# [erlang-questions] Interleave binaries

Bob Cowdery bob@REDACTED
Mon Aug 8 10:26:14 CEST 2011

```Jeff

Thanks very much for the suggestions. I will give them a whirl.

Bob

On 08/08/2011 02:33, Jeff Schultz wrote:
> On Fri, Aug 05, 2011 at 10:26:16PM +0100, Bob Cowdery wrote:
>> I'm wanting to interleave two binaries.
>>
>> Bin1 = <<0,1>>,
>> Bin2 = <<2,3>>,
>> << <<A,B,C,D>> || <<A,B>> <= Bin1, <<C,D>> <= Bin2 >>.
>> <<0,1,2,3>>
>>
>> Bin1 = <<0,1,2,3>>,
>> Bin2 = <<4,5,6,7>>,
>> << <<A,B,C,D>> || <<A,B>> <= Bin1, <<C,D>> <= Bin2 >>.
>> <<0,1,4,5,0,1,6,7,2,3,4,5,2,3,6,7>>
>>
>> whereas I want:
>> <<0,1,4,5,2,3,6,7>>
>>
>> I know what it does is what the documentation says but is there a way to
>> get a pure interleave efficiently. Perhaps there is a filter that would
>> work or a way to reduce the binary after merging without stepping
>> through the whole thing.
> I'd go via an intermediate list:
>
> 25> Bin1 = <<0,1,2,3>>, Bin2 = <<4,5,6,7>>.
> <<4,5,6,7>>
> 26> list_to_binary(
> 	[[binary:part(Bin1, I, 2), binary:part(Bin2, I, 2)]
> 	|| I <- lists:seq(0, byte_size(Bin1) - 1, 2)]).
> <<0,1,4,5,2,3,6,7>>
>
> Makes O(N) garbage.
>
>
> If you want a bit string comprehension, try
>
> 27> << <<P1/binary, P2/binary>>
>     || I <- lists:seq(0, byte_size(Bin1) - 1, 2),
>     P1 <- [binary:part(Bin1, I, 2)],
>     P2 <- [binary:part(Bin2, I, 2)]>>.
> <<0,1,4,5,2,3,6,7>>
>
> (I may be missing some syntax trick here.  Ugly, isn't it.)
>
> I have no idea how efficient that is.  The naive implementation would
> be O(N^2).  Even a clever one would make O(N) garbage, though unlike
> the list comprehension, it might not hang on to all of it at once.
>
>
>     Jeff Schultz

```

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