views:

437

answers:

2

Here I make a new column to indicate whether myData is above or below its median

### MedianSplits based on Whole Data
#create some test data
myDataFrame=data.frame(myData=runif(15),myFactor=rep(c("A","B","C"),5)) 

#create column showing median split
myBreaks= quantile(myDataFrame$myData,c(0,.5,1))
myDataFrame$MedianSplitWholeData = cut(
    myDataFrame$myData,
    breaks=myBreaks, 
    include.lowest=TRUE,
    labels=c("Below","Above"))

#Check if it's correct
myDataFrame$AboveWholeMedian = myDataFrame$myData > median(myDataFrame$myData)
myDataFrame

Works fine. Now I want to do the same thing, but compute the median splits within each level of myFactor.

I've come up with this:

#Median splits within factor levels
byOutput=by(myDataFrame$myData,myDataFrame$myFactor, function (x) {
     myBreaks= quantile(x,c(0,.5,1))
     MedianSplitByGroup=cut(x,
       breaks=myBreaks, 
       include.lowest=TRUE,
       labels=c("Below","Above"))
     MedianSplitByGroup
     })

byOutput contains what I want. It categorizes each element of factors A, B, and C correctly. However I'd like to create a new column, myDataFrame$FactorLevelMedianSplit, that shows the newly-computed median split.

How do you convert the output of the "by" command into a useful data-frame column?

I think perhaps the "by" command is not R-like way to do this ...

Update:

With Thierry's example of how to use factor() cleverly, and upon discovering the "ave" function in Spector's book, I've found this solution, which requires no additional packages.

myDataFrame$MediansByFactor=ave(
    myDataFrame$myData,
    myDataFrame$myFactor,
    FUN=median)

myDataFrame$FactorLevelMedianSplit = factor(
    myDataFrame$myData>myDataFrame$MediansByFactor, 
    levels = c(TRUE, FALSE), 
    labels = c("Above", "Below"))
+1  A: 

Here is a hack-ish way. Hadley may come with something more elegant:

To start, we simple concatenate the by output:

 R> do.call(c,byOutput)
A1 A2 A3 A4 A5 B1 B2 B3 B4 B5 C1 C2 C3 C4 C5 
 1  2  2  1  1  1  1  2  1  2  1  2  1  1  2

and what matters that we get the factor levels 1 and 2 here which we can use to re-index a new factor with those levels:

R> c("Below","Above")[do.call(c,byOutput)]
 [1] "Below" "Above" "Above" "Below" "Below" "Below" "Below" "Above" 
 [8] "Below" "Above" "Below" "Above" "Below" "Below" "Above"
R> as.factor(c("Below","Above")[do.call(c,byOutput)])
[1] Below Above Above Below Below Below Below Above Below Above 
[11] Below Above Below Below Above
Levels: Above Below

which we can then assign into the data.frame you wanted to modify:

R> myDataFrame$FactorLevelMedianSplit <- 
      as.factor(c("Below","Above")[do.call(c,byOutput)])

Update: Never mind, we'd need to reindex myDataFrame to be sorted A A ... A B ... B C ... C as well before we add the new column. Left as an exercise...

Dirk Eddelbuettel
+2  A: 

Here is a solution using the plyr package.

myDataFrame <- data.frame(myData=runif(15),myFactor=rep(c("A","B","C"),5))
library(plyr)
ddply(myDataFrame, "myFactor", function(x){
    x$Median <- median(x$myData)
    x$FactorLevelMedianSplit <- factor(x$myData <= x$Median, levels = c(TRUE, FALSE), labels = c("Below", "Above"))
    x
})
Thierry
This worked great. See also the update to the post for a packageless way.
Dan Goldstein