Recursive algorithm for all k-way partitions of a list

Question

I am writing a function that should output all k-way partitions of a list A. This problem is clearly recursive, and the implementation should be straightforward:

def gen_partition_k_group( A, k):
    #
    if len(A) == 0 :
        # EDITED FOLLOWING SUGGESTION
        yield [ [] for _ in xrange(k) ]
    #
    else :
        # all k-partitions of the list of size N-1
        for ss in gen_partition_k_group(A[:-1], k) :
            assert( sum( len(gg) for gg in ss ) == len(A) -1 )
            for ii in xrange(k) :
                tt = list(ss)
                print tt
                tt[ ii ].append( A[ -1 ] )
                print tt
                assert( sum( len(gg) for gg in tt ) == len(A) )
                yield tt

A = range(3)
k = 2
[ xx for xx in gen_partition_k_group( A, k) ]

Output

AssertionError:

[[], []]

[[0], [0]]

I don't understand the output. It should be [[0], []] instead of [[0], [0]]. What am I missing?

NB: I know how to write a different function without append which outputs the correct result. Iterator over all partitions into k groups? (first answer)

What I don't understand is the behaviour of this particular function.

Answer 1

Ok so the problem was the line tt = list(ss) which is only making shallow copies of the list. Using tt = copy.deepcopy(ss) solved the problem.

Answer 2

One problem is probably that [ [] ] * k doesn't do what you think it does. That doesn't make k empty lists, it makes one new empty list and k references to it. For example:

>>> [[]]*3
[[], [], []]
>>> a = [[]]*3
>>> a
[[], [], []]
>>> a[0].append(1)
>>> a
[[1], [1], [1]]
>>> id(a[0]), id(a[1]), id(a[2])
(25245744, 25245744, 25245744)
>>> a[0] is a[1]
True
>>> a[0] is a[2]
True

To make multiple new lists, you could do something like

>>> a = [[] for _ in xrange(3)]
>>> a
[[], [], []]
>>> id(a[0]), id(a[1]), id(a[2])
(41563560, 41564064, 41563056)

I don't think this by itself will fix your program -- I still get an assert tripping -- but it should help.