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