[R] replacement functions for subsets

[David Winsemius]

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]]
>> 
>> ______________________________________________
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> 
> David Winsemius
> Alameda, CA, USA
> 

David Winsemius
Alameda, CA, USA