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Re: Binary Vector Manipulation

  • To: mathgroup at smc.vnet.net
  • Subject: [mg84112] Re: Binary Vector Manipulation
  • From: thomas <thomas.muench at gmail.com>
  • Date: Mon, 10 Dec 2007 20:39:28 -0500 (EST)
  • References: <fjj183$frv$1@smc.vnet.net>

One of several possibilities is

With[{newB = Clip[vectorA + vectorB]},
 ReplacePart[newB,
  RandomSample[Position[Normal@newB, 1], Total[newB] - Total[vectorB]]
->
   0]]

where newB is your new vector with "more than 50%" 1's. It is achieved
by adding vectorA and vectorB, yielding some "2" entries where the "1"
entries between A and b overlapped. Clip[...] corrects for that.

In the second part you randomly replace the appropriate number of 1's
with 0's. You get this number with Total[newB] - Total[vectorB], which
is the number of excess 1's in newB. You need the "Normal@" in
"Position[Normal@newB, 1]" only if your original vectors were
SparseArrays, because Position doesn't work with SparseArrays.

another option [maybe faster, depending on the size of your vectors]:

SparseArray[
 RandomSample[
   SparseArray@Clip[vectorA + vectorB]/.
    SparseArray[_, _, _, p_] :> Flatten@p[[2, 2]], Total[vectorB]] ->
  Table[1, {Total[vectorB]}]]


Hope that helps,
Thomas

On Dec 10, 10:34 am, Tara.Ann.Lor... at gmail.com wrote:
> I am need of assistance programming the following scenario:
>
> I have two vectors composed of 0's and 1's.  Vector "A" has 5% 1's
> (and 95% 0's) while Vector "B" has 50% 1's and 50% 0's.
>
> First, I would like to change Vector B to have a "1" in every place
> that Vector A also has a "1" (in other words, I will then have more
> than 50% 1's in Vector B once this step is completed).
>
> Then, I would like to *randomly* return Vector B back to a 50/50
> distribution of 1's and 0's.
>
> I greatly appreciate any proposed programming methods.
>
> Thank you,
> Tara


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