V.Petretto - 2 years ago 86

Python Question

using the

`itertools`

`List=[0,0,0,0,3,6,0,0,5,0,0]`

`itertools`

`List=[0,3,0,0,0,6,0,0,5,0,0]`

List=[0,3,0,0,0,6,0,0,5,0,0]

they are the same but

`itertools`

The question is: how can I iterate only some selected numbers and left alone others such like zero ? it can be with or without

`itertools`

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Answer Source

Voilá - it works now - after getting the permutations on the "meat", I further get all possible combnations for the "0"s positions and yield one permutation for each possible set of "0 positions" for each permutation of the non-0s:

```
from itertools import permutations, combinations
def permut_with_pivot(sequence, pivot=0):
pivot_indexes = set()
seq_len = 0
def yield_non_pivots():
nonlocal seq_len
for i, item in enumerate(sequence):
if item != pivot:
yield item
else:
pivot_indexes.add(i)
seq_len = i + 1
def fill_pivots(permutation):
for pivot_positions in combinations(range(seq_len), len(pivot_indexes)):
sequence = iter(permutation)
yield tuple ((pivot if i in pivot_positions else next(sequence)) for i in range(seq_len))
for permutation in permutations(yield_non_pivots()):
for filled_permutation in fill_pivots(permutation):
yield filled_permutation
```

(I've used Python's 3 "nonlocal" keyword - if you are still on Python 2.7,
you will have to take another approach, like making `seq_len`

be a list with a single item you can then repplace on the inner function)

**My second try** (the working one is actually the 3rd)

This is a naive approach that just keeps a cache of the already "seen" permutations - it saves on the work done to each permutation but notonthe work to generate all possible permutations:

```
from itertools import permutations
def non_repeating_permutations(seq):
seen = set()
for permutation in permutations(seq):
hperm = hash(permutation)
if hperm in seen:
continue
seen.add(hperm)
yield permutation
```

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