how to update a value in a list with lapply

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

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

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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