Question

I have the following code:

def produceCombinations(text, maximumWindowWidth)

  combinations = []

  for windowWidth in range(1,maximumWindowWidth+1):
    for startingIndex in range(len(text)-windowWidth+1):
      combinations.append((text[startingIndex:startingIndex+windowWidth]))

  return combinations

produceCombinations(("1","2","3","4","5","6",),3)

which gives the following output:

('1',)
('2',)
('3',)
('4',)
('5',)
('6',)
('1', '2')
('2', '3')
('3', '4')
('4', '5')
('5', '6')
('1', '2', '3')
('2', '3', '4')
('3', '4', '5')
('4', '5', '6')

However I also want this algorithm to give me the additional combinations:

('1', '3') # hop of 1
('2', '4')
('3', '5')
('4', '6')
('1', '4') # hop of 2
('2', '5')
('3', '6')
('4', '7')
('1', '3', '4') # hop of 1 and 0
('2', '4', '5')
('3', '5', '6')
('1', '2', '4') # hop of 0 and 1
('2', '3', '5') 
('3', '4', '6')
('1', '3', '5') # hop of 1 and 1
('2', '4', '6')
('1', '2', '5') # hop of 0 and 2
('2', '3', '6')
('1', '3', '6') # hop of 1 and 2
('1', '4', '5') # hop of 2 and 0
('2', '5', '6')
('1', '4', '6') # hop of 2 and 1

where my function will have a new argument called maximumHop so as to limit the number of these extra combinations. For the above example the maximumHop is two since the combination (‘1’, ‘5’) is not possible.

Any suggestions regarding a nice way to do this?

Thanks,

Barry

Answer 1

use itertools
see http://docs.python.org/library/itertools.html#itertools.combinations