Make multiple modifications to string: how to, being inmutable in Python?

Question

I'm new to Python, so maybe I'm asking for something very easy but I can't think of the problem in a Python way.

I have a compressed string. The idea is, if a character gets repeated 4-15 times, I make this change:

'0000' ---> '0|4'

If more than 15 times, I use a slash and two digits to represent the amount (working with hexadecimal values):

'00...(16 times)..0' ---> '0/10'

So, accustomed to other languages, my approach is the following:

def uncompress(line):
    verticalBarIndex = line.index('|')
    while verticalBarIndex!=-1:
        repeatedChar = line[verticalBarIndex-1:verticalBarIndex]
        timesRepeated = int(line[verticalBarIndex+1:verticalBarIndex+2], 16)
        uncompressedChars = [repeatedChar]
        for i in range(timesRepeated):
            uncompressedChars.append(repeatedChar)
        uncompressedString = uncompressedChars.join()
        line = line[:verticalBarIndex-1] + uncompressedString + line[verticalBarIndex+2:]
        verticalBarIndex = line.index('|') #next one

    slashIndex = line.index('/')
    while slashIndex!=-1:
        repeatedChar = line[slashIndex-1:slashIndex]
        timesRepeated = int(line[slashIndex+1:verticalBarIndex+3], 16)
        uncompressedChars = [repeatedChar]
        for i in range(timesRepeated):
            uncompressedChars.append(repeatedChar)
        uncompressedString = uncompressedChars.join()
        line = line[:slashIndex-1] + uncompressedString + line[slashIndex+3:]
        slashIndex = line.index('/') #next one
    return line

Which I know it is wrong, since strings are inmutable in Python, and I am changing line contents all the time until no '|' or '/' are present.

I know UserString exists, but I guess there is an easier and more Pythonish way of doing it, which would be great to learn.

Any help?

you're just creating a new object and naming that object line (the same as was the name of the old object). There's nothing wrong w/ that. All str remain immutable. — SilentGhost
– SilentGhost, Commented Oct 31, 2012 at 12:15
Oh, I guess you are right and my question is stupid then! Sorry! — Roman Rdgz
– Roman Rdgz, Commented Oct 31, 2012 at 12:17
could you use compress = lambda s: s.encode('zip') and uncompress = lambda z: z.decode('zip')? It is inefficient to compress/uncompress byte-by-byte in pure Python — jfs
– jfs, Commented Oct 31, 2012 at 12:31

Tim · Accepted Answer · 2012-10-31 12:22:39Z

2

The changes necessary to get your code running with the sample strings:

Change .index() to .find(). .index() raises an exception if the substring isn't found, .find() returns -1.

Change uncompressedChars.join() to ''.join(uncompressedChars).

Change timesRepeated = int(line[slashIndex+1:verticalBarIndex+3], 16) to timesRepeated = int(line[slashIndex+1:slashIndex+3], 16)

Set uncompressedChars = [] to start with, instead of uncompressedChars = [repeatedChar].

This should get it working properly. There are a lot of places where the code an be tidied and otpimised, but this works.

edited Oct 31, 2012 at 12:22

answered Oct 31, 2012 at 12:16

Tim

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Krumelur · Accepted Answer · 2012-10-31 12:12:44Z

0

The most common pattern I have seen is to use a list of characters. Lists are mutable and work as you describe above.

To create a list from a string

mystring = 'Hello'
mylist = list(mystring)

To create a string from a list

mystring = ''.join(mylist)

answered Oct 31, 2012 at 12:12

Krumelur

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Pavel Anossov · Accepted Answer · 2012-10-31 12:37:13Z

You should build a list of substrings as you go and join them at the end:

def uncompress(line):
  # No error checking, sorry. Will crash with empty strings.
  result = []
  chars = iter(line)
  prevchar = chars.next()   # this is the previous character
  while True:
    try:
      curchar = chars.next()  # and this is the current character
      if curchar == '|':
        # current character is a pipe.
        # Previous character is the character to repeat
        # Get next character, the number of repeats
        curchar = chars.next()
        result.append(prevchar * int(curchar, 16))
      elif curchar == '/':
        # current character is a slash.
        # Previous character is the character to repeat
        # Get two next characters, the number of repeats
        curchar = chars.next()
        nextchar = chars.next()
        result.append(prevchar * int(curchar + nextchar, 16))
      else:
        # No need to repeat the previous character, append it to result.
        result.append(curchar)
      prevchar = curchar
    except StopIteration:
      # No more characters. Append the last one to result.
      result.append(curchar)
      break
  return ''.join(result)

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