[R] on "do.call" function

[Zhang,Yanwei]

Hi, 

Your use of "do.call" is essentially equal to "dpois(lt$y0, exp(rowSums(t(X[lt$i,])*B[,1])))". You do not need to use "do.call", "sapply" or "apply" will do, e.g.,   

>  sapply(1:nrow(lt), function(x) fc(lt[x,2],lt[x,1]))
[1] 0.1891356 0.1859965 0.3149658 0.3128512 0.2622549 0.2631122

Wayne (Yanwei) Zhang
Statistical Research 
>CNA

-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf Of Kathie
Sent: Monday, August 08, 2011 10:16 AM
To: r-help at r-project.org
Subject: [R] on "do.call" function

Dear all,

Even though one of R users answered my question, I cannot understand, so I
re-ask this question.

I am trying to use "do.call", but I don't think I totally understand this
function.

Here is an simple example.

--------------------------------------------

> B <- matrix(c(.5,.1,.2,.3),2,2)
> B
     [,1] [,2]
[1,]  0.5  0.2
[2,]  0.1  0.3
> x <- c(.1,.2)
> X <- cbind(1,x)
> X
         x
[1,] 1 0.1
[2,] 1 0.2
>
> lt <- expand.grid(i=seq(1,2), y0=seq(0,2))
> lt
  i y0
1 1  0
2 2  0
3 1  1
4 2  1
5 1  2
6 2  2
>
> fc <- function(y0,i) dpois(y0, exp(rowSums(t(X[i,])*B[,1])))
>
> do.call(fc,lt)
[1] 1.892179e-09 3.348160e-01 3.800543e-08 3.663470e-01 3.816797e-07
2.004237e-01

--------------------------------------------

Unfortunately, what I want to get is

dpois(0, exp(rowSums(t(X[1,])*B[,1]))) = 0.1891356
dpois(0, exp(rowSums(t(X[2,])*B[,1]))) = 0.1859965
dpois(1, exp(rowSums(t(X[1,])*B[,1]))) = 0.3149658
dpois(1, exp(rowSums(t(X[2,])*B[,1]))) = 0.3128512
dpois(2, exp(rowSums(t(X[1,])*B[,1]))) = 0.2622549
dpois(2, exp(rowSums(t(X[2,])*B[,1]))) = 0.2631122

--------------------------------------------

Would you plz tell me why these two results are different?? and how do I get
what I want to using "do.call" function??


Regards,

Kathryn Lord 

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