Run length encoding in python3

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Given a string, print the run-length encoded output.
Input Size : N <= 100000
Sample Testcase :
INPUT
aaab
OUTPUT
a3b1

the code I tried executed for above sample case but my problem is private cases didn't pass
I am sorry if this is a naive doubt I am a beginner for coding
thank you in advance

What I have tried:

from collections import Counter
text=input()
if(text==None):
    print(" ")
if(text.isalpha()):
    
        dict1=dict(Counter(text))
        for i,j in dict1.items():
            print(i,j,end="",sep="")

Posted 10-Oct-19 4:09am

_gvsss

Updated 10-Oct-19 7:14am

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Comments

Richard MacCutchan 10-Oct-19 10:31am

We cannot guess what tests failed or how they failed.

_gvsss 10-Oct-19 10:50am

yeah I know that part, my intention was am I missing anything if any please let me know

Richard MacCutchan 10-Oct-19 10:55am

Yes, you have not explained what the problem is.

k5054 10-Oct-19 11:00am

What happens when you try the input "aaabbaa"? The output should be a3b2a2

_gvsss 10-Oct-19 11:28am

no according to my code it will be
aaabbaa
a5b2

Richard MacCutchan 10-Oct-19 11:35am

Yes and that is correct because you are using the Counter class which counts the total occurrences of each unique item. If you want RLE then you need to write the code to calculate each part separately.

k5054 10-Oct-19 11:40am

Since this is "Run Length Encoding", one would expect that you would be able to "decode" the output in some way, and get the original text back. What you have produces the same output for any combination of 5 a's and 2 b's, as long as the first character is an a. That means you've lost information when "encoding", which is probably not what you want.

Patrice T 10-Oct-19 12:55pm

But correct answer is a3b2a2

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This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

CPallini · Accepted Answer · 2019-10-10T07:15:00

Solution 2

The Python implementation of the C programmer :-O

Python

s = 'aaabbaa'
c = s[0]
n = 1
o = c
for k in range(1, len(s)):
  o = s[k]
  if o == c:
    n = n + 1
  else:
    print(c,n, sep='', end='')
    n = 1 
    c = o
print(o,n, sep='')

Posted 10-Oct-19 7:15am

CPallini

Comments

_gvsss 11-Oct-19 2:43am

thank you

Run length encoding in python3

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Run length encoding in python3

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