5
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I had to write a code for finding all non-empty sublists partitionings of a list:

def f(s):
    if s:
        for j in range(1,len(s)+1):
            for p in f(s[j:]):
                yield [s[:j]]+p
    else:
        yield []

I managed to shorten it (for Python 3.4+) to:

def f(s, p=[]):
    if s:
        for j in range(1,len(s)+1):
            yield from f(s[j:],p+[s[:j]])
    else:
        yield p

using yield from and an accumulator argument p (prefix).

Any suggestions on how to shorten it even more?

Example output:

>>> for p in f([1,2,3,4]): print(p)
...
[[1], [2], [3], [4]]
[[1], [2], [3, 4]]
[[1], [2, 3], [4]]
[[1], [2, 3, 4]]
[[1, 2], [3], [4]]
[[1, 2], [3, 4]]
[[1, 2, 3], [4]]
[[1, 2, 3, 4]]
>>>
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4
  • 3
    \$\begingroup\$ For starters, you can use single char variable names and drop a bunch of whitespace \$\endgroup\$
    – Sp3000
    Jul 13, 2015 at 13:25
  • 1
    \$\begingroup\$ Can we answer with code in other languages, too? This would work pretty well as a language agnostic challenge. \$\endgroup\$ Jul 13, 2015 at 13:37
  • \$\begingroup\$ @JanDvorak This is asking for help in golfing the python program. \$\endgroup\$
    – Okx
    May 18, 2017 at 15:24
  • \$\begingroup\$ @okx hence my trying to convert this into a regular challenge via my comment. The answer would carry over just fine. \$\endgroup\$ May 18, 2017 at 15:28

1 Answer 1

3
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66 bytes

p=lambda l:[q+[l[i:]]for i in range(len(l))for q in p(l[:i])]or[l]

If the last part is changed from [l] to [[]], then it can also be used with tuples or strings, in addition to lists.

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1

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