function in aggregate applied to specific columns only

That makes eight solutions. Any others?  :)

Just for the fun of it, here are two more: by and ave.

with(basicSub, by(score, student, mean))

student: 1
[1] 55
------------------------------------------------------------
student: 2
[1] 60
------------------------------------------------------------
student: 3
[1] 67.5

Not my favorite print method;  to return a vector, do instead

as.vector(with(basicSub, by(score, student, mean)))

[1] 55.0 60.0 67.5
You can cbind the unique student IDs to get a matrix result.

ave() is used to map the average (or comparable summary) to each
observation.
By itself, it returns a vector of the same length as the number of
observations:

with(basicSub, ave(score, student))

[1] 55.0 60.0 67.5 67.5 55.0

It's more useful if you want to add the means to the data frame:

transform(basicSub, avg = ave(score, student))

 student gender score  avg
1       1      m    50 55.0
2       2      m    60 60.0
3       3      f    70 67.5
4       3      f    65 67.5
5       1      m    60 55.0

That makes eight solutions. Any others?  :)

Dennis


On Sun, Jan 3, 2010 at 8:14 PM, Gabor Grothendieck
<ggrothendieck at gmail.com>wrote:

Here are 6 ways:

1. aggregate

aggregate(basicSub["score"], basicSub["student"], mean)

 student score
1       1  55.0
2       2  60.0
3       3  67.5

2. tapply

with(basicSub, tapply(score, student, mean))

  1    2    3
55.0 60.0 67.5

3. summaryBy in doBy package

library(doBy)
summaryBy(. ~ student, basicSub)

 student score.mean
1       1       55.0
2       2       60.0
3       3       67.5

4. sqldf in sqldf package.  Uses SQL:

library(sqldf)
sqldf("select student, avg(score) from basicSub group by student")

 student avg(score)
1       1       55.0
2       2       60.0
3       3       67.5

5. summary.formula in Hmisc

summary(score ~ student, basicSub)

score    N=5

+-------+-+-+-----+
|       | |N|score|
+-------+-+-+-----+
|student|1|2|55.0 |
|       |2|1|60.0 |
|       |3|2|67.5 |
+-------+-+-+-----+
|Overall| |5|61.0 |
+-------+-+-+-----+

6. plyr (see Dennis Murphy's solution in this thread)


On Sun, Jan 3, 2010 at 10:46 PM, david hilton shanabrook
<dhshanab at acad.umass.edu> wrote:

I want to use aggregate with the mean function on specific columns

gender <- factor(c("m", "m", "f", "f", "m"))
student <- c(0001, 0002, 0003, 0003, 0001)
score <- c(50, 60, 70, 65, 60)
basicSub <- data.frame(student, gender, score)
basicSubMean <- aggregate(basicSub, by=list(basicSub$student), 
FUN=mean,

na.rm=TRUE)

This doesn't work, one cannot take the mean of a factor (gender).  Is

there any way of specifying which columns to use for the mean?  I want to
aggregate by student, obtaining mean scores, and assume any other factors
are unchanging in a specific student, ie. gender.

Thanks
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