Count and store in a file

Question

counts the occurrences of letter a in the first 200 characters in the file characters.txt the result should get stored inside a new folder with a txt file

Example: characters.txt: abcdefghijklmnopqerstuvwxzy

so there is 1 occurrence of g

then "1" should be stored in foulder/file.txt

 file = open(filename, "r")
 text = file.read()
 count = 0
 for char in text:
 if char == letter:
 count += 1
os.mkdir("g")
f = open("res.txt", mode = "w")
f.write(count)
f.close

• Almos: Where do you call the function?

Answer 1

Your code works, but in the samples provided you dont call it.

I made a local version without your file code.

def letterFrequency(letter):
 count = 0
 for char in 'abcdefghijklmnopqerstuvwxzy':
 if char == letter:
 count += 1
 return count
print(letterFrequency('g'))

If you only want to search the first 200 character of a file you should use a while loop. Also you will need to account for rows with less than 200 characters.

• Lucxgjs: Thank you for your response, can i make the same without a function?
• Almos: sure, you just need to delete the def and return. Then you need to fix the indentation. Count will then be global so you can just use it.
• Lucxgjs: I tried to specifie the range of but I didn't work. text = file.readlines()[0,119] any idea?
• Almos: you can need to itterate through the rows and check the length of the rows If row 1 has 180 character you need to substrac them from you maximum and so on until you have your 200 chars

Answer 2

I modified your given example and added some improvements. The code below is a minimal working example:

import os
file = open("./Desktop/text.txt", "r")
text = file.read()
count = 0
letter = "g"
if len(text) < 200:
 text = text[0:199]
for char in text:
 if char == letter:
 count += 1
 
try: 
 os.mkdir("./Desktop/DIR")
except FileExistsError:
 print("Dir already exists")
f = open("./Desktop/DIR/res.txt", "w")
f.write(str(count))