迭代列表中没有两个重复的行

Question

I would like to iterate over lists of integers in a similar way to:

itertools.product(range(n), repeat=5)

If n = 3 this gives:

[(0, 0, 0, 0, 0),
 (0, 0, 0, 0, 1),
 (0, 0, 0, 0, 2),
 (0, 0, 0, 1, 0),
 (0, 0, 0, 1, 1),
 (0, 0, 0, 1, 2),
 (0, 0, 0, 2, 0),
 (0, 0, 0, 2, 1),
 (0, 0, 0, 2, 2),
 (0, 0, 1, 0, 0),
[...]

However I want only those tuples that don't have the same number twice in a row. So (0,0,1,0,0) would be excluded as would many others.

How can you do this?

Answer 1

It's probably more efficient to generate the sequences without consecutive duplicates yourself, rather than generating all of the sequences with itertools.product and filtering them. I'd use a recursive generator like this:

def gen(seq, n, prefix=()):
    if n == 0:
        yield prefix
        return

    for x in seq:
        if not prefix or x != prefix[-1]:
             yield from gen(seq, n-1, prefix+(x,))

example output:

>>> list(gen(range(3), 5))
[(0, 1, 0, 1, 0),
 (0, 1, 0, 1, 2),
 (0, 1, 0, 2, 0),
 (0, 1, 0, 2, 1),
 (0, 1, 2, 0, 1),
 (0, 1, 2, 0, 2),
 (0, 1, 2, 1, 0),
 (0, 1, 2, 1, 2),
 (0, 2, 0, 1, 0),
 (0, 2, 0, 1, 2),
 (0, 2, 0, 2, 0),
 (0, 2, 0, 2, 1),
 (0, 2, 1, 0, 1),
 (0, 2, 1, 0, 2),
 (0, 2, 1, 2, 0),
 (0, 2, 1, 2, 1),
 (1, 0, 1, 0, 1),
 (1, 0, 1, 0, 2),
 (1, 0, 1, 2, 0),
 (1, 0, 1, 2, 1),
 (1, 0, 2, 0, 1),
 (1, 0, 2, 0, 2),
 (1, 0, 2, 1, 0),
 (1, 0, 2, 1, 2),
 (1, 2, 0, 1, 0),
 (1, 2, 0, 1, 2),
 (1, 2, 0, 2, 0),
 (1, 2, 0, 2, 1),
 (1, 2, 1, 0, 1),
 (1, 2, 1, 0, 2),
 (1, 2, 1, 2, 0),
 (1, 2, 1, 2, 1),
 (2, 0, 1, 0, 1),
 (2, 0, 1, 0, 2),
 (2, 0, 1, 2, 0),
 (2, 0, 1, 2, 1),
 (2, 0, 2, 0, 1),
 (2, 0, 2, 0, 2),
 (2, 0, 2, 1, 0),
 (2, 0, 2, 1, 2),
 (2, 1, 0, 1, 0),
 (2, 1, 0, 1, 2),
 (2, 1, 0, 2, 0),
 (2, 1, 0, 2, 1),
 (2, 1, 2, 0, 1),
 (2, 1, 2, 0, 2),
 (2, 1, 2, 1, 0),
 (2, 1, 2, 1, 2)]

Answer 2

Try:

output = [
          lst for lst in itertools.product(range(3), repeat=3) 
          if not any(i==j for i, j in zip(lst[:-1],lst[1:]))
         ]

Answer 3

You can use a comprensive list with a guard:

n = 3
repeat = 5
[l for l in itertools.product(range(n), repeat=repeat) if not any(l[n-1] == l[n] for n in range(1, repeat))]

This will filter out any line with at least one value identical to the previous one.

Answer 4

If you want to use higher-level functions you can check that the number of groups in a groupby is that same as the repeat.

n = 3
r = 5
out_gen = filter(
    lambda x: len(tuple(itertools.groupby(x)))==r, 
    itertools.product(range(n), repeat=r)
)

Answer 5

Here's another take on the recursive generator approach that is based on excluding the value from the calling recursion rather than passing down the whole result:

def comb(A,n,x=None):
    yield from ([v]+r for v in A if v!=x for r in comb(A,n-1,v)) if n else [[]]

Output:

for p in comb(range(3),5):print(p)
[0, 1, 0, 1, 0]
[0, 1, 0, 1, 2]
[0, 1, 0, 2, 0]
[0, 1, 0, 2, 1]
[0, 1, 2, 0, 1]
[0, 1, 2, 0, 2]
[0, 1, 2, 1, 0]
[0, 1, 2, 1, 2]
[0, 2, 0, 1, 0]
[0, 2, 0, 1, 2]
[0, 2, 0, 2, 0]
[0, 2, 0, 2, 1]
[0, 2, 1, 0, 1]
[0, 2, 1, 0, 2]
[0, 2, 1, 2, 0]
[0, 2, 1, 2, 1]
[1, 0, 1, 0, 1]
[1, 0, 1, 0, 2]
[1, 0, 1, 2, 0]
[1, 0, 1, 2, 1]
[1, 0, 2, 0, 1]
[1, 0, 2, 0, 2]
[1, 0, 2, 1, 0]
[1, 0, 2, 1, 2]
[1, 2, 0, 1, 0]
[1, 2, 0, 1, 2]
[1, 2, 0, 2, 0]
[1, 2, 0, 2, 1]
[1, 2, 1, 0, 1]
[1, 2, 1, 0, 2]
[1, 2, 1, 2, 0]
[1, 2, 1, 2, 1]
[2, 0, 1, 0, 1]
[2, 0, 1, 0, 2]
[2, 0, 1, 2, 0]
[2, 0, 1, 2, 1]
[2, 0, 2, 0, 1]
[2, 0, 2, 0, 2]
[2, 0, 2, 1, 0]
[2, 0, 2, 1, 2]
[2, 1, 0, 1, 0]
[2, 1, 0, 1, 2]
[2, 1, 0, 2, 0]
[2, 1, 0, 2, 1]
[2, 1, 2, 0, 1]
[2, 1, 2, 0, 2]
[2, 1, 2, 1, 0]
[2, 1, 2, 1, 2]