Generating all possible permutations for a given length and set of characters

Question

I'm trying to convert this Python function, which uses the itertools library, to VB.Net:

permutations = itertools.product('ab', repeat=3)

The above function returns all permutations:

[['a','a','a'],
 ['a','a','b'],
 ['a','b','a'],
 ['a','b','b'],
 ['b','a','a'],
 ['b','a','b'],
 ['b','b','a'],
 ['b','b','b']]

Is there a nice clean way to do this in VB.Net?

score 1 · Answer 1 · edited May 23 '17 at 12:23

1

The answer to your question, if I'm not mistaken. - https://stackoverflow.com/a/21090635/2319909

edited May 23 '17 at 12:23

Community

1
1

answered Aug 19 '14 at 17:48

Sam Makin

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i'm looking for some sort of library or one liner native function or may be linq.. something simple – razz Aug 19 '14 at 17:52
OK... This will do it though.. Permutation.Create({"a"c, "b"c, "c"c}, sort:=True) not sweet enough for you? – Sam Makin Aug 19 '14 at 17:54
Add the class that dude provided, then reference it. – Sam Makin Aug 19 '14 at 17:57
it doesn't have option for string length – razz Aug 19 '14 at 17:59
@SamMakin +1 for referencing instead of c&p without attribution:) – Bjørn-Roger Kringsjå Aug 19 '14 at 18:02
Can you clarify what the repeat paramter does? I'm still not sure looking at the documentation. – Sam Makin Aug 19 '14 at 18:06
like say i have 'a' and 'b' and i want to permute them like 'aaaab' 'abaaa'... etc where the output length is 5 here – razz Aug 19 '14 at 18:08

razz · Answer 2 · 2016-02-27T17:42:29.090

I ended up converting the python itertools.product function to vb.net:

Function Permute (chars As List (Of Char), len As Integer) As List (Of List (Of Char))
    Dim pools As New List (Of List (Of Char))
    Dim result As New List (Of List (Of Char)) From {New List (Of Char)}
    For i = 0 To len - 1
        pools.Add (chars)
    Next
    For Each pool As List (Of Char) In pools
        Dim result2 As New List (Of List (Of Char))
        For Each x In result
            For Each y In pool
                Dim temp As New List (Of Char)(x)
                temp.Add (y)
                result2.Add (temp)
            Next
        Next
        result = New List (Of List (Of Char))(result2)
    Next
    Return result
End Function

Generating all possible permutations for a given length and set of characters

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