[R] bitwise addition

[Nameeta Lobo]

Hi again

Sorry I should have been more specific the first time. I am not actually trying
to generate this matrix but what am trying to do is to write a loop which takes
an input of 0000, does a bitwise addition, then checks to see if the new number
generated has two 1s(I just summed the values to get 2) and if it does outputs
that to a matrix. It goes on doing it till it reaches 1111.

So my final output for the above would read
0011
0101
0110
1001
1010
1100

For 2^6, I was looking for three 1s and so on. 

I wrote this in a for loop etc. Created the original matrix with expand.grid
command that you guys helped me out with before. 

My only concern was that for e.g. 2^25 or higher, not much space can be
allocated to the creation of the original matrix and then also because of the
loops there is the problem of computer time. So was just wondering if it was
easier to do it via bitwise addition.

thanks a lot for your help everyone.

any more suggestion!!!!

nameeta





Quoting "Marc Schwartz (via MN)" <mschwartz at mn.rr.com>:

> On Fri, 2006-05-12 at 14:05 -0500, Marc Schwartz (via MN) wrote:
> > On Fri, 2006-05-12 at 11:41 -0500, Nameeta Lobo wrote:
> > > 
> > > Hello all again,
> > > 
> > > I want to do bitwise addition in R. I am trying to generate a matrix
> > > 0000
> > > 0001
> > > 0010
> > > ....
> > > ....
> > > 1111
> > > 
> > > I know the other ways of generating this matrix but I need to look at
> bitwise
> > > addition. 
> > > 
> > > Any suggestions???
> > > 
> > > thanks a lot
> > > 
> > > Nameeta
> > 
> > Nameeta,
> > 
> > I may be misunderstanding what you are trying to do, so here are two
> > approaches that might be helpful:
> > 
> > 1. Presuming that each of the above rows is a binary representation of a
> > number x >= 0 (so we don't have to worry about two's complements) and
> > that you want to add the rows to get a total, you can do:
> > 
> > > mat
> >      [,1] [,2] [,3] [,4]
> > [1,]    0    0    0    0
> > [2,]    0    0    0    1
> > [3,]    0    0    1    0
> > [4,]    1    1    1    1
> > 
> > # This will convert each row to it's base 10 value
> > > apply(mat, 1, function(x) sum(x * (2 ^ ((length(x) - 1):0))))
> > [1]  0  1  2 15
> > 
> > # So just sum them
> > > sum(apply(mat, 1, function(x) sum(x * (2 ^ ((length(x) - 1):0)))))
> > [1] 18
> > 
> > 
> > 2. If you want to actually generate the above matrix as a sequence of
> > binary values from a sequence of base 10 integer values, you can use the
> > digitsBase() function in Martin's sfsmisc package on CRAN:
> > 
> > install.packages("sfsmisc")
> > library(sfsmisc)
> > 
> > > t(digitsBase(1:15))
> > Class 'basedInt'(base = 2) [1:4]
> >       [,1] [,2] [,3] [,4]
> >  [1,]    0    0    0    1
> >  [2,]    0    0    1    0
> >  [3,]    0    0    1    1
> >  [4,]    0    1    0    0
> >  [5,]    0    1    0    1
> >  [6,]    0    1    1    0
> >  [7,]    0    1    1    1
> >  [8,]    1    0    0    0
> >  [9,]    1    0    0    1
> > [10,]    1    0    1    0
> > [11,]    1    0    1    1
> > [12,]    1    1    0    0
> > [13,]    1    1    0    1
> > [14,]    1    1    1    0
> > [15,]    1    1    1    1
> 
> 
> One quick notation here. I just noted that Martin has an as.integer()
> method for the basedInt class above. This allows for converting the
> numbers back to base 10 representation. It presumes that the number to
> be converted is the column, not the row. Thus the transpose of the
> matrix above results in:
> 
> > as.integer(t(digitsBase(1:15)))
>     1     2     3     4
>   255  3855 13107 21845
> 
> If you of course leave the matrix as is (as Martin clearly intended),
> you get:
> 
> > digitsBase(1:15)
> Class 'basedInt'(base = 2) [1:15]
>      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
> [1,]    0    0    0    0    0    0    0    1    1     1     1     1
> [2,]    0    0    0    1    1    1    1    0    0     0     0     1
> [3,]    0    1    1    0    0    1    1    0    0     1     1     0
> [4,]    1    0    1    0    1    0    1    0    1     0     1     0
>      [,13] [,14] [,15]
> [1,]     1     1     1
> [2,]     1     1     1
> [3,]     0     1     1
> [4,]     1     0     1
> 
> and therefore:
> 
> > as.integer(digitsBase(1:15))
>  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15
>  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15
> 
> 
> This would make basic arithmetic operations easier. Generating 'mat'
> above for example:
> 
> > mat <- digitsBase(c(0:2, 15))
> 
> # Again, note that each number is a column
> # not the row
> > mat
> Class 'basedInt'(base = 2) [1:4]
>      [,1] [,2] [,3] [,4]
> [1,]    0    0    0    1
> [2,]    0    0    0    1
> [3,]    0    0    1    1
> [4,]    0    1    0    1
> 
> > as.integer(mat)
>  1  2  3  4
>  0  1  2 15
> 
> 
> Finally:
> 
> > sum(as.integer(mat))
> [1] 18
> 
> and back to binary:
> 
> > digitsBase(sum(as.integer(mat)))
> Class 'basedInt'(base = 2) [1:1]
>      [,1]
> [1,]    1
> [2,]    0
> [3,]    0
> [4,]    1
> [5,]    0
> 
> or even as a character vector:
> 
> > paste(digitsBase(sum(as.integer(mat))), collapse = "")
> [1] "10010"
> 
> 
> HTH,
> 
> Marc Schwartz
> 
> 
> 
> 
> 





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