[datatable-help] [Rd] speeding up perception

[Simon Urbanek]

Interesting, and I stand corrected:

> x = data.frame(a=1:n,b=1:n)
> .Internal(inspect(x))
@103511c00 19 VECSXP g0c2 [OBJ,NAM(2),ATT] (len=2, tl=0)
  @102c7b000 13 INTSXP g0c7 [] (len=100000, tl=0) 1,2,3,4,5,...
  @102af3000 13 INTSXP g0c7 [] (len=100000, tl=0) 1,2,3,4,5,...

> x[1,1]=42L
> .Internal(inspect(x))
@10349c720 19 VECSXP g0c2 [OBJ,NAM(2),ATT] (len=2, tl=0)
  @102c19000 13 INTSXP g0c7 [] (len=100000, tl=0) 42,2,3,4,5,...
  @102b55000 13 INTSXP g0c7 [] (len=100000, tl=0) 1,2,3,4,5,...

> x[[1]][1]=42L
> .Internal(inspect(x))
@103511a78 19 VECSXP g1c2 [OBJ,MARK,NAM(2),ATT] (len=2, tl=0)
  @102e65000 13 INTSXP g0c7 [] (len=100000, tl=0) 42,2,3,4,5,...
  @101f14000 13 INTSXP g1c7 [MARK] (len=100000, tl=0) 1,2,3,4,5,...

> x[[1]][1]=42L
> .Internal(inspect(x))
@10349c800 19 VECSXP g0c2 [OBJ,NAM(2),ATT] (len=2, tl=0)
  @102a2f000 13 INTSXP g0c7 [] (len=100000, tl=0) 42,2,3,4,5,...
  @102ec7000 13 INTSXP g0c7 [] (len=100000, tl=0) 1,2,3,4,5,...


I have R to release ;) so I won't be looking into this right now, but it's something worth investigating ... Since all the inner contents have NAMED=0 I would not expect any duplication to be needed, but apparently becomes so is at some point ...

Cheers,
Simon


On Jul 6, 2011, at 4:36 AM, Matthew Dowle wrote:

