Question

Follows on from this question Algorithm to generate (not quite) spanning set in Python

Given n items in a list, for example n = 7, so this input: [1,2,3,4,5,6,7]

I’d like to generate this set of sets in python:

[1] [2] [3] [4] [5] [6] [7]
[1] [2] [3] [4] [5] [6, 7]
[1] [2] [3] [4] [5, 6, 7]
...
...
[1, 2, 3, 4, 5] [6, 7]

But not:

[1, 2, 3, 4, 5, 6] [7]
or
[1, 2, 3, 4, 5, 6, 7]

From the previous question I have this great answer:

def span(lst):
  yield [lst]
  for i in range(1, len(lst)):
    for x in span(lst[i:]):
      yield [lst[:i]] + x

Is it possible to work within this existing code to produce this more specific output

Thanks

Answer 1

While you could of course just filter the result of the original function using a comprehension like (s for s in span(lst) if max(len(g) for g in s) <= 5), here I present you a recursive solution that doesn’t create the invalid results in the first place:

def span(lst, most=float("inf")):
  if not lst:
    yield []
    return

  for i in range(1, min(len(lst), most) + 1):
    for x in span(lst[i:], most):
      yield [lst[:i]] + x

lst = [1,2,3,4,5,6,7]
n = 5
spannings = list(span(lst, n))         
print '\n'.join(map(str, spannings))

# proof that it produces the correct result
assert sorted(spannings) == sorted(s for s in span_orig(lst) if max(map(len, s)) <= n)
# proof that it produces the same result as the old
# function if called without a second argument
assert sorted(span_orig(lst)) == sorted(span(lst))

The logic is very similar, although we can’t make use of the yield [lst] “trick” to avoid having to explicitly state the terminating condition len(lst) == 0. I actually think this is cleaner, overall 🙂

Note how this is a generalization of the original function: If you don’t give it a second argument, it will work the same way the old function did.