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.
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.
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
Related questions
Question
In python
Transcribed Image Text:Complete the function getLetterFrequency() to return
a list containing the number of times each letter of the
alphabet occurs in the string passed to it. The Oth
index corresponds to the letter "a" or "A", the 1st
corresponds to the letter "b" or "B", ..., the 25th
corresponds to "z" or "Z".
You should ignore any non-alphabet characters that
occur in the string. But make sure that both lowercase
and uppercase letters contribute to the same count.
You must use the ord() function to convert letters into
indexed based values. Note that you can offset a
number by ord("a") or ord("A") to get the index that
you need.
You should use at least some of the (or all of) following
string methods:
• isalpha()
lower()
• upper()
• isupper()
• islower()
and the following string functions:
ord()
len()
![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 = []
for i in range(26):
%3D
freq.append(0)
# TODO: count each letter here
return freq
print(getLetterFrequency("Hello World"))
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])
print("Done!")
5.
Expected Output
Example, the string "Hello World" returns list: [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]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbb7bc7a6-31dc-4260-a709-54a058cd3d65%2F5816be3a-8f17-4496-aa07-d7803235d476%2Fxfs0joe_processed.jpeg&w=3840&q=75)
Transcribed Image Text: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 = []
for i in range(26):
%3D
freq.append(0)
# TODO: count each letter here
return freq
print(getLetterFrequency("Hello World"))
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])
print("Done!")
5.
Expected Output
Example, the string "Hello World" returns list: [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]
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