Hi:
x <- matrix(20:35, ncol = 1)
u <- c(1, 4, 5, 6, 11) # 'x values'
m <- c(1, 3, 1, 1, 0.5)
# Function to compute the inner product of the multipliers with the
extracted
# elements of x determined by u
f <- function(mat, inputs, mults) crossprod(mat[inputs], mults)
f(x, u, mults = c(1, 3, 1, 1, 0.5))
[,1]
[1,] 153
20 + 23 * 3 + 24 + 25 + 30 * 0.5
[1] 153
The function is flexible enough to allow you to play with the input matrix
(although a vector would also work), the 'observation vector' inputs and the
set of multipliers. Here's one way (not necessarily the most efficient):
uv <- matrix(sample(1:15, 25, replace = TRUE), ncol = 5)
uv # like an X matrix, where each row provides the input values of the
vars
[,1] [,2] [,3] [,4] [,5]
[1,] 12 8 11 10 15
[2,] 15 11 14 14 8
[3,] 4 8 4 10 12
[4,] 10 5 2 1 7
[5,] 11 4 9 1 11
# Apply the function f to each row of uv:
apply(uv, 1, function(y) f(x, y, mults = c(1, 3, 1, 1, 0.5)))
[1] 188.0 203.5 171.5 155.0 162.0
The direct matrix version:
crossprod(t(matrix(x[uv], ncol = 5)), c(1, 3, 1, 1, 0.5))
[,1]
[1,] 188.0
[2,] 203.5
[3,] 171.5
[4,] 155.0
[5,] 162.0
Notice that the apply() call returns a vector whereas crossprod() returns a
matrix.
x[uv] selects the x values associated with the indices in uv and returns a
vector in column-major order. The crossprod() call transposes the reshaped
x[uv] and then 'matrix' multiplies it by the vector c(1, 3, 1, 1, 0.5).
HTH,
Dennis
On Fri, Oct 29, 2010 at 3:54 PM, M.Ribeiro <mresende...@yahoo.com.br> wrote:
>
> So, I am having a tricky reference file to extract information from.
>
> The format of the file is
>
> x 1 + 4 * 3 + 5 + 6 + 11 * 0.5
>
> So, the elements that are not being multiplied (1, 5 and 6) and the
> elements
> before the multiplication sign (4 and 11) means actually the reference for
> the row in a matrix where I need to extract the element from.
>
> The numbers after the multiplication sign are regular numbers
> Ex:
>
> > x<-matrix(20:35)
> > x
> [,1]
> [1,] 20
> [2,] 21
> [3,] 22
> [4,] 23
> [5,] 24
> [6,] 25
> [7,] 26
> [8,] 27
> [9,] 28
> [10,] 29
> [11,] 30
> [12,] 31
> [13,] 32
> [14,] 33
> [15,] 34
> [16,] 35
>
> I would like to read the rows 1,4,5,6 and 11 and sum then. However the
> numbers in the elements row 4 and 11 are multiplied by 3 and 0.5
>
> So it would be
> 20 + 23 * 3 + 24 + 25 + 30 * 0.5.
>
> And I have this format in different files so I can't do all by hand.
> Can anybody help me with a script that can differentiate this?
> Thanks
> --
> View this message in context:
> http://r.789695.n4.nabble.com/Differenciate-numbers-from-reference-for-rows-tp3019853p3019853.html
> Sent from the R help mailing list archive at Nabble.com.
>
> ______________________________________________
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