How to output a string with a space between each letter

Question

For example, if I entered I love dogs , it would need to look like this:

I  l o v e  d o g s

This code does not do what I need it to do:

def spaceitout(source):
    pile = ""
    for letter in source:
        pile = pile+letter
        print pile
    print pile

Answer 1

def evenly_spaced(string_,space_= 1):

    import re

    return (' '*space_).join([c for c in re.split(r'(\w)',string_) if c.isalpha()])

print(evenly_spaced(" This a long story ",2))

T  h  i  s  a  l  o  n  g  s  t  o  r  y

Answer 2

Does this do what you need?

pile = ' '.join(source)

This takes the elements of "source" and joins them with a single space as the connector.

If you need only the letters separated, then build the list of letters only, and join that:

pile = ' '.join([c for c in source if c.isalpha()])

Answer 3

Spaces between letters:

def spaceitout(source, count): 
    return (' '*count).join([letter for letter in source.replace(' ','')])

Spaces between words:

def spaceitout(source, count):
    return (' '*count).join(source.split())

Spaces between all chars:

def spaceitout(source, count): 
    return (''.join([c + (' '*count) for c in source]).strip())

Answer 4

Simple answer would be:

def spaceitout(source):
    pile = ""
    for letter in source:
        pile = pile + letter + " "
    pile = pile[:-1] #Strip last extraneous space.
    print pile

Answer 5

Allows you to specify spaces between words, and spaces between characters. Based on the answers provided by BigOldTree

def space_it(text, word_space=1, char_space=0):
    return (' '*word_space).join([(' '*char_space).join(x) for x in text.split(' ')])

Note: This will treat two spaces in the input text as having an "invisible word" between them, change text.split(' ') to text.split() if this isn't desired.

Answer 6

i think that this is what you was looking for:

line = 'I love dogs'
for i in line:
 if i != ' ':
  print i,
 else:
  print '',

Answer 7

using itertools:

import itertools

def space_word(word, spaces_count=1):
    if spaces_count < 1:
        raise ValueError("spaces_count < 1")

    def space_word_wrapped(word, spaces_count):
        letters = itertools.chain.from_iterable(zip(word, itertools.cycle(" ")))
        skipped = 0  # have to keep track of how many letter skips
                     # or else from_word starts breaking!
                     # TODO : better implementation of this
        for i, lett in enumerate(letters, start=1):
            from_word = (i + skipped) % 2
            if lett.isspace() and from_word:
                if spaces_count == 1:
                    next(letters)
                    skipped += 1
                    continue       # and skip the space itself
                elif spaces_count == 2:
                    continue       # just count the space
                elif spaces_count == 3:
                    pass           # count everything
                else:          # spaces_count >= 4
                    yield from [" "] * (spaces_count - 3)
            yield lett

    return ''.join(space_word_wrapped(word, spaces_count)).rstrip()

It's probably cheaper to just use an iterator here, but I like the generator in a nested function approach. So sue me! :)

Answer 8

This lists your word( 'his' = ['h','i','s'] and then joins it with a space instead of a comma.

def space(word):
   return ' '.join(list(word))