R help - Nov 2012 - Multiplying elements of a list by rows of a matrix

Hi all,
I have the following code:

set.seed(1)
x1 <- matrix(sample(1:12), ncol=3)
x2 <- matrix(sample(1:12), ncol=3)
x3 <- matrix(sample(1:12), ncol=3)
X <- list(x1,x2,x3)
tt <- matrix(round(runif(5*4),2), ncol=4)

Is there a way I can construct a new list where
newlist[[i]] = tt[i,] %*% X[[i]]
without using a for loop? Each element of newlist will be 3 x 1 vector.

Thanks
--
Tina Alexander

Hello,

Thanks for the data example. Try

lapply(seq_along(X), function(i) tt[i,] %*% X[[i]])

Hope this helps,

Rui Barradas
Em 11-11-2012 16:33, Clemontina Davenport escreveu:
> Hi all,
> I have the following code:
>
> set.seed(1)
> x1 <- matrix(sample(1:12), ncol=3)
> x2 <- matrix(sample(1:12), ncol=3)
> x3 <- matrix(sample(1:12), ncol=3)
> X <- list(x1,x2,x3)
> tt <- matrix(round(runif(5*4),2), ncol=4)
>
> Is there a way I can construct a new list where
> newlist[[i]] = tt[i,] %*% X[[i]]
> without using a for loop? Each element of newlist will be 3 x 1 vector.
>
> Thanks
> --
> Tina Alexander
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.

Clemontina:

As you have seen, the answer is yes, but my question is why bother?
What's wrong with a for() loop?
lapply() should offer no advantage in speed over a loop. Vectorization
could, but lapply() is not vectorization.

-- Bert

On Sun, Nov 11, 2012 at 8:33 AM, Clemontina Davenport <ckalexa2 at
ncsu.edu> wrote:
> Hi all,
> I have the following code:
>
> set.seed(1)
> x1 <- matrix(sample(1:12), ncol=3)
> x2 <- matrix(sample(1:12), ncol=3)
> x3 <- matrix(sample(1:12), ncol=3)
> X <- list(x1,x2,x3)
> tt <- matrix(round(runif(5*4),2), ncol=4)
>
> Is there a way I can construct a new list where
> newlist[[i]] = tt[i,] %*% X[[i]]
> without using a for loop? Each element of newlist will be 3 x 1 vector.
>
> Thanks
> --
> Tina Alexander
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.


-- 

Bert Gunter
Genentech Nonclinical Biostatistics

Internal Contact Info:
Phone: 467-7374
Website:
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm

Hi,
In this case, you could try:

res<-lapply(mapply(c,X,lapply(data.frame(t(tt[1:3,])),function(x)
x),SIMPLIFY=FALSE),function(x) x[13:16]%*% matrix(x[1:12],ncol=3) )

res

#[[1]]
?# ??? [,1]? [,2]? [,3]
#[1,] 14.27 16.65 10.12

#[[2]]
#????? [,1] [,2]? [,3]
#[1,] 10.14 5.17 18.28
#
#[[3]]
?# ??? [,1]? [,2] [,3]
#[1,] 22.66 10.34 8.31


A.K.


----- Original Message -----
From: Clemontina Davenport <ckalexa2 at ncsu.edu>
To: r-help at r-project.org
Cc: 
Sent: Sunday, November 11, 2012 11:33 AM
Subject: [R] Multiplying elements of a list by rows of a matrix

Hi all,
I have the following code:

set.seed(1)
x1 <- matrix(sample(1:12), ncol=3)
x2 <- matrix(sample(1:12), ncol=3)
x3 <- matrix(sample(1:12), ncol=3)
X <- list(x1,x2,x3)
tt <- matrix(round(runif(5*4),2), ncol=4)

Is there a way I can construct a new list where
newlist[[i]] = tt[i,] %*% X[[i]]
without using a for loop? Each element of newlist will be 3 x 1 vector.

Thanks
--
Tina Alexander

______________________________________________
R-help at r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Thank you all for your responses, the lapply works perfectly. The
example I gave IS odd. I only realized after I posted that tt had 5
rows and should have only had 3, so I apologize for that (I initially
had X to have 5 list elements).

Bert, I tend to use for loops in excess and often unnecessarily, so I
am trying to learn to code without the mental crutch that for loops
solve all problems.
Thanks again,
Tina