Sign Up

Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.

Have an account? Sign In

Have an account? Sign In Now

Sign In

Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.

Sign Up Here

Forgot Password?

Don't have account, Sign Up Here

Forgot Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Have an account? Sign In Now

You must login to ask a question.

Forgot Password?

Need An Account, Sign Up Here

Please briefly explain why you feel this question should be reported.

Please briefly explain why you feel this answer should be reported.

Please briefly explain why you feel this user should be reported.

Sign InSign Up

The Archive Base

The Archive Base Logo The Archive Base Logo

The Archive Base Navigation

  • SEARCH
  • Home
  • About Us
  • Blog
  • Contact Us
Search
Ask A Question

Mobile menu

Close
Ask a Question
  • Home
  • Add group
  • Groups page
  • Feed
  • User Profile
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Buy Points
  • Users
  • Help
  • Buy Theme
  • SEARCH
Home/ Questions/Q 3355166
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: May 18, 20262026-05-18T02:21:43+00:00 2026-05-18T02:21:43+00:00

I have been working on the problems presented in Python Challenge . One of

  • 0

I have been working on the problems presented in Python Challenge. One of the problems asks to sift through a mess of characters and pick out the rarest character/s.

My methodology was to read the characters from a text file, store the characters/occurrence as a key/value pair in a dictionary. Sort the dictionary by value and invert the dictionary where the occurrence is the key and the string of characters is the value. Assuming that the rarest character occurs only once, I return the values where the key of this inverted dictionary equals one.

The input(funkymess.txt) is like this:

%%$@$^_#)^)&!_+]!*@&^}@@%%+$&[(_@%+%$*^@$^!+]!&#)*}{}}!}]$[%}@[{@#_^{*……

The code is as follows:

from operator import itemgetter
characterDict = dict()

#put the characters in a dictionary
def putEncounteredCharactersInDictionary(lineStr):
    for character in lineStr:
        if character in characterDict:
            characterDict[character] = characterDict[character]+1
        else:
            characterDict[character] = 1

#Sort the character dictionary
def sortCharacterDictionary(characterDict):
    sortCharDict = dict()
    sortsortedDictionaryItems = sorted(characterDict.iteritems(),key = itemgetter(1))
    for key, value in sortsortedDictionaryItems:
        sortCharDict[key] = value
    return sortCharDict 

#invert the sorted character dictionary
def inverseSortedCharacterDictionary(sortedCharDict):
    inv_map = dict()
    for k, v in sortedCharDict.iteritems():
        inv_map[v] = inv_map.get(v, [])
        inv_map[v].append(k)
    return inv_map


f = open('/Users/Developer/funkymess.txt','r')
for line in f:
    #print line
    processline = line.rstrip('\n')
    putEncounteredCharactersInDictionary(processline)
f.close()

sortedCharachterDictionary = sortCharacterDictionary(characterDict)
#print sortedCharachterDictionary
inversedSortedCharacterDictionary = inverseSortedCharacterDictionary(sortedCharachterDictionary)
print inversedSortedCharacterDictionary[1]r

Can somebody take a look and provide me with some pointers on whether I am on the right track here and if possible provide some feedback on possible optimizations/best-practices and potential refactorings both from the language as well as from an algorithmic standpoint.

Thanks

  • 1 1 Answer
  • 0 Views
  • 0 Followers
  • 0
Share
  • Facebook
  • Report

Leave an answer
Cancel reply

You must login to add an answer.

Forgot Password?

Need An Account, Sign Up Here

1 Answer

  • Voted
  • Oldest
  • Recent
  • Random
  1. Editorial Team
    Editorial Team
    2026-05-18T02:21:44+00:00Added an answer on May 18, 2026 at 2:21 am

    Refactoring: A Walkthrough

    I want to walk you through the process of refactoring. Learning to program is not just about knowing the end result, which is what you usually get when you ask a question on Stack Overflow. It’s about how to get to that answer yourself. When people post short, dense answers to a question like this it’s not always obvious how they arrived at their solutions.

    So let’s do some refactoring and see what we can do to simplify your code. We’ll rewrite, delete, rename, and rearrange code until no more improvements can be made.

    Simplify your algorithms

    Python need not be so verbose. It is usually a code smell when you have explicit loops operating over lists and dicts in Python, rather than using list comprehensions and functions that operate on containers as a whole.

