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I've seen many questions on getting all the possible substrings (i.e., adjacent sets of characters), but none on generating all possible strings including the combinations of its substrings.

For example, let:

x = 'abc'

I would like the output to be something like:

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

The main point is that we can remove multiple characters that are not adjacent in the original string (as well as the adjacent ones).

Here is what I have tried so far:

def return_substrings(input_string):
    length = len(input_string)
    return [input_string[i:j + 1] for i in range(length) for j in range(i, length)]

print(return_substrings('abc'))

However, this only removes sets of adjacent strings from the original string, and will not return the element 'ac' from the example above.

Another example is if we use the string 'abcde', the output list should contain the elements 'ace', 'bd' etc.

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  • 2
    Do you really want to do this the hard way? It's very easy to do it with the standard itertools.combinations function, and it will run faster, too. Commented Jul 26, 2018 at 11:57
  • 4
    This is basically a string version of whats a good way to find power_set. stackoverflow.com/questions/1482308/… Commented Jul 26, 2018 at 12:14
  • Those are combinations, not "iterations". A combination is basically a subset where order does not matter. If order matters, it's a permutation. Commented Jul 26, 2018 at 12:17
  • @MikeHousky Thanks for the correction, I'll change the title. Commented Jul 26, 2018 at 12:17
  • 1
    If you ever want to take care of strings like "AAAA" and not have unnecessary duplicates, look at my answer. Commented Jul 26, 2018 at 13:01

5 Answers 5

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28

You can do this easily using itertools.combinations

>>> from itertools import combinations
>>> x = 'abc'
>>> [''.join(l) for i in range(len(x)) for l in combinations(x, i+1)]
['a', 'b', 'c', 'ab', 'ac', 'bc', 'abc']

If you want it in the reversed order, you can make the range function return its sequence in reversed order

>>> [''.join(l) for i in range(len(x),0,-1) for l in combinations(x, i)]
['abc', 'ab', 'ac', 'bc', 'a', 'b', 'c']
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3 Comments

@PM2Ring.. Updated the answer to get the o/p in reversed order too
Also, shouldn't it be len(x) and not len(s) ?
@scharette. Thanks... It should have been len(x).
9

This is a fun exercise. I think other answers may use itertools.product or itertools.combinations. But just for fun, you can also do this recursively with something like

def subs(string, ret=['']):
    if len(string) == 0:
        return ret
    head, tail = string[0], string[1:]
    ret = ret + list(map(lambda x: x+head, ret))
    return subs(tail, ret)

subs('abc')
# returns ['', 'a', 'b', 'ab', 'c', 'ac', 'bc', 'abc']
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2 Comments

Although Python comes with "batteries included", it's certainly good for people to see how to do this task recursively. Now see if you can get the output in the order shown in the question without using sorting. :evil grin:
@PM2Ring Maybe a merge between my answer and his could do it ;)
2

@Sunitha answer provided the right tool to use. I will just go and suggest an improved way while using your return_substrings method. Basically, my solution will take care of duplicates.

I will use "ABCA" in order to prove validity of my solution. Note that it would include a duplicate 'A' in the returned list of the accepted answer.

Python 3.7+ solution,

x= "ABCA"
def return_substrings(x):
    all_combnations = [''.join(l) for i in range(len(x)) for l in combinations(x, i+1)]
    return list(reversed(list(dict.fromkeys(all_combnations))))
    # return list(dict.fromkeys(all_combnations)) for none-reversed ordering

print(return_substrings(x))
>>>>['ABCA', 'BCA', 'ACA', 'ABA', 'ABC', 'CA', 'BA', 'BC', 'AA', 'AC', 'AB', 'C', 'B', 'A']

Python 2.7 solution,

You'll have to use OrderedDict instead of a normal dict. Therefore,

 return list(reversed(list(dict.fromkeys(all_combnations))))

becomes 

return list(reversed(list(OrderedDict.fromkeys(all_combnations))))

Order is irrelevant for you ?

You can reduce code complexity if order is not relevant,

x= "ABCA"
def return_substrings(x):
    all_combnations = [''.join(l) for i in range(len(x)) for l in combinations(x, i+1)]
    return list(set(all_combnations))
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Comments

0 
def return_substrings(s):
    all_sub = set()
    recent = {s}

    while recent:
        tmp = set()
        for word in recent:
            for i in range(len(word)):
                tmp.add(word[:i] + word[i + 1:])
        all_sub.update(recent)
        recent = tmp

    return all_sub
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0

For an overkill / different version of the accepted answer (expressing combinations using https://docs.python.org/3/library/itertools.html#itertools.product ):

["".join(["abc"[y[0]] for y in x if y[1]]) for x in map(enumerate, itertools.product((False, True), repeat=3))]

For a more visual interpretation, consider all substrings as a mapping of all bitstrings of length n.

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