im writing a code for counting the frequency of adjacent letters in python

Question

line=input()
#taking a sentence

d1=1
#count variable

ln=len(line)
for i in range(0,ln):

    if line[i]==line[i+1]:

        d1+=1
    else:

        print(line[i],d1)
        d1=1

**sample test cases1

aaabcaaaa
sample output1 a 3 b 1 c 1 a 4

**

**my wrong output

aaabcaaaa
sample output1 a 3 b 1 c 1

** in my output, I'm not getting the last frequency of the last character

Does this answer your question? [Reading and Compressing a picture with RLE](https://stackoverflow.com/questions/22456480/reading-and-compressing-a-picture-with-rle) — Idea O., Jun 15 '20 at 11:22
first of all, `range(ln)` is the exact same as `range(0, ln)`, so you can remove the 0 — Lwi, Jun 15 '20 at 12:47

score 0 · Answer 1 · answered Jun 15 '20 at 11:29

0

You only print the number if the next char is different. but the last char has no next char. change your if statement to:

if i != ln - 1 and line[i] == line[i + 1]:

answered Jun 15 '20 at 11:29

Creepsy

321
3
13

No it is not working its throwing "string index out of range" error – Santosh Burada Jun 16 '20 at 14:09

score 0 · Answer 2 · answered Jun 15 '20 at 11:30

You need to add one last statement at the end, because it no longer goes into the else statement to print out the last letter frequency

Try


d1=1
#count variable

ln=len(line)
for i in range(0,ln-1):
    if line[i]==line[i+1]:

        d1+=1
    else:
        print(line[i],d1)
        d1=1

if d1 > 1:
    print(line[i],d1)

score 0 · Answer 3 · answered Jun 16 '20 at 14:28

enter code here
line=input()
d1=1
ln=len(line)
output=""
for i in range(0,ln-1):
    if line[i]==line[i+1] and i!=ln-1:
        d1+=1

    else:
        output+=line[i]+str(d1)
        d1=1

output+=line[i+1]+str(d1)#this extra line solves my question#
print(output)

**Thanks to both of you for your respons **

im writing a code for counting the frequency of adjacent letters in python

3 Answers3