I"m looking for the Python equivalent of
String str = "many fancy word
hello hi";
String whiteSpaceRegex = "s";
String[] words = str.split(whiteSpaceRegex);
["many", "fancy", "word", "hello", "hi"]
Split string on whitespace in Python split: Questions
How do you split a list into evenly sized chunks?
5 answers
I have a list of arbitrary length, and I need to split it up into equal size chunks and operate on it. There are some obvious ways to do this, like keeping a counter and two lists, and when the second list fills up, add it to the first list and empty the second list for the next round of data, but this is potentially extremely expensive.
I was wondering if anyone had a good solution to this for lists of any length, e.g. using generators.
I was looking for something useful in itertools but I couldn"t find anything obviously useful. Might"ve missed it, though.
Related question: What is the most “pythonic” way to iterate over a list in chunks?
Answer #1
Here"s a generator that yields the chunks you want:
def chunks(lst, n):
"""Yield successive n-sized chunks from lst."""
for i in range(0, len(lst), n):
yield lst[i:i + n]
import pprint
pprint.pprint(list(chunks(range(10, 75), 10)))
[[10, 11, 12, 13, 14, 15, 16, 17, 18, 19],
[20, 21, 22, 23, 24, 25, 26, 27, 28, 29],
[30, 31, 32, 33, 34, 35, 36, 37, 38, 39],
[40, 41, 42, 43, 44, 45, 46, 47, 48, 49],
[50, 51, 52, 53, 54, 55, 56, 57, 58, 59],
[60, 61, 62, 63, 64, 65, 66, 67, 68, 69],
[70, 71, 72, 73, 74]]
If you"re using Python 2, you should use xrange() instead of range():
def chunks(lst, n):
"""Yield successive n-sized chunks from lst."""
for i in xrange(0, len(lst), n):
yield lst[i:i + n]
Also you can simply use list comprehension instead of writing a function, though it"s a good idea to encapsulate operations like this in named functions so that your code is easier to understand. Python 3:
[lst[i:i + n] for i in range(0, len(lst), n)]
Python 2 version:
[lst[i:i + n] for i in xrange(0, len(lst), n)]
Answer #2
If you want something super simple:
def chunks(l, n):
n = max(1, n)
return (l[i:i+n] for i in range(0, len(l), n))
Use xrange() instead of range() in the case of Python 2.x
Answer #3
Directly from the (old) Python documentation (recipes for itertools):
from itertools import izip, chain, repeat
def grouper(n, iterable, padvalue=None):
"grouper(3, "abcdefg", "x") --> ("a","b","c"), ("d","e","f"), ("g","x","x")"
return izip(*[chain(iterable, repeat(padvalue, n-1))]*n)
The current version, as suggested by J.F.Sebastian:
#from itertools import izip_longest as zip_longest # for Python 2.x
from itertools import zip_longest # for Python 3.x
#from six.moves import zip_longest # for both (uses the six compat library)
def grouper(n, iterable, padvalue=None):
"grouper(3, "abcdefg", "x") --> ("a","b","c"), ("d","e","f"), ("g","x","x")"
return zip_longest(*[iter(iterable)]*n, fillvalue=padvalue)
I guess Guido"s time machine works—worked—will work—will have worked—was working again.
These solutions work because [iter(iterable)]*n (or the equivalent in the earlier version) creates one iterator, repeated n times in the list. izip_longest then effectively performs a round-robin of "each" iterator; because this is the same iterator, it is advanced by each such call, resulting in each such zip-roundrobin generating one tuple of n items.