> 
> On Tue, 2011-07-05 at 21:11 -0400, Simon Urbanek wrote:
>> No subassignment function satisfies that condition, because you can always call them directly. However, that doesn't stop the default method from making that assumption, so I'm not sure it's an issue.
>> 
>> David, Just to clarify - the data frame content is not copied, we are talking about the vector holding columns.
> 
> If it is just the vector holding the columns that is copied (and not the
> columns themselves), why does n make a difference in this test (on R
> 2.13.0)?
> 
>> n = 1000
>> x = data.frame(a=1:n,b=1:n)
>> system.time(for (i in 1:1000) x[1,1] <- 42L)
>   user  system elapsed 
>  0.628   0.000   0.628 
>> n = 100000
>> x = data.frame(a=1:n,b=1:n)      # still 2 columns, but longer columns
>> system.time(for (i in 1:1000) x[1,1] <- 42L)
>   user  system elapsed 
> 20.145   1.232  21.455 
>> 
> 
> With $<- :
> 
>> n = 1000
>> x = data.frame(a=1:n,b=1:n)
>> system.time(for (i in 1:1000) x$a[1] <- 42L)
>   user  system elapsed 
>  0.304   0.000   0.307 
>> n = 100000
>> x = data.frame(a=1:n,b=1:n)
>> system.time(for (i in 1:1000) x$a[1] <- 42L)
>   user  system elapsed 
> 37.586   0.388  38.161 
>> 
> 
> If it's because the 1st column needs to be copied (only) because that's
> the one being assigned to (in this test), that magnitude of slow down
> doesn't seem consistent with the time of a vector copy of the 1st
> column : 
> 
>> n=100000
>> v = 1:n
>> system.time(for (i in 1:1000) v[1] <- 42L)
>   user  system elapsed 
>  0.016   0.000   0.017 
>> system.time(for (i in 1:1000) {v2=v;v2[1] <- 42L})
>   user  system elapsed 
>  1.816   1.076   2.900
> 
> Finally, increasing the number of columns, again only the 1st is
> assigned to :
> 
>> n=100000
>> x = data.frame(rep(list(1:n),100))
>> dim(x)
> [1] 100000    100
>> system.time(for (i in 1:1000) x[1,1] <- 42L)
>   user  system elapsed 
> 167.974  50.903 219.711 
>> 
> 
> 
> 
>> 
>> Cheers,
>> Simon
>> 
>> Sent from my iPhone
>> 
>> On Jul 5, 2011, at 9:01 PM, David Winsemius <dwinsemius at comcast.net> wrote:
>> 
>>> 
>>> On Jul 5, 2011, at 7:18 PM, <luke-tierney at uiowa.edu> <luke-tierney at uiowa.edu> wrote:
>>> 
>>>> On Tue, 5 Jul 2011, Simon Urbanek wrote:
>>>> 
>>>>> 
>>>>> On Jul 5, 2011, at 2:08 PM, Matthew Dowle wrote:
>>>>> 
>>>>>> Simon (and all),
>>>>>> 
>>>>>> I've tried to make assignment as fast as calling `[<-.data.table`
>>>>>> directly, for user convenience. Profiling shows (IIUC) that it isn't
>>>>>> dispatch, but x being copied. Is there a way to prevent '[<-' from
>>>>>> copying x?
>>>>> 
>>>>> Good point, and conceptually, no. It's a subassignment after all - see R-lang 3.4.4 - it is equivalent to
>>>>> 
>>>>> `*tmp*` <- x
>>>>> x <- `[<-`(`*tmp*`, i, j, value)
>>>>> rm(`*tmp*`)
>>>>> 
>>>>> so there is always a copy involved.
>>>>> 
>>>>> Now, a conceptual copy doesn't mean real copy in R since R tries to keep the pass-by-value illusion while passing references in cases where it knows that modifications cannot occur and/or they are safe. The default subassign method uses that feature which means it can afford to not duplicate if there is only one reference -- then it's safe to not duplicate as we are replacing that only existing reference. And in the case of a matrix, that will be true at the latest from the second subassignment on.
>>>>> 
>>>>> Unfortunately the method dispatch (AFAICS) introduces one more reference in the dispatch chain so there will always be two references so duplication is necessary. Since we have only 0 / 1 / 2+ information on the references, we can't distinguish whether the second reference is due to the dispatch or due to the passed object having more than one reference, so we have to duplicate in any case. That is unfortunate, and I don't see a way around (unless we handle subassignment methods is some special way).
>>>> 
>>>> I don't believe dispatch is bumping NAMED (and a quick experiment
>>>> seems to confirm this though I don't guarantee I did that right). The
>>>> issue is that a replacement function implemented as a closure, which
>>>> is the only option for a package, will always see NAMED on the object
>>>> to be modified as 2 (because the value is obtained by forcing the
>>>> argument promise) and so any R level assignments will duplicate.  This
>>>> also isn't really an issue of imprecise reference counting -- there
>>>> really are (at least) two legitimate references -- one though the
>>>> argument and one through the caller's environment.
>>>> 
>>>> It would be good it we could come up with a way for packages to be
>>>> able to define replacement functions that do not duplicate in cases
>>>> where we really don't want them to, but this would require coming up
>>>> with some sort of protocol, minimally involving an efficient way to
>>>> detect whether a replacement funciton is being called in a replacement
>>>> context or directly.
>>> 
>>> Would "$<-" always satisfy that condition. It would be big help to me if it could be designed to avoid duplication the rest of the data.frame.
>>> 
>>> -- 
>>> 
>>>> 
>>>> There are some replacement functions that use C code to cheat, but
>>>> these may create problems if called directly, so I won't advertise
>>>> them.
>>>> 
>>>> Best,
>>>> 
>>>> luke
>>>> 
>>>>> 
>>>>> Cheers,
>>>>> Simon
>>>>> 
>>>>> 
>>>>> 
>>>> 
>>>> -- 
>>>> Luke Tierney
>>>> Statistics and Actuarial Science
>>>> Ralph E. Wareham Professor of Mathematical Sciences
>>>> University of Iowa                  Phone:             319-335-3386
>>>> Department of Statistics and        Fax:               319-335-3017
>>>> Actuarial Science
>>>> 241 Schaeffer Hall                  email:      luke at stat.uiowa.edu
>>>> Iowa City, IA 52242                 WWW:  http://www.stat.uiowa.edu______________________________________________
>>>> R-devel at r-project.org mailing list
>>>> https://stat.ethz.ch/mailman/listinfo/r-devel
>>> 
>>> David Winsemius, MD
>>> West Hartford, CT
>>> 
>>> 
> 
> 
>