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
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