39

Say I have a string s = 'BINGO'; I want to iterate over the string to produce 'B I N G O'.

This is what I did:

result = ''
for ch in s:
   result = result + ch + ' '
print(result[:-1])    # to rid of space after O

Is there a more efficient way to go about this?

Improve this question
10
  • 4
    On my machine, this takes about 500ns. Is that really a bottleneck in your program? If not, don't ask about efficiency; ask about simplicity, readability, etc.—things that actually matter. Commented Aug 14, 2013 at 0:44
  • 3
    @abarnert perhaps efficient means something particular to you, but it doesn't actually necessarily mean machine efficiency. Commented Aug 14, 2013 at 0:45
  • @kojiro: "efficient" means something particular in the disciple/profession/hobby/etc. of computer programming, which is what this site is about, so that's the definition that matters. That's why this site has an efficiency tag that's intended and used for exactly that purpose. Commented Aug 14, 2013 at 0:49
  • 1
    Yes, @abarnert's right -- the problem with this solution isn't first of all that it's inefficient, it's that it's not Pythonic and not simple. ' '.join(s) is the simple, Pythonic way. However, efficiency is a concern, as the above solution will be O(N^2), whereas the join is O(N) -- this won't matter for 'BINGO' but will matter for long strings. Commented Aug 14, 2013 at 2:08
  • 1
    @BenHoyt: The reason it looks linear with smallish strings is that copying a string is basically just a call to memmove—which still loops, of course, but it does so in C code that's usually highly optimized for the platform (especially on x86, which has opcodes specifically designed to speed up memmove). So, the constant multiplier on the second N is orders of magnitude smaller than the one on the first, which makes it hard to see until N gets very large. Commented Aug 15, 2013 at 18:19

5 Answers 5

Reset to default
76 
s = "BINGO"
print(" ".join(s))

Should do it.

Improve this answer
Sign up to request clarification or add additional context in comments.

2 Comments

explanation from str.join documentation: S.join(iterable) -> string Return a string which is the concatenation of the strings in the iterable. The separator between elements is S.
Right, exactly true. Python treats individual strings as an iterable so it will join each character (element in the iterable) with a space.
33 
s = "BINGO"
print(s.replace("", " ")[1: -1])

Timings below

$ python -m timeit -s's = "BINGO"' 's.replace(""," ")[1:-1]'
1000000 loops, best of 3: 0.584 usec per loop
$ python -m timeit -s's = "BINGO"' '" ".join(s)'
100000 loops, best of 3: 1.54 usec per loop
Improve this answer

8 Comments

This is interesting, any idea why the replace method is faster?
@MikeVella At a guess, it might have to do with the fact that Python's source code has a special case replace handler for interleaving when the from string is an empty string.
By the way, I think this answer would be better if it gave timeit values for a few different string sizes.
how can I convert this into putting space every n digits?
@MikeSchem, the [1:-1] doesn't add anything - it removes stuff. If you try without it, you'll see there is are extra spaces at the beginning and end. You could also use .strip() depending on your use case.
|
3

The Pythonic way

A very pythonic and practical way to do it is by using the string join() method:

str.join(iterable)

The official Python documentations says:

Return a string which is the concatenation of the strings in iterable... The separator between elements is the string providing this method.

How to use it?

Remember: this is a string method.

This method will be applied to the str above, which reflects the string that will be used as separator of the items in the iterable.

Let's have some practical example!

iterable = "BINGO"
separator = " " # A whitespace character.
                # The string to which the method will be applied
separator.join(iterable)
> 'B I N G O'

In practice you would do it like this:

iterable = "BINGO"    
" ".join(iterable)
> 'B I N G O'

But remember that the argument is an iterable, like a string, list, tuple. Although the method returns a string.

iterable = ['B', 'I', 'N', 'G', 'O']    
" ".join(iterable)
> 'B I N G O'

What happens if you use a hyphen as a string instead?

iterable = ['B', 'I', 'N', 'G', 'O']    
"-".join(iterable)
> 'B-I-N-G-O'
Improve this answer

Comments

1

The most efficient way is to take input make the logic and run

so the code is like this to make your own space maker

need = input("Write a string:- ")
result = ''
for character in need:
   result = result + character + ' '
print(result)    # to rid of space after O

but if you want to use what python give then use this code

need2 = input("Write a string:- ")

print(" ".join(need2))
Improve this answer

Comments

0 

def space_the_chars(string):
    string = string.upper().replace(' ','')
    return string.replace('','  ')[2:-2]


space_the_chars(“I’m going to be spaced”)

the function spaces the string’s characters

Improve this answer

Comments

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.