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Still trying to avoid loops

11 messages · Rui Barradas, Bert Gunter, Tom +3 more

Original
Tom
# 
Given a dataframe:
dat<-data.frame(S=factor(c(rep('a',2),rep('b',1),rep('c',3)),levels=c('b','a','c')),
		D=c(5,1,3,2,3,4))

where S is a subject identifier and D a visit (actually a date in my
real dataset). I would like to generate another column giving the visit
number

R=c(2,1,1,1,2,3)

My current solution uses nested loops and is slow and ugly. I've looked
at by() but can't see how to keep the order of R correct.

Thanks,
Tom
Rui Barradas
# 
Hello,

Aren't the levels of your example wrong? If the levels are 
levels=c('a','b','c'), not c('b', 'a', 'c'), then the following will do 
the job.

unname(unlist(tapply(dat$D, dat$S, order)))


Hope this helps,

Rui Barradas

Em 04-02-2015 19:34, Tom Wright escreveu:
Bert Gunter
# 
tapply() (of which by() is essentially a wrapper) **is** a (disguised)
loop (at the R level, of course).

Cheers,
Bert



Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374

"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
Clifford Stoll
On Wed, Feb 4, 2015 at 11:49 AM, Rui Barradas <ruipbarradas at sapo.pt> wrote:
Tom
# 
Thanks, I was not aware of order().
I did deliberately mess up the order of S. The following example breaks
your solution
dat_2<-data.frame(S=factor(c('a','c','a','b','c','c')),
		  D=c(5,3,1,3,2,4))

which should give the answer c(2,2,1,1,2,3)

Your solution does indicate that sorting the data correctly before
starting might solve the problem.
On Wed, 2015-02-04 at 19:49 +0000, Rui Barradas wrote:
Tom
# 
No problem with disguise, I'm looking for pretty.
On Wed, 2015-02-04 at 12:06 -0800, Bert Gunter wrote:
jim holtman
# 
dat<-data.frame(S=factor(c(rep('a',2),rep('b',1),rep('c',3)),levels=c('b','a','c')),
+                 D=c(5,1,3,2,3,4))
S D
1 a 5
2 a 1
3 b 3
4 c 2
5 c 3
6 c 4
S D visit
1 a 5     1
2 a 1     2
3 b 3     1
4 c 2     1
5 c 3     2
6 c 4     3



Jim Holtman
Data Munger Guru

What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
On Wed, Feb 4, 2015 at 3:08 PM, Tom Wright <tom at maladmin.com> wrote:

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        

      
      
    

        

      
      
    

        

  
  


  


      

    

    

      
      

        


  

        

      Tom
    

    

      
      #
    

  


  

      
Sorry Jim,
That messes up on S=='a'. Should be 2,1 not 1,2

Neat answer though and looks like it should be pretty quick after I
apply some sorting.

        
On Wed, 2015-02-04 at 15:37 -0500, jim holtman wrote:

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        

      
      
    

  
  


  


      

    

    

      
      

        


  

        

      William Dunlap
    

    

      
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A useful technique when it is easy to compute a vector from an ordered
data.frame but you need to do it for an unordered one is to compute the
order
vector 'ord', compute the vector from df[ord,], and use df[ord,...] <-
vector
to reorder the vector.  In your case you could do:
  > dat_2<-data.frame(S=factor(c('a','c','a','b','c','c')),
  +                   D=c(5,3,1,3,2,4))
  > ord <- with(dat_2, order(S, D)) # order by subject, break ties by date
  > dat_2$visitNo <- integer(nrow(dat_2)) # will fill this in next
  > dat_2$visitNo[ord] <- with(dat_2[ord,], ave(visitNo, S, FUN=seq_along))
  > dat_2
    S D visitNo
  1 a 5       2
  2 c 3       2
  3 a 1       1
  4 b 3       1
  5 c 2       1
  6 c 4       3

Now this is different from your answer, c(2,2,1,1,2,3).  Which is correct?

You can also do the reordering of the result from the ordered dataset by
subscripting the right hand side with [order(ord)], but I find using [ord]
on left side easier to remember.
  with(dat_2[ord,], ave(visitNo, S, FUN=seq_along))[order(ord)]
  [1] 2 2 1 1 1 3



Bill Dunlap
TIBCO Software
wdunlap tibco.com

        
On Wed, Feb 4, 2015 at 12:07 PM, Tom Wright <tom at maladmin.com> wrote:

        

            

      
      
      
    

        

      
      
    

        

      
      
    

        

  
  


  


      

    

    

      
      

        


  

        

      David L Carlson
    

    

      
      #
    

  


  

      
How about?

            

      
      
      
    

        
[1] 2 1 1 1 2 3

            

      
      
      
    

        
[1] 2 2 1 1 1 3

Note, your answer for the second example is incorrect since row 2 (c, 3) and row 5 (c, 2) are both assigned 2.

-------------------------------------
David L Carlson
Department of Anthropology
Texas A&M University
College Station, TX 77840-4352

-----Original Message-----
From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Tom Wright
Sent: Wednesday, February 4, 2015 2:08 PM
To: Rui Barradas
Cc: r-help at stat.math.ethz.ch
Subject: Re: [R] Still trying to avoid loops

Thanks, I was not aware of order().
I did deliberately mess up the order of S. The following example breaks
your solution
dat_2<-data.frame(S=factor(c('a','c','a','b','c','c')),
		  D=c(5,3,1,3,2,4))

which should give the answer c(2,2,1,1,2,3)

Your solution does indicate that sorting the data correctly before
starting might solve the problem.

        
On Wed, 2015-02-04 at 19:49 +0000, Rui Barradas wrote:

            

      
      
      
    

        

      
      
    

        
______________________________________________
R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
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.

  
  


  


      

    

    

      
      

        


  

        

      William Dunlap
    

    

      
      #
    

  


  

      
A potential problem with
   ave(dat_2$D, dat_2$S, FUN=order)
is that it will silently give the wrong answer
or give an error if dat_2$D is not numeric.

E.g., if D is a Date vector we get
  > dat_3 <- dat_2[,1:2]
  > dat_3$D <- as.Date(paste0("2015-02-", dat_2$D))
  > with(dat_3, ave(D, S, FUN=order))
  Error in as.Date.numeric(value) : 'origin' must be supplied

Another problem is that it may take a lot more time than
is required if you have a lot of small groups in your data.

Both of those are avoided if you sort the entire dataset first
and 'unsort' the results when putting them into dataset.




Bill Dunlap
TIBCO Software
wdunlap tibco.com

        
On Wed, Feb 4, 2015 at 12:53 PM, David L Carlson <dcarlson at tamu.edu> wrote:

        

            

      
      
      
    

        

      
      
    

        

      
      
    

        

  
  


  


      

    

    

      
      

        


  

        

      Tom
    

    

      
      #
    

  


  

      
Of course you are correct the second answer should be
 c(2,2,1,1,1,3)

Thanks everyone.

        
On Wed, 2015-02-04 at 20:53 +0000, David L Carlson wrote: