How to replace every third word in a string with the # length equivalent

user_18506239 0
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Input:

string = "My dear adventurer, do you understand the nature of the given discussion?"

expected output:

string = 'My dear ##########, do you ########## the nature ## the given ##########?'

How can you replace the third word in a string of words with the # length equivalent of that word while avoiding counting special characters found in the string such as apostrophes('), quotations("), full stops(.), commas(,), exclamations(!), question marks(?), colons(:) and semicolons (;).

I took the approach of converting the string to a list of elements but am finding difficulty filtering out the special characters and replacing the words with the # equivalent. Is there a better way to go about it?

filter list python replace string
5 Answers

With help of some regex. Explanation in the comments.

import re


imp = "My dear adventurer, do you understand the nature of the given discussion?"
every_nth = 3  # in case you want to change this later

out_list = []

# split the input at spaces, enumerate the parts for looping
for idx, word in enumerate(imp.split(' ')):

    # only do the special logic for multiples of n (0-indexed, thus +1)
    if (idx + 1) % every_nth == 0:
        # find how many special chars there are in the current segment
        len_special_chars = len(re.findall(r'[.,!?:;\'"]', word))  
                                            # ^ add more special chars here if needed
        
        # subtract the number of special chars from the length of segment
        str_len = len(word) - len_special_chars
        
        # repeat '#' for every non-special char and add the special chars
        out_list.append('#'*str_len + word[-len_special_chars] if len_special_chars > 0 else '')
    else:
        # if the index is not a multiple of n, just add the word
        out_list.append(word)
        

print(' '.join(out_list))
user_144476140

There are more efficient ways to solve this question, but I hope this is the simplest!

My approach is:

  1. Split the sentence into a list of the words
  2. Using that, make a list of every third word.
  3. Remove unwanted characters from this
  4. Replace third words in original string with # times the length of the word.

Here's the code (explained in comments) :

# original line
line = "My dear adventurer, do you understand the nature of the given discussion?"

# printing original line
print(f'\n\nOriginal Line:\n"{line}"\n')

# printing somehting to indicate that next few prints will be for showing what is happenning after each lone
print('\n\nStages of parsing:')

# splitting by spaces, into list
wordList = line.split(' ')

# printing wordlist
print(wordList)

# making list of every third word
thirdWordList = [wordList[i-1] for i in range(1,len(wordList)+1) if i%3==0]

# pritning third-word list
print(thirdWordList)

# characters that you don't want hashed
unwantedCharacters = ['.','/','|','?','!','_','"',',','-','@','\n','\\',':',';','(',')','<','>','{','}','[',']','%','*','&','+']

# replacing these characters by empty strings in the list of third-words
for unwantedchar in unwantedCharacters:
    for i in range(0,len(thirdWordList)):
        thirdWordList[i] = thirdWordList[i].replace(unwantedchar,'')

# printing third word list, now without punctuation 
print(thirdWordList)

# replacing with #
for word in thirdWordList:
    line = line.replace(word,len(word)*'#')

# Voila! Printing the result:
print(f'\n\nFinal Output:\n"{line}"\n\n')

Hope this helps!

user_170488340

Following works and does not use regular expressions

special_chars = {'.','/','|','?','!','_','"',',','-','@','\n','\\'}

def format_word(w, fill):
    if w[-1] in special_chars:
        return fill*(len(w) - 1) + w[-1]
    else:
        return fill*len(w)


def obscure(string, every=3, fill='#'):
    return ' '.join(
        (format_word(w, fill) if (i+1) % every == 0 else w)
        for (i, w) in enumerate(string.split())
    )

Here are some example usage

In [15]: obscure(string)
Out[15]: 'My dear ##########, do you ########## the nature ## the given ##########?'

In [16]: obscure(string, 4)
Out[16]: 'My dear adventurer, ## you understand the ###### of the given ##########?'

In [17]: obscure(string, 3, '?')
Out[17]: 'My dear ??????????, do you ?????????? the nature ?? the given ???????????'
user_61535620

I solved it with:

s  = "My dear adventurer, do you understand the nature of the given discussion?"

def replace_alphabet_with_char(word: str, replacement: str) -> str:
    new_word = []
    alphabet = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ'
    for c in word:
        if c in alphabet:
            new_word.append(replacement)
        else:
            new_word.append(c)
    return "".join(new_word)

every_nth_word = 3
s_split = s.split(' ')
result = " ".join([replace_alphabet_with_char(s_split[i], '#') if i % every_nth_word == every_nth_word - 1 else s_split[i] for i in range(len(s_split))])
print(result)

Output: My dear ##########, do you ########## the nature ## the given ##########?

user_54732670

A mixed of regex and string manipulation

import re
string = "My dear adventurer, do you understand the nature of the given discussion?"

new_string = []
for i, s in enumerate(string.split()):
    if (i+1) % 3 == 0:
        s = re.sub(r'[^\.:,;\'"!\?]', '#', s)
    new_string.append(s)

new_string = ' '.join(new_string)
print(new_string)
user_164628780
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