Combinations without using “itertools.combinations

问题:

What I'd need is to create combinations for two elements a time.

if a list contains: seq = ['A', 'B', 'C'] the output would be com = [['A', 'B'], ['A', 'C'], ['B', 'C']]

all this without "itertools.combinations" method.

I was used to use this code for the permutations. But how could I modify the code to make it work with the combinations?

def permute(seq):

    if len(seq) <= 1:
        perms = [seq]
    else:
        perms = []
        for i in range(len(seq)):
            sub = permute(seq[:i]+seq[i+1:]) 
            for p in sub:    
                perms.append(seq[i:i+1]+p)

return perms

回答1:

If you don't want to use itertools, then use the documented pure-Python equivalent:

def combinations(iterable, r):
    # combinations('ABCD', 2) --> AB AC AD BC BD CD
    # combinations(range(4), 3) --> 012 013 023 123
    pool = tuple(iterable)
    n = len(pool)
    if r > n:
        return
    indices = range(r)
    yield tuple(pool[i] for i in indices)
    while True:
        for i in reversed(range(r)):
            if indices[i] != i + n - r:
                break
        else:
            return
        indices[i] += 1
        for j in range(i+1, r):
            indices[j] = indices[j-1] + 1
        yield tuple(pool[i] for i in indices)

回答2:

This is trivial to do directly:

def comb2(s):
    for i, v1 in enumerate(s):
        for j in range(i+1, len(s)):
            yield [v1, s[j]]

Then:

print list(comb2(['A', 'B', 'C']))

displays:

[['A', 'B'], ['A', 'C'], ['B', 'C']]

The only thing that makes combinations at all tricky is catering to a known-only-at-runtime number of elements to take at a time. So long as you know that number in advance, a fixed number of loops nested that deep does the job in an obvious way.

回答3:

def combo2(lst,n):
    if n==0:
        return [[]]
    l=[]
    for i in range(0,len(lst)):
        m=lst[i]
        remLst=lst[i+1:]
        for p in combo2(remLst,n-1):
            l.append([m]+p)
    return l

Input:

combo2(list('ABC'),2)

Output:

[['A', 'B'], ['A', 'C'], ['B', 'C']]

回答4:

Here's a loose translation of the recursive C++ code in an answer to a similar question:

def combinations(sequence, length, NULL=object()):
    """ Find all combinations of the specified length from a sequence. """
    if length <= 0:
        combos = [NULL]
    else:
        combos = []
        for i, item in enumerate(sequence, 1):
            rem_items = sequence[i:]
            rem_combos = combinations(rem_items, length-1)
            combos.extend(item if combo is NULL else [item, combo]
                            for combo in rem_combos)
    return combos

print(combinations(['A', 'B', 'C'], 2))

Output:

[['A', 'B'], ['A', 'C'], ['B', 'C']]