You can do the timing yourself on a dataset which you feel is typical of
your usage.
E.g., define a function the implements each algorithm
> f1 <- function(foo) lapply(foo, function(x) { if (x[1] == 1) x[2] <- 0
; x })
> f2 <- function(foo) { for(i in seq_along(foo)) if (foo[[i]][1] == 1)
foo[[i]][2] <- 0 ; foo }
and compare the times (and return values) on various datasets.
> foo1 <- rep(list(c(1,2,1:100)), length=1e5) # every element needs
changing
> system.time(v1 <- f1(foo1))
user system elapsed
0.28 0.01 0.29
> system.time(v2 <- f2(foo1))
user system elapsed
0.26 0.03 0.30
> identical(v1, v2)
[1] TRUE
> foo2 <- rep(list(c(0,2,1:100)), length=1e5) # no element needs changing
> system.time(v1 <- f1(foo2))
user system elapsed
0.09 0.00 0.09
> system.time(v2 <- f2(foo2))
user system elapsed
0.07 0.00 0.07
> identical(v1, v2)
[1] TRUE
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Sun, May 10, 2015 at 6:11 AM, ce <zadi...@excite.com> wrote:
>
> yes indeed :
>
> foo <- lapply(foo, function(x) if(x[1] == 1 ) {x[2] <- 0; x }else{x} )
>
> would work. But if the list is too long, would it be time consuming
> rather than just updating elements that meet the if condition?
>
> thx
> ce
>
>
> -----Original Message-----
> From: "David Winsemius" [dwinsem...@comcast.net]
> Date: 05/09/2015 08:00 PM
> To: "ce" <zadi...@excite.com>
> CC: r-help@r-project.org
> Subject: Re: [R] how to update a value in a list with lapply
>
>
> On May 9, 2015, at 4:35 PM, ce wrote:
>
> > Dear All,
> >
> > I have a list, using lapply I find some elements of the list, and then I
> want to change the values I find. but it doesn't work:
> >
> > foo<-list(A = c(1,3), B =c(1, 2), C = c(3, 1))
> > lapply(foo, function(x) if(x[1] == 1 ) x )
> > $A
> > [1] 1 3
> >
> > $B
> > [1] 1 2
> >
> > $C
> > NULL
> >
> > lapply(foo, function(x) if(x[1] == 1 ) x[2] <- 0 )
> > $A
> > [1] 0
> >
> > $B
> > [1] 0
> >
> > $C
> > NULL
> >
> >> lapply(foo, function(x) if(x[1] == 1 ) x )
> > $A
> > [1] 1 3
> >
> > $B
> > [1] 1 2
> >
> > $C
> > NULL
> >
> >
> > how to do it correctly ?
>
> I find it useful to think of the `if` function as `if(cond){cons}else{alt}`
>
> lapply(foo, function(x) if(x[1] == 1 ) {x[2] <- 0; x }else{x} )
> #-----
> $A
> [1] 1 0
>
> $B
> [1] 1 0
>
> $C
> [1] 3 1
>
>
> You were not supply an alternative which was the cause of the NULL (and
> you were not returning a value which meant that the value returned was the
> value on the RHS of the assignment).
>
> --
>
> David Winsemius
> Alameda, CA, USA
>
> ______________________________________________
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
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