Create a compress function in Python?

Question

I need to create a function called compress that compresses a string by replacing any repeated letters with a letter and number. My function should return the shortened vers

Answer 1

Short version with generators:

from itertools import groupby
import re
def compress(string):
    return re.sub(r'(?<![0-9])[1](?![0-9])', '', ''.join('%s%s' % (char, sum(1 for _ in group)) for char, group in groupby(string)))

---

(1) Grouping by chars with groupby(string)

(2) Counting length of group with sum(1 for _ in group) (because no len on group is possible)

(3) Joining into proper format

(4) Removing 1 chars for single items when there is a no digit before and after 1

Answer 2

x="mississippi"
res = ""
count = 0
while (len(x) > 0):
    count = 1
    res= ""
    for j in range(1, len(x)):
        if x[0]==x[j]:
            count= count + 1
        else:
            res = res + x[j]
    print(x[0], count, end=" ")
    x=res

Answer 3

You can simply achieve that by:

gstr="aaabbccccffffdee"
last=gstr[0]
count=0
rstr=""
for i in gstr:
    if i==last:
        count=count+1
    elif i!=last:
        rstr=rstr+last+str(count)
        count=1
        last=i
rstr=rstr+last+str(count)
print ("Required string for given string {} after conversion is {}.".format(gstr,rstr))

Answer 4

This is a solution to the problem. But keep in mind that this method only effectively works if there's a lot of repetition, specifically if consecutive characters are repetitive. Otherwise, it will only worsen the situation.

e.g.,
AABCD --> A2B1C1D1
BcDG ---> B1c1D1G1

def compress_string(s):
    result = [""] * len(s)
    visited = None

    index = 0
    count = 1

    for c in s:
        if c == visited:
            count += 1
            result[index] = f"{c}{count}"
        else:
            count = 1
            index += 1
            result[index] = f"{c}{count}"
            visited = c

    return "".join(result)

Answer 5

string = 'aabccccd' output = '2a3b4c4d'

new_string = " "
count = 1
for i in range(len(string)-1):
    if string[i] == string[i+1]:
        count = count + 1
    else:         
        new_string =  new_string + str(count) + string[i]
        count = 1 
new_string = new_string + str(count) + string[i+1]    
print(new_string)

Answer 6

This Is The Modification Of Patrick Yu's Code. His Code Fails In The Below Test Cases.

SAMPLE INPUT:
c
aaaaaaaaaabcdefgh

EXPECTED OUTPUT:
c1
a10b1c1d1e1f1g1h1

OUPUT OF Patrick's Code:
c
a10bcdefgh

Below Is The Modification Of His Code:

def Compress(S):
    Ans = S[0]
    count = 1
    for i in range(len(S)-1):
        if S[i] == S[i+1]:
            count += 1
        else:
            if count >= 1:
                Ans += str(count)
            Ans += S[i+1]
            count = 1
    if count>=1:
        Ans += str(count)
    return Ans

Just The Condition Must Be Changed From Greater(">") To Greater Than Equal To(">=") When Comparing The Count With 1.