neogeek23 - 4 months ago 10x

Python Question

I'm trying to get a list of permutations of x unique characters repeated y times each. So that would look something like this:

`x = ['0', '1']`

y = 2

permutation_list = ['0011','0101','1001','1010','0110','1100']

I don't want extra stuff ('0001', '1110', etc) and I don't want duplicates - does anyone know of a neat way to do this?

I've tried using itertools but I end up with duplicates.

Answer

Use all *unique* permutations of your list, repeated `y`

times:

```
from itertools import permutations
permutation_list = set(permutations(x * y))
```

Demo:

```
>>> from itertools import permutations
>>> x = ['0', '1']
>>> y = 2
>>> set(permutations(x * 2))
{('0', '1', '1', '0'), ('0', '1', '0', '1'), ('1', '0', '1', '0'), ('1', '1', '0', '0'), ('1', '0', '0', '1'), ('0', '0', '1', '1')}
```

These can be mapped back to a list:

```
[''.join(combo) for combo in set(permutations(x * 2))]
```

Source (Stackoverflow)

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