[R] replacement functions for subsets

Bert Gunter gunter.berton at gene.com
Wed Jul 10 22:10:50 CEST 2013


I think the OP may perhaps want to define a method for "[<-" .

e.g. try:

methods("[<-")

If this is not it ... ??

Cheers,
Bert

On Wed, Jul 10, 2013 at 12:51 PM, David Winsemius
<dwinsemius at comcast.net> wrote:
>
> On Jul 10, 2013, at 12:17 PM, Harry Mamaysky wrote:
>
>> As I understand it rownames(aa) returns a copy of an attribute of aa. So changing the value of this vector should make the change to the copy of the row.names attribute. I would then have to set the original row.names equal to this copy to effect the change.
>>
>> So my question is why "rownames(aa)[2:4] <-" changes the original attribute rather than its copy?
>
> I'm not sure how you decide that was happening. Your first paragraph seemed correct:
>
> aa <- data.frame( a=1:10,b=101:110 )
> str(aa)
> attributes(aa)
> dput(aa)
> `rownames<-`
>
>> trace(`rownames<-`)
>> rownames(aa)[2:4] <- c('row2','row3','row4')
> trace: `rownames<-`(`*tmp*`, value = c("1", "row2", "row3", "row4",
> "5", "6", "7", "8", "9", "10"))
>
> You can see that R first builds a full length vector with the second argumens to `rownames<-` fully expanded before doing the assignment to the 'row.names' attribute.
>
>>
>> And the follow on question is whether it's possible to have "f(x)[2:4] <-" operate in the same way for some user defined replacement function f.
>
> Take a look at the code:
>
> `row.names<-.data.frame`
>
> --
> David.
>>
>> Sent from my iPhone
>>
>> On Jul 10, 2013, at 3:05 PM, David Winsemius <dwinsemius at comcast.net> wrote:
>>
>>
>> On Jul 10, 2013, at 11:47 AM, Harry Mamaysky wrote:
>>
>>> I know how to define replacement functions in R (i.e. ‘foo<-‘ <- function(x,value) x<-value, etc.), but how do you define replacement functions that operate on subsets of arrays (i.e. how do you pass an index into foo)?
>>> For example, why does the following use of “rownames” work?
>>
>> `rownames` of a dataframe is a vector, so indexing with "[" and a single vector of indices is adequate. I cannot really tell what your conceptual "why"-difficulty might be. This is just assignment within a vector. That is not really a "replacement function operating on a subset of an array" since rownames are not values of the dataframe .... and it's not an "array". (Careful use of terms is needed here.)
>>
>>
>>>
>>>> aa <- data.frame( a=1:10,b=101:110 )
>>>
>>>> aa
>>>
>>>  a   b
>>>
>>> 1   1 101
>>>
>>> 2   2 102
>>>
>>> 3   3 103
>>>
>>> 4   4 104
>>>
>>> 5   5 105
>>>
>>> 6   6 106
>>>
>>> 7   7 107
>>>
>>> 8   8 108
>>>
>>> 9   9 109
>>>
>>> 10 10 110
>>>
>>>> rownames(aa)[2:4] <- c('row2','row3','row4')
>>>
>>>> aa
>>>
>>>    a   b
>>>
>>> 1     1 101
>>>
>>> row2  2 102
>>>
>>> row3  3 103
>>>
>>> row4  4 104
>>>
>>> 5     5 105
>>>
>>> 6     6 106
>>>
>>> 7     7 107
>>>
>>> 8     8 108
>>>
>>> 9     9 109
>>>
>>> 10   10 110
>>>
>>>
>>>
>>>
>>> Thanks,
>>>
>>> Harry
>>>
>>>
>>>   [[alternative HTML version deleted]]
>>>
>>> ______________________________________________
>>> R-help at r-project.org mailing list
>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>>> and provide commented, minimal, self-contained, reproducible code.
>>
>> David Winsemius
>> Alameda, CA, USA
>>
>
> David Winsemius
> Alameda, CA, USA
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.



-- 

Bert Gunter
Genentech Nonclinical Biostatistics

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