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the first and last observation for each subject

9 messages · gallon li, PIKAL Petr, Carlos J. Gil Bellosta +5 more

Original
PIKAL Petr
# 
Hi

r-help-bounces at r-project.org napsal dne 02.01.2009 10:20:23:
time.
sapply(split(test$y, test$ID), function(x) tail(x, 1)-head(x,1))

I am leaving formating to the resulting table to you. Hint: aggregate

Best regards
Petr
http://www.R-project.org/posting-guide.html
Carlos J. Gil Bellosta
# 
Hello,

First, order your data by ID and time.

The columns you want in your output dataframe are then

unique(ID),

tapply( x, ID, function( z ) z[ 1 ] )

and

tapply( y, ID, function( z ) z[ lenght( z ) ] - z[ 1 ] )

Best regards,

Carlos J. Gil Bellosta
http://www.datanalytics.com
On Fri, 2009-01-02 at 17:20 +0800, gallon li wrote:
Gabor Grothendieck
# 
Try this:
+ 1  10 20 0
+ 1  10 30 1
+ 1 10 40 2
+ 2 12 23 0
+ 2 12 25 1
+ 2 12 28 2
+ 2 12 38 3
+ 3 5 10 0
+ 3 5 15 2"
ID  x  y
1  3  5  5
2  1 10 20
3  2 12 15
On Fri, Jan 2, 2009 at 4:20 AM, gallon li <gallon.li at gmail.com> wrote:
Hadley Wickham
#
On Fri, Jan 2, 2009 at 3:20 AM, gallon li <gallon.li at gmail.com> wrote:
One approach is to use the plyr package, as documented at
http://had.co.nz/plyr.  The basic idea is that your problem is easy to
solve if you have a subset for a single subject value:

one <- subset(DF, ID == 1)
with(one, y[length(y)] - y[1])

The difficulty is splitting up the original dataset in to subjects,
applying the solution to each piece and then joining all the results
back together.  This is what the plyr package does for you:

library(plyr)

# ddply is for splitting up data frames and combining the results
# into a data frame.  .(ID) says to split up the data frame by the subject
# variable
ddply(DF, .(ID), function(one) with(one, y[length(y)] - y[1]))

# if you want a more informative variable name in the result
# return a named vector:
ddply(DF, .(ID), function(one) c(diff = with(one, y[length(y)] - y[1])))

# plyr takes care of labelling the result for you.

You don't say why you want to include x, or what to do if x is not
invariant, but here are couple of options:

# Split up by ID and x
ddply(DF, .(ID, x), function(one) c(diff = with(one, y[length(y)] - y[1])))

# Return the first x value
ddply(DF, .(ID), function(one) {
  with(one, c(
    x = x[1],
    diff = y[length(y)] - y[1]
  ))
})

# Throw an error is x is not unique

ddply(DF, .(ID), function(one) {
  stopifnot(length(unique(one$x)) == 1)
  with(one, c(
    x = x[1],
    diff = y[length(y)] - y[1]
  ))
})

Regards,

Hadley
Frank E Harrell Jr
# 
Here is a fast approach using the Hmisc package's summarize function.

 > g <- function(w) {
+  time <- w[,'time']; y <- w[,'y']
+  c(y[which.min(time)], y[which.max(time)])}
 >
 > with(DF, summarize(DF, ID, g, stat.name=c('first','last')))
   ID first last
1  1    20   40
2  2    23   38
3  3    10   15

summarize converts DF to a matrix for speed, as subscripting matrices is 
much faster than subscripting data frames.

Frank
hadley wickham wrote:

  
    

      
      
    

  


  


      

    

    

      
      

        


  

        

      Stavros Macrakis
    

    

      
      #
    

  


  

      
I think there's a pretty simple solution here, though probably not the
most efficient:

t(sapply(split(a,a$ID),
    function(q) with(q,c(ID=unique(ID),x=unique(x),y=max(y)-min(y)))))

Using 'unique' instead of min or [[1]] has the advantage that if x is
in fact not time-invariant, this gives an error rather than silently
ignore inconsistencies.

Trying to package up this idiom into a function leads to:

select <-
  function(df, groupby, selection)
   {
     pf <- parent.frame()
     fields <- substitute(selection)
     t(sapply(split(df,eval(substitute(groupby),df,enclos=pf)),
             function(q) eval(fields,q,enclos=pf)))  }

which I admit is rather ugly (and does no error-checking), but it does work:

            

      
      
      
    

        
[,1] [,2] [,3]
  1 1    10   20
  2 2    12   15
  3 3    5    5

Perhaps some of the more experienced people on the list could show me
how to write this more cleanly.

           -s

        
On Fri, Jan 2, 2009 at 4:20 AM, gallon li <gallon.li at gmail.com> wrote:

            

      
      
      
    

        

      
      
    

  
  


  


      

    

    

      
      

        


  

        

      Carlos J. Gil Bellosta
    

    

      
      #
    

  


  

      
Hello,

Is is truly

y=max(y)-min(y)

what you want below?

Best regards,

Carlos J. Gil Bellosta
http://www.datanalytics.com

        
On Fri, 2009-01-02 at 13:16 -0500, Stavros Macrakis wrote: