Insert a newline character every 64 characters using Python - regex

Question

Using Python I need to insert a newline character into a string every 64 characters. In Perl it's easy:
s/(.{64})/$1\n/
How could this be done using regular expressions in Python?
Is there a more pythonic way to do it?

Answer 1

Same as in Perl, but with a backslash instead of the dollar for accessing groups:
s = "0123456789"*100 # test string
import re
print re.sub("(.{64})", "\\1\n", s, 0, re.DOTALL)
re.DOTALL is the equivalent to Perl's s/ option.

Answer 2

without regexp:
def insert_newlines(string, every=64):
lines = []
for i in xrange(0, len(string), every):
lines.append(string[i:i+every])
return '\n'.join(lines)
shorter but less readable (imo):
def insert_newlines(string, every=64):
return '\n'.join(string[i:i+every] for i in xrange(0, len(string), every))
The code above is for Python 2.x. For Python 3.x, you want to use range and not xrange:
def insert_newlines(string, every=64):
lines = []
for i in range(0, len(string), every):
lines.append(string[i:i+every])
return '\n'.join(lines)
def insert_newlines(string, every=64):
return '\n'.join(string[i:i+every] for i in range(0, len(string), every))

Answer 3

I'd go with:
import textwrap
s = "0123456789"*100
print('\n'.join(textwrap.wrap(s, 64)))

Answer 4

Taking #J.F. Sebastian's solution one step further (this is nearly criminal! :-) ):
import textwrap
s = "0123456789"*100
print textwrap.fill(s, 64)
Look ma... no regexes! because as you know... http://regex.info/blog/2006-09-15/247
Thanks for introducing us to textwrap module... although it's been in Python since 2.3, I wasn't aware of it until now (yes, i'll admit that publically)!!

Answer 5

Tiny, not nice:
"".join(s[i:i+64] + "\n" for i in xrange(0,len(s),64))

Answer 6

I suggest the following method:
"\n".join(re.findall("(?s).{,64}", s))[:-1]
This is, more-or-less, the non-RE method taking advantage of the RE engine for the loop.
On a very slow computer I have as a home server, this gives:
$ python -m timeit -s 's="0123456789"*100; import re' '"\n".join(re.findall("(?s).{,64}", s))[:-1]'
10000 loops, best of 3: 130 usec per loop
AndiDog's method:
$ python -m timeit -s "s='0123456789'*100; import re" 're.sub("(?s)(.{64})", r"\1\n", s)'
1000 loops, best of 3: 800 usec per loop
gurney alex's 2nd/Michael's method:
$ python -m timeit -s "s='0123456789'*100" '"\n".join(s[i:i+64] for i in xrange(0, len(s), 64))'
10000 loops, best of 3: 148 usec per loop
I don't consider the textwrap method to be correct for the specification of the question, so I won't time it.
EDIT
Changed answer because it was incorrect (shame on me!)
EDIT 2
Just for the fun of it, the RE-free method using itertools. It rates third in speed, and it's not Pythonic (too lispy):
"\n".join(
it.imap(
s.__getitem__,
it.imap(
slice,
xrange(0, len(s), 64),
xrange(64, len(s)+1, 64)
)
)
)
$ python -m timeit -s 's="0123456789"*100; import itertools as it' '"\n".join(it.imap(s.__getitem__, it.imap(slice, xrange(0, len(s), 64), xrange(64, len(s)+1, 64))))'
10000 loops, best of 3: 182 usec per loop

Answer 7

itertools has a nice recipe for a function grouper that is good for this, particularly if your final slice is less than 64 chars and you don't want a slice error:
def grouper(iterable, n, fillvalue=None):
"Collect data into fixed-length chunks or blocks"
# grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx
args = [iter(iterable)] * n
return izip_longest(fillvalue=fillvalue, *args)
Use like this:
big_string = <YOUR BIG STRING>
output = '\n'.join(''.join(chunk) for chunk in grouper(big_string, 64))