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

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)]

Another solution:

def get_all_substrings(string):
   length = len(string)+1
   return [string[x:y] for x in range(length) for y in range(length) if string[x:y]]

print get_all_substring('abcde')

I’ve never been fond of range(len(seq)), how about using enumerate and just using the index value:

def indexes(seq, start=0):
    return (i for i,_ in enumerate(seq, start=start))

def gen_all_substrings(s):
    return (s[i:j] for i in indexes(s) for j in indexes(s[i:], i+1))

def get_all_substrings(string):
    return list(gen_all_substrings(string))

print(get_all_substrings('abcde'))

Use itertools.permutations to generate all pairs of possible start and end indexes,
and filter out only those where the start index is less than then end index. Then
use these pairs to return slices of the original string.

from itertools import permutations

def gen_all_substrings(s):
    lt = lambda pair: pair[0] < pair[1]
    index_pairs = filter(lt, permutations(range(len(s)+1), 2))
    return (s[i:j] for i,j in index_pairs)

def get_all_substrings(s):
    return list(gen_all_substrings(s))

print(get_all_substrings('abcde'))

Python 3

s='abc'
list(s[i:j+1] for i in range (len(s)) for j in range(i,len(s)))

['a', 'ab', 'abc', 'b', 'bc', 'c']

can be done concisely with itertools.combinations

from itertools import combinations

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

Another solution using 2-D matrix approach

p = "abc"
a = list(p)
b = list(p)
c = list(p)
count = 0
for i in range(0,len(a)):
       dump = a[i]
            for j in range(0, len(b)):
                if i < j:
                    c.append(dump+b[j])
                    dump = dump + b[j]  

If you want to get the substrings sorted by the length:

s = 'abcde'
def allSubstrings(s: str) -> List[str]:
    length = len(s)
    mylist = []
    for i in range(1, length+1):
        for j in range(length-i+1):
            mylist.append(s[j:j+i])
    return mylist

print(allSubstrings(s))

['a', 'b', 'c', 'd', 'e', 'ab', 'bc', 'cd', 'de', 'abc', 'bcd', 'cde', 'abcd', 'bcde', 'abcde']