[R] processing matrix equation derived from rows of two matrices

arun smartpink111 at yahoo.com
Fri Apr 12 19:57:30 CEST 2013 

Hi Janis,
For example 
The goal here is to get results per row 
BTW, your original posting had tb[,1] 
tb[,1]%*%(((val-rep(meansb79[1,],5))^2)/6) 
#Error in tb[, 1] %*% (((val - rep(meansb79[1, ], 5))^2)/6) : 
 # non-conformable arguments
I guess it should be:

tb[1,]%*%(((val-rep(meansb79[1,],5))^2)/6) 
#        [,1]
#[1,] 1.47619

If you can split the codes  in the first solution:
seq_len(nrow(tb))
# [1]  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
#[26] 26 27 28 29 30 31 32 33 34 35 36
 sapply(seq_len(nrow(tb)),function(i) i)
 #[1]  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
#[26] 26 27 28 29 30 31 32 33 34 35 36
#you can also do with lapply (makes it more easy to understand)
lapply(seq_len(nrow(tb)),function(i) i)[1:2]
#[[1]]
#[1] 1

#[[2]]
#[1] 2

 lapply(seq_len(nrow(tb)),function(i) tb[i,])[1:2]
#[[1]]
#[1] 1 1 1 4 0

#[[2]]
#[1] 1 1 1 3 1
 tb[1:2,]
 #    [,1] [,2] [,3] [,4] [,5]
#[1,]    1    1    1    4    0
#[2,]    1    1    1    3    1

lapply(seq_len(nrow(tb)),function(i) (((val-rep(meansb79[i,],5))^2/6)))[1:2]
#[[1]]
#[1] 0.765306122 0.217687075 0.003401361 0.122448980 0.574829932

#[[2]]
#[1] 0.87074830 0.27551020 0.01360544 0.08503401 0.48979592

(((val-rep(meansb79[1:2,],5))^2/6))
# [1] 0.765306122 0.275510204 0.003401361 0.085034014 0.574829932 0.870748299
 #[7] 0.217687075 0.013605442 0.122448980 0.489795918

lapply(seq_len(nrow(tb)),function(i)tb[i,] %*%(((val-rep(meansb79[i,],5))^2/6)))[1:2]
#[[1]]
 #       [,1]
#[1,] 1.47619

#[[2]]
 #        [,1]
#[1,] 1.904762
sapply(seq_len(nrow(tb)),function(i)tb[i,] %*%(((val-rep(meansb79[i,],5))^2/6)))[1:2]
#[1] 1.476190 1.904762

Hope it helps.
A.K.





----- Original Message -----
From: "Beckstrand, Janis, NCOD" <Janis.Beckstrand at va.gov>
To: smartpink111 at yahoo.com
Cc: 
Sent: Friday, April 12, 2013 1:05 PM
Subject: FW: [R] processing matrix equation derived from rows of two matrices


Thanks so much Arun.  If you have time, what reference(s) do you use to
learn the techniques like "seq_len(nrow" etc and other aspects of your
code?

Again. Thanks very much.  Jan Beckstrand



From: arun kirshna [via R]
[mailto:ml-node+s789695n4664037h93 at n4.nabble.com]
Sent: Friday, April 12, 2013 2:01 AM
To: Beckstrand, Janis, NCOD
Subject: Re: processing matrix equation derived from rows of two
matrices