    Use defaultdict to store character counts

    A defaultdict(int) will generate entries when they are accessed if they do not exist. This let’s us eliminate the if/else branch when counting characters.

    from collections import defaultdict
    characterDict = defaultdict(int)
    
    def putEncounteredCharactersInDictionary(lineStr):
        for character in lineStr:
            characterDict[character] += 1
    

    Sorting dicts

    Dictionaries don’t guarantee any ordering on their keys. You cannot assume that the items are stored in the same order that you insert them. So sorting the dict entries and then putting them right back into another dict just scrambles them right back up.

    This means that your function is basically a no-op. After you sort the items you will need to keep them as a list of tuples to retain their sorting order. Removing that code we can then reduce this method down to a single line.

    def sortCharacterDictionary(characterDict):
        return sorted(characterDict.iteritems(), key=itemgetter(1))
    

    Inverting dicts

    Given the previous comment you won’t actually have a dict any more after sorting. But assuming you did, this function is one of those cases where explicit looping is discouraged. In Python, always be thinking how you can operate over collections all at once rather than one item at a time.

    def inverseSortedCharacterDictionary(sortedCharDict):
        return dict((v, k) for k, v in sortedCharDict.iteritems())
    

    All in one line we (1) iterate over the key/value pairs in the dict; (2) switch them and create inverted value/key tuples; (3) create a dict out of these inverted tuples.

    Comment and name wisely

    Your method names are long and descriptive. There’s no need to repeat the same information in comments. Use comments only when your code isn’t self-descriptive, such as when you have a complex algorithm or an unusual construct that isn’t immediately obvious.

    On the naming front, your names are unnecessarily long. I would stick with far less descriptive names, and also make them more generic. Instead of inverseSortedCharacterDictionary, try just invertedDict. That’s all that method does, it inverts a dict. It doesn’t actually matter if it’s passed a sorted character dict or any other type of dict.

    As a rule of thumb, try to use the most generic names possible so that your methods and variables can be as generic as possible. More generic means more reusable.

    characters = defaultdict(int)
    
    def countCharacters(string):
        for ch in string:
            characters[ch] += 1
    
    def sortedCharacters(characters):
        return sorted(characters.iteritems(), key=itemgetter(1))
    
    def invertedDict(d):
        return dict((v, k) for k, v in d.iteritems())
    

    Reduce volume

    Using temporary variables and helper methods is a good programming practice, and I applaud you for doing so in your program. However, now that we have them simple enough that each one is only one or two lines we probably don’t even need them any more.

    Here’s your program body after changing the functions as above:

    f = open('funkymess.txt', 'r')
    
    for line in f:
        countCharacters(line.rstrip('\n'))
    
    f.close()
    
    print sortedCharacters(characters)[0]
    

    And then let’s just go ahead and inline those helper methods since they’re so simple. Here’s the final program after all the refactoring:

    Final program

    #!/usr/bin/env python
    
    from operator import itemgetter
    from collections import defaultdict
    
    characters = defaultdict(int)
    
    f = open('funkymess.txt','r')
    
    for line in f:
        for ch in line.rstrip('\n'):
            characters[ch] += 1
    
    f.close()
    
    print sorted(characters.iteritems(), key=itemgetter(1))[0]
    
    • 0
    • Reply
    • Share
      Share
      • Share on Facebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
      • Report

Sidebar

Related Questions

I'm working on http://lobounce.com.s138810.gridserver.com/ and have been having some problems with the slider which
Okay this one has me stumped.. mainly because i have been working on this
I have been working on this problem for 2 days now and it's an
I have been working on this problem for a while now. I am trying
I have been working on the following problem from this book . A certain
I have been working for a couple of days on a problem with my
I've been working on this problem for a few hours and have used many
I have a homework problem here I've been working on; here is the description:
I've been working with R just a few months, I have a problem with
Here's a tricky iPhone problem I've been working on. I have three UIScrollViews on

Explore

  • Home
  • Add group
  • Groups page
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Users
  • Help
  • SEARCH

Footer

© 2021 The Archive Base. All Rights Reserved
With Love by The Archive Base

Insert/edit link

Enter the destination URL

Or link to existing content

    No search term specified. Showing recent items. Search or use up and down arrow keys to select an item.