[R] how to update a value in a list with lapply

[ce]

Thanks a lot Bill and David. 
Very few elements will be updated in the list. I think I will go for for loop in this case. As a classic programmer I still feel uncomfortable with lapply anyway. 
ce

-----Original Message-----
From: "William Dunlap" [wdunlap at tibco.com]
Date: 05/10/2015 06:00 PM
To: "ce" <zadig_1 at excite.com>
CC: "David Winsemius" <dwinsemius at comcast.net>, r-help at r-project.org
Subject: Re: [R] how to update a value in a list with lapply

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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