,0, 0, 1, 1, 1, 1, 0, 2, 0, 1, 0]) assert( getLetterFrequency("The quick brown fox jumps over the lazy dog.") == [1, 1, 1, 1, 3, 1, 1, 2, 1, 1, 1, 1, 1, 1, 4, 1, 1, 2, 1, 2, 2, 1, 1, 1, 1, 1]) %3%3D

Count all the letters in a sentence. Both A and a should count for the same letter. The best way to get an indexed letter value is subtract ord("a") or ord("A") from your current letter. String methods: isalpha(), lower(), upper(), islower(), isupper() String functions: ord(), len() def getletterFrequency(sentence): freq = 0 for i in range(26): freq.append(0) # TODO: count each letter here return freq print(getLetterFrequency("Hello World")) # remove the when you're ready to rest fully print("Testing") assert( getLetterFrequency("Hello World!") == [0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 3, 0, 0, 2, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0]) assert( getLetterFrequency("Wow, such tricky..") == [0, 0, 2, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1,0, 0, 1, 1, 1, 1, 0, 2, 0, 1, 0]) assert( getLetterFrequency("The quick brown fox jumps over the lazy dog.") == [1, 1, 1, 1, 3, 1, 1, 2, 1, 1, 1, 1, 1, 1, 4, 1, 1, 2, 1, 2, 2, 1, 1, 1, 1, 1]) %3%3D

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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