CODEWITHSUNDEEP
rmatrixbinary
Adam Warner
Adam Warner

Reputation: 1354

Updating Matrix Using Data From Coerced Matrix

Here is the code that I am working with:

  y <- c(0,5,10)
n <- 9 

R <- t(c(2.05, 2.05, 2.05, 2.55, 2.55, 2.55, 2.95, 2.95, 2.95))
R <- (replicate(3,R))
R <- (matrix(R, nrow=3))
R <- t(apply(R,1,sort))

mat <- t(sapply(y, function(test) pmax(seq(test, (test-n+1), -1), 0) ))
mat




P <- replicate(ncol(R),(c(6447.88,6447.88,6447.88)))

EnvTest <- new.env()
EnvTest$Orig <- 548453.5

FuncTest <- function(pp){
  EnvTest$Orig <- EnvTest$Orig-(P[,pp]-EnvTest$Orig*R[,pp]/1200)
  return(EnvTest$Orig)
}

Test<- rbind(EnvTest$Orig,
             do.call(rbind,lapply(1:9, FuncTest)))
x <- t(Test)
x

This produces:

     [,1]     [,2]     [,3]     [,4]     [,5]     [,6]     [,7]     [,8]     [,9]    [,10]
[1,] 548453.5 542942.6 537422.2 531892.4 526574.8 521245.9 515905.7 510726.1 505533.7 500328.6
[2,] 548453.5 542942.6 537422.2 531892.4 526574.8 521245.9 515905.7 510726.1 505533.7 500328.6
[3,] 548453.5 542942.6 537422.2 531892.4 526574.8 521245.9 515905.7 510726.1 505533.7 500328.6

Which is what I am telling the function to do so that is correct. But now I want to use some information from the matrix R*(mat==0) which is: 

     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
[1,] 2.05 2.05 2.05 2.55 2.55 2.55 2.95 2.95 2.95
[2,] 0.00 0.00 0.00 0.00 0.00 2.55 2.95 2.95 2.95
[3,] 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

So I tried to do something like this... 

 FuncTest2 <- function(pp){
          EnvTest$Orig <- ifelse(R*(mat==0)==0,EnvTest$Orig,EnvTest$Orig-(P[,pp]-EnvTest$Orig*R[,pp]/1200))
                 return(EnvTest$Orig)
        }
        Test2<- rbind(EnvTest$Orig,
                     do.call(rbind,lapply(1:9, FuncTest2)))
        x2 <- t(Test2)
        x2

But that did not produce the desired outcome which is:

      [,1]     [,2]     [,3]     [,4]     [,5]     [,6]     [,7]     [,8]     [,9]    [,10]
[1,] 548453.5 542942.6 537422.2 531892.4 526574.8 521245.9 515905.7 510726.1 505533.7 500328.6
[2,] 548453.5 548453.5 548453.5 548453.5 548453.5 543171.1 538058.5 532933.3 527795.5 522645.1
[3,] 548453.5 548453.5 548453.5 548453.5  548453.5 548453.5 548453.5 548453.5  548453.5 548453.5

So basically if R*(mat==0) = 0 then EnvTest$Orig <- 548453.5 does not change until the R*(mat==0) is greater than zero the EnvTest$Orig does update.

Another Attempt

I also tried to do the following, but it did not work but I think that something along these lines will prove fruitful. 

    maybe <- R*(mat==0)
    FuncTest2 <- function(pp, maybe){
        EnvTest$Orig <- if(maybe == 0) (EnvTest$Orig-0) else
        (EnvTest$Orig-(P[,pp]-EnvTest$Orig*R[,pp]/1200))
      return(EnvTest$Orig)
    }

Test2<- rbind(EnvTest$Orig,
              do.call(rbind,lapply(1:9, FuncTest2)))
x2 <- t(Test2)
x2

Upvotes: 3

Views: 57

Answers (1)

bgoldst
bgoldst

Reputation: 35324

Firstly, with respect to your FuncTest() and FuncTest2() computations, R is a constant, as it is not modified during those function calls or by the surrounding data manipulation (referring to the do.call()/rbind()/t() stuff). This is R:

R;
##      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
## [1,] 2.05 2.05 2.05 2.55 2.55 2.55 2.95 2.95 2.95
## [2,] 2.05 2.05 2.05 2.55 2.55 2.55 2.95 2.95 2.95
## [3,] 2.05 2.05 2.05 2.55 2.55 2.55 2.95 2.95 2.95

As you can see, none of its elements are zero. This means that the expression

R*(mat==0)==0

is equivalent to

mat!=0

since multiplication by R cannot turn a non-zero value into zero, or turn a zero into a non-zero value.

If you still want to parameterize the ifelse() conditional on R, perhaps because you may want to run this code for different values of R that can contain zeroes, then it should be rewritten as

R==0|mat!=0

Secondly, your code has a bit of a suspicious pattern with respect to the dimensionality of the state variable EnvTest$Orig. You initialize it to a one-element vector (548453.5), but on first run of FuncTest() it is immediately overwritten with a three-element vector, where the initial scalar was cycled 3 times to accommodate the three-element vectors P[,pp] and R[,pp]. From that point forward it is permanently a three-element vector. In FuncTest2() (first attempt), since mat and R are 3x9 matrices, the true and false arguments of the ifelse() vectorized ternary will be cycled to cover that dimensionality, and then EnvTest$Orig will be overwritten by the resulting 3x9 matrix. From that point forward it is permanently a 3x9 matrix.

You should probably think carefully about what dimensionality you want for EnvTest$Orig and stick to that from beginning to end. Judging from your expected output, it looks like you want it to be a vector of length 3, and thus you should be indexing the matrix in the ifelse() conditional:

FuncTest2 <- function(pp) {
    EnvTest$Orig <- ifelse(R[,pp]==0|mat[,pp]!=0,EnvTest$Orig,EnvTest$Orig-(P[,pp]-EnvTest$Orig*R[,pp]/1200));
    return(EnvTest$Orig);
};
EnvTest$Orig <- 548453.5;
Test2 <- rbind(EnvTest$Orig,do.call(rbind,lapply(1:9,FuncTest2)));
x2 <- t(Test2);
x2;
##          [,1]     [,2]     [,3]     [,4]     [,5]     [,6]     [,7]     [,8]     [,9]    [,10]
## [1,] 548453.5 542942.6 537422.2 531892.4 526574.8 521245.9 515905.7 510726.1 505533.7 500328.6
## [2,] 548453.5 548453.5 548453.5 548453.5 548453.5 548453.5 543171.1 538058.5 532933.3 527795.6
## [3,] 548453.5 548453.5 548453.5 548453.5 548453.5 548453.5 548453.5 548453.5 548453.5 548453.5

However, this result does not exactly match your expected output in the second row, and I'm not sure why.

Upvotes: 1

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