Splitting a set of vectors in a list (Solved )

From: Bert Gunter

?"["  ?InternalMethods

x[i,j] is just shorthand for  "["(x,i,j) .
(AFAIK)**All** operators
(+,-,...,subscripting,...) in R are functions,
stemming from 
its LISP-like
heritage, and can actually called by the usual
functional 
syntax, f(...),
instead of the operator syntax.
That is true even for assignment:

R> "<-"(junk, 1:3)
R> junk
[1] 1 2 
Okay I think I've got this one but
and "{":

R> "{"(1, 2, 3)
this defeats me. I see what it is doing but I have not
the slightest idea why . 

I had a look at ?"{" and if I am understanding the
example {2+3; 4+5}  what is happening is that anything
within the {} is being executed as separate statments
but I have not the slightest idea of what is happening
when "{"(1, 2, 3) returns 3.  

The other thing is, is it worth trying to figure out
what appears to be rather esotheric coding if I can do
the same with more intuitively understood albeit
clumsier code?
I believe this is in the (draft) R Language
Definition, part of the
official manuals that shipped with R.

Andy

Not sure where this is explicitly discussed within
R's 
documentation, but
you can find info on it in V&R's "S Programming",
esp. p.24 and 4.3,
"Extracting or replacing coefficients".

No doubt, other S/R  books explain it also.

Cheers,

Bert Gunter
Genentech Nonclinical Statistics
47374

-----Original Message-----
From: r-help-bounces at r-project.org 
[mailto:r-help-bounces at r-project.org] On
Behalf Of John Kane
Sent: Thursday, March 13, 2008 11:53 AM
To: Henrique Dallazuanna
Cc: R R-help
Subject: Re: [R] Splitting a set of vectors in a
list (Solved )
My thanks to Henrique Dallazuanna and Phil
Spector. 
Both solutions worked well. 
Phil suggested that an alterative to my function
would
be 
vect1 = sapply(mylist,'[[',1)
and I see that Henrique used `[` in his solution.

Can you point me to some documentation that
discusses
these usages. I have seen them before but I have
never
actually figured out how to use them.? 

Thanks.

Problem and solutions

========================================================
mylist <- list(aa=c("cat","peach" ), bb=c("dog",
"apple", "iron"), 
         cc = c("rabbit", "orange", "zinc",
"silk"))
myfun <- function(dff) dff[1]       
vect1  <- unlist(lapply(mylist, myfun))

# Desired output
t(cbind( c("cat" ,  "peach" , NA, NA), bbb  <-
c("dog"
,  "apple" ,"iron", NA),
ccb <- c("rabbit" ,"orange" ,"zinc" ,  "silk" ))) 

# Phil Spector's approach
mlen = max(sapply(mylist,length))
eqlens = lapply(mylist,function(x)if(length(x) <
mlen)

c(x,rep('',mlen-length(x))) else x)
do.call(rbind,eqlens)

# 	"Henrique Dallazuanna" <wwwhsd at gmail.com>
#    I added the t()
t(as.data.frame(lapply(mylist, `[`,
1:max(unlist(lapply(mylist,
 length))))))

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Splitting a set of vectors in a list (Solved )

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