Python letter frequency project

Question

I wrote a code to solve the letter frequency project ( python data structure project 1 )
And got the 1,4,5 cases right and the rest have wrong answers so i need someone to tell me whats the problem or what went wrong in my program

#your code goes here
text=input()
letter=input()
count=0
freq=0

for letter in text:
count=text.count(letter)

x=(count/len(text))
freq= x*100
print (int(freq))

Answer 1

You use the count() method, it has already calculated everything for you:
#your code goes here
text=input()
letter=input()
count=text.count(letter)
print(count)

If you do not understand this, then you have not mastered the material covered and you definitely need to repeat it from the very beginning.

Happy learning! ☺️

Answer 2

No need of loop there. Remove it.
Loop making count last letter only but you need to count for input letter..

edit: post full question if not solved..

Answer 3

Ohh thanks i got it ,my mistake was
For >> counts every letter
If >> count a specific letter
So i should use if not for

Answer 4

for i in text:
if i == letter:
count += 1

sets variable count to how many times variable letter is in variable text

while your code

for letter in text:
count=text.count(letter)

count is set to how many times the last letter appears in variable text
ex.
missisipi

count is set to 4 as the last letter is "i" and it appears 4 times

so while for loop can be used, you used it wrong, i would recommend to re-read the course again to hopefully get better at it

Answer 5

text = input()
search_letter = input()
count = 0
freq = 0

for letter in text:
if letter == search_letter:
count += 1

x = count / len(text)
freq = x * 100
print(int(freq))

This took me a while while trying different things, but here's what I got for the Letter Frequency project in Python.

Answer 6

I hope you understand that this code can be reduced to a minimum by writing everything in one line? Do it ☺️

Answer 7

w = input()
l = input()
print(int(w.count(l)/len(w)*100))

The shortest one liner for this solution, i suggest