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" [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
> 
> 

David Winsemius
Alameda, CA, USA

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