How to replace repeated instances of a character with a single instance of that character in python

Question

I want to replace repeated instances of the \"*\" character within a string with a single instance of \"*\". For example if the string is \"*

Answer 1

Well regular expressions wise I would do exactly as JoshD has suggested. But one improvement here.

Use -

regex  = re.compile('\*+')
result = re.sub(regex, "*", string)

This would essentially cache your regex. So subsequent usage of this in a loop would make your regex operations fast.

Answer 2

I'd suggest using the re module sub function:

import re

result = re.sub("\*+", "*", "***abc**de*fg******h")

I highly recommend reading through the article about RE and good practices. They can be tricky if you're not familiar with them. In practice, using raw strings is a good idea.

Answer 3

Lets assume for this sake of this example, your character is a space.

You can also do it this way:

while True:
    if "  " in pattern: # if two spaces are in the variable pattern
        pattern = pattern.replace("  ", " ") # replace two spaces with one
    else: # otherwise
        break # break from the infinite while loop

This:

File Type                       : Win32 EXE
File Type Extension             : exe
MIME Type                       : application/octet-stream
Machine Type                    : Intel 386 or later, and compatibles
Time Stamp                      : 2017:04:24 09:55:04-04:00

Becomes:

File Type : Win32 EXE
File Type Extension : exe
MIME Type : application/octet-stream
Machine Type : Intel 386 or later, and compatibles
Time Stamp : 2017:04:24 09:55:04-04:00

I find this is a little easier than having to muck around with the re module, which can get a little annoying sometimes (I think).

Hope that was helpful.

Answer 4

You wrote:

pattern.replace("*"\*, "*")

You meant:

pattern.replace("\**", "*")
#                ^^^^

You really meant:

pattern_after_substitution= re.sub(r"\*+", "*", pattern)

which does what you wanted.

Answer 5

I timed all the methods in the current answers (with Python 3.7.2, macOS High Sierra).

b() was the best overall, c() was best when no matches are made.

def b(text):
    re.sub(r"\*\*+", "*", text)

# aka squeeze()
def c(text):
    while "*" * 2 in text:
        text = text.replace("*" * 2, "*")
    return text

Input 1, no repeats: 'a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*'

• a) 10000 loops, best of 5: 24.5 usec per loop
• b) 100000 loops, best of 5: 3.17 usec per loop
• c) 500000 loops, best of 5: 508 nsec per loop
• d) 10000 loops, best of 5: 25.4 usec per loop
• e) 5000 loops, best of 5: 44.7 usec per loop
• f) 500000 loops, best of 5: 522 nsec per loop

Input 2, with repeats: 'a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*****************************************************************************************************a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*'

• a) 5000 loops, best of 5: 46.2 usec per loop
• b) 50000 loops, best of 5: 5.21 usec per loop
• c) 20000 loops, best of 5: 13.4 usec per loop
• d) 5000 loops, best of 5: 47.4 usec per loop
• e) 2000 loops, best of 5: 103 usec per loop
• f) 20000 loops, best of 5: 13.1 usec per loop

The methods:

#!/usr/bin/env python
# encoding: utf-8
"""
See which function variants are fastest. Run like:
python -mtimeit -s"import time_functions;t='a*'*100" "time_functions.a(t)"
python -mtimeit -s"import time_functions;t='a*'*100" "time_functions.b(t)"
etc.
"""
import re


def a(text):
    return re.sub(r"\*+", "*", text)


def b(text):
    re.sub(r"\*\*+", "*", text)


# aka squeeze()
def c(text):
    while "*" * 2 in text:
        text = text.replace("*" * 2, "*")
    return text


regex = re.compile(r"\*+")


# like a() but with (premature) optimisation
def d(text):
    return re.sub(regex, "*", text)


def e(text):
    return "".join(c for c, n in zip(text, text[1:] + " ") if c + n != "**")


def f(text):
    while True:
        if "**" in text:  # if two stars are in the variable pattern
            text = text.replace("**", "*")  # replace two stars with one
        else:  # otherwise
            break  # break from the infinite while loop
    return text

Answer 6

text = "aaaaaaaaaabbbbbbbbbbcccccccffffdffffdaaaaaa"

result = " "

for char in text:

if len(result) > 0 and result[-1] == char:
    continue
else:
    result += char

print(result) # abcda