使用递归计算长度为n的列表中的长度k的组合

Question

I need to generate all the combinations with length k from a list of length n , and I must do it using recursion.

For Example:

INPUT:  choose_sets([1,2,3,4],3)
OUTPUT: [[1,2,3],[1,2,4],[1,3,4],[2,3,4]]

INPUT:  choose_sets([1,2,3,4],2)
OUTPUT: [[1,2],[1,3],[1,4],[2,3],[2,4],[3,4]]

I'm stuck implementing this in code, so I would be happy for some help. This is my code so far (I'm missing something just don't know what):

def choose_sets(lst,k):

    if k == len(lst):
        return lst
    if k == 0:
        return []
    if k > len(lst):
        return []

    sets=[]
    sub_lst=lst[:]
    sub_lst.remove(sub_lst[0])

    a= choose_sets(sub_lst,k-1)
    for i in a:
        i.append(lst[0])
    sets.append(a)

    b= choose_sets(sub_lst,k)
    sets.append(b)


    return sets

Answer 1

You can get solution from Generator for permutations, combinations, selections of a sequence (Python recipe)

def xuniqueCombinations(items, n):
    if n==0: yield []
    else:
        for i in xrange(len(items)):
            for cc in xuniqueCombinations(items[i+1:],n-1):
                yield [items[i]]+cc



>>> def xuniqueCombinations(items, n):
...     if n==0: yield []
...     else:
...         for i in xrange(len(items)):
...             for cc in xuniqueCombinations(items[i+1:],n-1):
...                 yield [items[i]]+cc
... 
>>> for x in xuniqueCombinations( [1,2,3,4],2):
...     print x
[1, 2]
[1, 3]
[1, 4]
[2, 3]
[2, 4]
[3, 4]

Edited 4 year later (7/12/2015)

To run it on Python3 just change xrange to range , Python3's range is Python2's xrange. . Thanks @ederollora to notice me.

Answer 2

This is in Java, and I can't guarantee it works 100% properly, but based on quick prototyping seemed to work ok. Hope this helps a bit in any case.

public void choose_sets(int values[], int count) {
    int perm[] = new int[count];
    choose_sets(values, 0, perm, 0, count);
}

public void choose_sets(int[] values, int valuesIdx, int[] perm,
                        int permIdx, int count) {
    if (permIdx == count) {
        // At this point perm -array contains single permutation
        // of length ´count´.
    } else {
        for (int i = valuesIdx; i < values.length; ++i) {
            perm[permIdx] = values[i];
            choose_sets(values, i + 1, perm, permIdx + 1, count);
        }
    }
}

Answer 3

Give a look to this solution:

def choose_sets(mylist,length):
    mylen = len(mylist)

    if length == mylen:
        return [mylist]
    if length == 1:
        return [[i] for i in mylist]
    if length > mylen:
        return []

    ToRet = []
    for k in xrange(mylen): 
        if mylen - k + 1> length :
            for j in choose_sets(mylist[k+1:],length-1):
                New = [mylist[k]]
                New.extend(j)
                ToRet.append(New)
    return ToRet

print choose_sets([1,2,3,4,5],3)

there are more elegant ways, but this should be ok as homework...

Answer 4

You are almost there, just a few minor things. The algorithm is basically correct, but

if k == len(lst):
    return lst

This has the wrong type. The return type is not a list of thing , but a list of (list of thing ), so that should be

if k == len(lst):
    return [lst]

Next,

if k == 0:
    return []

Every list has exactly one nonempty sublist, the empty list, so that ought to be

if k == 0:
    return [[]]

For the rest,

if k > len(lst):
    return []

is completely correct.

sets=[]
sub_lst=lst[:]
sub_lst.remove(sub_lst[0])

That is correct but could be put more succinctly as

sub_lst = lst[1:]

Now, another type mix-up:

a= choose_sets(sub_lst,k-1)
for i in a:
    i.append(lst[0])
sets.append(a)

That sets.append(a) puts a into one slot of sets , you want to concatenate the two lists, sets = sets + a . And if you would like the combinations in the order in which elements appear in the list, instead of i.append(lst[0]) , you should append [lst[0]] + i to sets in the loop, but that's a matter of inclination.

b= choose_sets(sub_lst,k)
sets.append(b)

Again, do not append, but concatenate here,

sets = sets + b

Answer 5

basically you need to use the following recursion:

f(k,n) = append_to_each( f(k-1,n-1), n) | f(k,n-1)

def combinations(lst,k):
    n = len(lst)
    if n == k:
        return [set(lst)]
    if k == 1:
        return [set([lst[i]]) for i in range(n)]
    v1 = combinations(lst[:-1], k-1)
    v1new = [ i.add(lst[n-1]) for i in v1]
    v2 = combinations(lst[:-1], k)
    return v1+v2