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Subset and order at the same time?

6 messages · Noah Silverman, Berend Hasselman, jim holtman +2 more

Noah Silverman
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Is there a way to order data and subset it at the same time??



I want to sort all the members of group A by their values in column 3. (I'll then do the same for each subsequent group.)  This could be done in a loop building up another vector, but I like to avoid loops in R.
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a <- temp[temp$group=="A",]
a <- a[order(a[,3]),]
temp[temp$group=="A",] <- a
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Iid like to do this in a single step for each group.  However, I can't figure out how to order and subset at the same time.

This *does not work* but should illustrate what I'm trying to do

temp[temp$group=="A",] <- temp[ temp$group=="A" & order(temp[temp$group=="A",3]) , ]


Suggestions?

--
Noah Silverman
UCLA Department of Statistics
8208 Math Sciences Building
Los Angeles, CA 90095
Berend Hasselman
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On 12-05-2012, at 20:04, Noah Silverman wrote:

            

      
      
      
    

        
set.seed(413)

temp <- data.frame(group=rep(c("A","B","C"), rep=5), tt=1:15,val=round(runif(15),2), stringsAsFactors=FALSE)
idx <- order(temp$group,temp$val)
# or   idx <- order(temp[,1],temp[,3])
idx
z2 <- temp[idx,]
rownames(z2) <- NULL
z2

Possible NA's etc. not taken into account.

Berend

  
  


  


      

    

    

      
      

        
      

    

    

      
      

        


  

        

      jim holtman
    

    

      
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Just write a function so that you have a "one-liner" in your script.
It will probably be a lot simpler than trying to type some convoluted
one-liner.

        
On Sat, May 12, 2012 at 2:58 PM, Noah Silverman <noahsilverman at ucla.edu> wrote:

            

      
      
      
    

        

      
      
    

        

  
    

      
      
    

  


  


      

    

    

      
      

        


  

        

      Neal Fultz
    

    

      
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can be done in one line, but it is annoying and ugly, so you probably
shouldn't be doing it that way:

            

      
      
      
    

        
extra group ID
1   -1.6     1  2
2   -1.2     1  4
3   -0.2     1  3
4   -0.1     1  5
5    0.0     1  9
6    0.7     1  1
7    0.8     1  8
8    2.0     1 10
9    3.4     1  6
10   3.7     1  7
11   1.9     2  1
12   0.8     2  2
13   1.1     2  3
14   0.1     2  4
15  -0.1     2  5
16   4.4     2  6
17   5.5     2  7
18   1.6     2  8
19   4.6     2  9
20   3.4     2 10

        
On 5/12/12, jim holtman <jholtman at gmail.com> wrote:

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        

      
      
    

  
  


  


      

    

    

      
      

        


  

        

      Joshua Wiley
    

    

      
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Hi Noah,

I think it is hard to say what is best without your real example.  Is
the goal elegance or speed?  I have not tried it, but if you are
ordering say 9/10 groups, I would think you are better off calling
order once, even though you will not use it for one group.

I also think if you are ordering multiple groups, you should take
advantage of that and do it all at once rather than in a series of one
liners.

## A
sleep$index <- order(sleep$extra)
sleep[sleep$group == 1, ] <- with(subset(sleep, group == 1), sleep[index, ])

## A "one line"
sleep$index <- order(sleep$extra); sleep[sleep$group == 1, ] <-
with(subset(sleep, group == 1), sleep[index, ])

## B
sleep[sleep$group == 1, ] <- with(subset(sleep, group == 1),
sleep[order(extra), ])

I agree with Neal though, the one line solutions are probably
something you can do but probably should not do.  At an extreme:

a <- b <- 10; dat <- local(data.frame(first = b, second = a <- 0 -> b,
third = a <- rnorm(10, mean = a), fourth = a / 2, fifth = a <-
rnorm(10, mean = a <- b), sixth = a * rnorm(10, b) -> a, seventh = b
<<- a^2, eigth = {rm(b); 1:10}, ninth = a * b, tenth = b <<- a -> b,
eleventh = {rm(a); b <- 10:1}, twelth = a, thirteenth = local(a <<- b
-> a)))

Cheers,

Josh

        
On Sat, May 12, 2012 at 1:04 PM, Neal Fultz <nfultz at gmail.com> wrote: