How To Get All The Contiguous Substrings Of A String In Python?

The only improvement I could think of is, to use list comprehension like this

def get_all_substrings(input_string):  length = len(input_string)  return [input_string[i:j+1] for i in xrange(length) for j in xrange(i,length)]print get_all_substrings('abcde')

The timing comparison between, yours and mine

def get_all_substrings(string):  length = len(string)  alist = []  for i in xrange(length):    for j in xrange(i,length):      alist.append(string[i:j + 1])   return alistdef get_all_substrings_1(input_string):  length = len(input_string)  return [input_string[i:j + 1] for i in xrange(length) for j in xrange(i,length)]from timeit import timeitprint timeit("get_all_substrings('abcde')", "from __main__ import get_all_substrings")# 3.33308315277print timeit("get_all_substrings_1('abcde')", "from __main__ import get_all_substrings_1")# 2.67816185951

can be done concisely with itertools.combinations

from itertools import combinationsdef get_all_substrings_2(string):    length = len(string) + 1    return [string[x:y] for x, y in combinations(range(length), r=2)]

You could write it as a generator to save storing all the strings in memory at once if you don't need to

def get_all_substrings(string):    length = len(string)    for i in xrange(length):        for j in xrange(i + 1, length + 1):            yield(string[i:j]) for i in get_all_substrings("abcde"):    print i

you can still make a list if you really need one

alist = list(get_all_substrings("abcde"))

The function can be reduced to return a generator expression

def get_all_substrings(s):    length = len(s)    return (s[i: j] for i in xrange(length) for j in xrange(i + 1, length + 1))

Or of course you can change two characters to return a list if you don't care about memory

def get_all_substrings(s):    length = len(s)    return [s[i: j] for i in xrange(length) for j in xrange(i + 1, length + 1)]