How to create a sequential combined list in python?

Question

I've got a list ['a', 'b', 'c', 'd'] and I need a list ['a', 'ab', 'abc', 'abcd', 'b', 'bc', 'bcd', 'c', 'cd', 'd'].

I've been looking at itertools, but I'm not seeing how to make this work.

For all combinations, the code would be:

from itertools import permutations
stuff = ['a','b','c','d']
for i in range(0, len(stuff)+1):
    for subset in permutations(stuff, i):
           print(subset)

What would I need to do to return only sequential combinations? I guess I could check the order for each permutation as I go, but that doesn't seem to be the best way.

Do you need 'abc' and 'd' there as well? Because otherwise I don't see any logic in your list. — user707650
– user707650, Commented Oct 2, 2015 at 2:55
something like print([''.join(stuff[i:j]) for i in range(len(stuff)) for j in range(i+1, len(stuff)+1)]) — njzk2
– njzk2, Commented Oct 2, 2015 at 2:56
you want combinations, not permutations, or rather "ordered" combinations. See itertools.combinations — Wyrmwood
– Wyrmwood, Commented Oct 2, 2015 at 2:58
I lied, combinations are like [('a', 'b', 'c'), ('a', 'b', 'd'), ('a', 'c', 'd'), ('b', 'c', 'd')] — Wyrmwood
– Wyrmwood, Commented Oct 2, 2015 at 3:04

njzk2 · Accepted Answer · 2015-10-02 03:00:33Z

5

Quite simply:

stuff = ['a','b','c','d']
print([''.join(stuff[i:j]) for i in range(len(stuff)) for j in range(i+1, len(stuff)+1)])

Gives

['a', 'ab', 'abc', 'abcd', 'b', 'bc', 'bcd', 'c', 'cd', 'd']

answered Oct 2, 2015 at 3:00

njzk2

39.4k7 gold badges72 silver badges111 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Joseph Over a year ago

The stuff in the list will be more than one character... Will that cause a problem?

njzk2 Over a year ago

@Joseph I don't understand? the number of character in individual item does not matter.

Joseph Over a year ago

sorry... went brain dead for a sec... saw len() and thought len(characters)

user707650 · Accepted Answer · 2015-10-02 03:00:27Z

1

You could do this with a list in-comprehension:

>>> [''.join(['a', 'b', 'c', 'd'])[i:j+1] for i in range(4) for j in range(i, 4)]
['a', 'ab', 'abc', 'abcd', 'b', 'bc', 'bcd', 'c', 'cd', 'd']

Not sure if you want to do it this way though.

answered Oct 2, 2015 at 3:00

user707650

Comments

dursk · Accepted Answer · 2015-10-02 03:00:58Z

1

I think this should do the trick:

items = ['a', 'b', 'c', 'd']
combinations = []
for i, x in enumerate(items):
    combinations.append(x)
    accum = x
    for y in items[i+1:]:
        accum += y
        combinations.append(accum)

answered Oct 2, 2015 at 3:00

dursk

4,4452 gold badges22 silver badges30 bronze badges

Comments

BrianO · Accepted Answer · 2015-10-02 03:10:50Z

1

This function does it:

def subsequences(lst):
    return [''.join(lst[i: j+1])
            for i in range(len(lst)) 
            for j in range(i, len(lst))]

>>> subsequences(['a', 'b', 'c'])
['a', 'ab', 'abc', 'b', 'bc', 'c']
>>> subsequences(['a', 'b', 'c', 'd'])
['a', 'ab', 'abc', 'abcd', 'b', 'bc', 'bcd', 'c', 'cd', 'd']

edited Oct 2, 2015 at 3:10

answered Oct 2, 2015 at 3:02

BrianO

1,5449 silver badges12 bronze badges

Comments

Óscar López · Accepted Answer · 2015-10-02 03:35:30Z

1

Yet another possible solution (without using itertools), this time using a helper procedure for clarity:

def combine(lst):
    return [''.join(lst[0:i+1]) for i in xrange(len(lst))]

lst = ['a', 'b', 'c', 'd']
sum([combine(lst[i:]) for i in xrange(len(lst))], [])
=> ['a', 'ab', 'abc', 'abcd', 'b', 'bc', 'bcd', 'c', 'cd', 'd']

edited Oct 2, 2015 at 3:35

answered Oct 2, 2015 at 3:18

Óscar López

237k38 gold badges321 silver badges391 bronze badges

Collectives™ on Stack Overflow

How to create a sequential combined list in python?

5 Answers 5

3 Comments

Comments

Comments

Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

3 Comments

Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related