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" [dwinsemius at comcast.net]
Date: 05/09/2015 08:00 PM
To: "ce" <zadig_1 at excite.com>
CC: r-help at 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
On May 10, 2015, at 6:11 AM, ce 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?Any change to an object will require copying the entire object. That is the computing model that R uses. If you had presented a modification strategy that used logical or numeric indexing to effect only targeted nodes of a list, it still would have ended up copying the whole object. The data.table package was invented in large part to get around that design concern. -- David.> > thx > ce > > > -----Original Message----- > From: "David Winsemius" [dwinsemius at comcast.net] > Date: 05/09/2015 08:00 PM > To: "ce" <zadig_1 at excite.com> > CC: r-help at 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 > >David Winsemius Alameda, CA, USA
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 <zadig_1 at 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" [dwinsemius at comcast.net]
> Date: 05/09/2015 08:00 PM
> To: "ce" <zadig_1 at excite.com>
> CC: r-help at 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
>
> ______________________________________________
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