Hi, 

May be this helps: 

tb[1,]%*%(((val-rep(meansb79[1,],5))^2)/6) 
#        [,1] 
#[1,] 1.47619
tryvarb<-c(1,2,3,4,4,4,4)
var(tryvarb)
#[1] 1.47619 

tb[2,]%*%(((val-rep(meansb79[2,],5))^2)/6) 
#         [,1] 
#[1,] 1.904762 

sapply(seq_len(nrow(tb)),function(i)
tb[i,]%*%(((val-rep(meansb79[i,],5))^2/6)))
# [1] 1.4761905 1.9047619 1.9047619 1.9047619 1.9047619 2.2857143
1.9047619 # [8] 1.9047619 2.2857143 2.2857143 1.9047619 1.9047619
2.2857143 2.2857143 #[15] 2.2857143 1.9523810 1.9523810 2.2857143
2.2857143 2.2857143 2.2857143 #[22] 1.6190476 1.6190476 1.9047619
1.9047619 1.9047619 1.9047619 1.8095238 #[29] 0.9047619 0.9047619
1.1428571 1.1428571 1.1428571 1.1428571 1.0000000 #[36] 0.4761905 

#or
mat1<-matrix(((val-rep(meansb79,each=5))^2/6),ncol=5,byrow=TRUE) 

diag(t(apply(tb,1,function(x) x))%*% apply(mat1,1,function(x) x)) # [1]
1.4761905 1.9047619 1.9047619 1.9047619 1.9047619 2.2857143 1.9047619
#[8] 1.9047619 2.2857143 2.2857143 1.9047619 1.9047619 2.2857143
2.2857143 #[15] 2.2857143 1.9523810 1.9523810 2.2857143 2.2857143
2.2857143 2.2857143 #[22] 1.6190476 1.6190476 1.9047619 1.9047619
1.9047619 1.9047619 1.8095238 #[29] 0.9047619 0.9047619 1.1428571
1.1428571 1.1428571 1.1428571 1.0000000 #[36] 0.4761905 

#or
diag(apply(tb,1,function(x) x%*% t(mat1))) # [1] 1.4761905 1.9047619
1.9047619 1.9047619 1.9047619 2.2857143 1.9047619  #[8] 1.9047619
2.2857143 2.2857143 1.9047619 1.9047619 2.2857143 2.2857143 #[15]
2.2857143 1.9523810 1.9523810 2.2857143 2.2857143 2.2857143 2.2857143
#[22] 1.6190476 1.6190476 1.9047619 1.9047619 1.9047619 1.9047619
1.8095238 #[29] 0.9047619 0.9047619 1.1428571 1.1428571 1.1428571
1.1428571 1.0000000 #[36] 0.4761905 



A.K. 


>I have a matrix (tb) that represents all samples of size n=7 from a 
>group of N=9, with scores c(1,2,3,4,4,4,4,5,5).  I want to calculate 
>the variance for every sample.  Here is my code.  The
bottom shows the matrix equations and an attempt to process it for each
row. I got >the strategies from reading the r-help, but neither works.
(I  do not understand the syntax well enough.) Any suggestions.  (I need
to  do may :>additional matrices in the same way.) Thanks. Jan 

>require(combinat)
>n=7
>base79<-combn(c(1,2,3,4,4,4,4,5,5), n, tabulate,nbins=5) #find
all samples,n=7 (gives cols of  7 cases  in sample like 1 1 1 4 0 ) 
>tb<-t(base79)
>val<-c(1,2,3,4,5)  #values on the scale meansb79<-t(base79)%*% (val/7)

>tb[ ,1])%*%(((val-rep(meansb79[1,],5))^2)/6)   #computes the sample
variance for the first sample 
#check 
>tryvarb<-c(1,2,3,4,4,4,4)
>var(tryvarb)
  
#Now I try to get the variance for each sample (row) in tb, but neither
of the following attempts work. 

>trybase <- apply(tb,1,function(i)
  >t(base79[,i])%*%(((val-rep(meansb79[i,],5))^2)/6)) 

#or 

>domatrix.f <- function(tb, meansb79) { 
  >   a <- nrow(A); b <- nrow(B); 
    >C <- matrix(NA, nrow=a, ncol=b); 
    >for (i in 1:a) 
     > for (j in 1:b) 
       > C[i,j] <- t(A[,i])%*%(((val-rep(B[i,],5))^2)/6) } 
>domatrix.f(tb, meansb79) 

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