Generate kth combination without generating/iterating previous

问题

Given a set of items, for example:

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

I'd like to generate all possible combinations of a certain length with repetition. The twist is I'd like to start at a predetermined combination (a sort of offset into the list of combinations).

For example, starting with this:

[ 1, 5, 6 ]

The first (next) combination would be:

[ 1, 6, 6 ]

I've had success using itertools.combinations_with_replacement() to generate the combinations, but the project this is for will require working with a set that generates way too many combinations - creating them all first and iterating to the correct point is not possible.

I've found this example for generating kth combination which doesn't seem to be working very well for me. This answer seemed another possibility, but I can't seem to port it from C to Python.

Here's my code so far using the kth combination example:

import operator as op
items = [ 1,2,3,4,5,6 ]

# https://stackoverflow.com/a/4941932/1167783
def nCr(n, r):
    r = min(r, n-r)
    if r == 0: 
        return 1
    numer = reduce(op.mul, xrange(n, n-r, -1))
    denom = reduce(op.mul, xrange(1, r+1))
    return numer // denom

# https://stackoverflow.com/a/1776884/1167783
def kthCombination(k, l, r):
    if r == 0:
        return []
    elif len(l) == r:
        return l
    else:
        i = nCr(len(l)-1, r-1)
        if k < i:
            return l[0:1] + kthCombination(k, l[1:], r-1)
        else:
            return kthCombination(k-i, l[1:], r)

# get 1st combination of 3 values from list 'items' 
print kthCombination(1, items, 3)

# returns [ 1, 2, 4 ]

Any help would be great!

回答1:

Instead of inventing the wheel time number 37,289,423,987,239,489,826,364,653 (that's counting only human beings), you can map the numbers. itertools will return first combination [1,1,1], but you want [1,5,6]. Simply add [0,4,5] mod 6 to each position. You can also map back and forth between numbers, objects, and modulo, of course.

This works even if the number of elements in each position is different.

You will have more fun with what you started, though.

回答2:

If you assume that all values in the array are digits in a base-n numbering system where n is the length of the array, the k-th combination will be the equivalent of k expressed in base-n.

If you are wanting to start with a given combination (i.e. [1,6,5]) and continue from there, simply read this starting point as a number in base-n. You can then start iterating through successive combinations by incrementing.

EDIT: Further explanation:

Let's start with the array. The array contains 6 values, so we are working in base-6. We will assume the index of each element in the array is the element's base-6 value.

Values in base-6 range from 0 to 5. This may get confusing because our example uses digits, but we could do this with combinations of anything. I will put 'quote' marks around the digits we are combining.

Given a combination ['1', '6', '5'], we first need to convert this to a base-6 value. '1' becomes 0, '6' becomes 5 and '5' becomes 4. Using their positions in the starting value as their powers in base-6, we get:

(0 * 6^0) + (5 * 6^1) + (4 * 6^2) = 174 (decimal)

If we want to know the next combination, we can add 1. If we want to know 20 combinations ahead, we add 20. We can also subtract to go backwards. Let's add 1 to 174 and convert it back to base-6:

175 (decimal) = (1 + 6^0) + (5 * 6^1) + (4 * 6^2) = 451 (base-6) = ['2', '6', '5'] (combination)

For more on number bases, see http://en.wikipedia.org/wiki/Radix and http://en.wikipedia.org/wiki/Base_%28exponentiation%29

来源：https://stackoverflow.com/questions/18523641/generate-kth-combination-without-generating-iterating-previous

标签

python

combinations

itertools