Looking for a sort of tapply() to data frames
On 12/15/05, January Weiner <january at uni-muenster.de> wrote:
Hello again, On 12/14/05, Thomas Lumley <tlumley at u.washington.edu> wrote:
You want by(df[,-1], df$Day, function.that.means.each.column)
OK, slowly :-) I don't understand it. - why df[,-1] and not df? don't we loose the df$Day entries?
You don't get them as a column but you get them as the
component labels.
by(df, df$Day, function(x) colMeans(x[,-1]))
If you convert it to a data frame you get them as the rownames:
do.call("rbind", by(df, df$Day, function(x) colMeans(x[,-1])))
(by the way, why does typeof(df) show "list"? I thought that read.table() returns a data frame?)
I think you want class(df) which shows its a data frame.
so all you need to do is write function.that.means.each.column() In this case there is a built-in function, colMeans, so you don't even have to write it.
Hmmmmm, I tried it and it did not work. That is, it works - but not as intended :-). Fake example:
df <- data.frame(Day=c("Tue","Tue","Tue", "Wed", "Wed"), val1=seq(1,5), val2=3*seq(1,5))
df
Day val1 val2 1 Tue 1 3 2 Tue 2 6 3 Tue 3 9 4 Wed 4 12 5 Wed 5 15
ddf <- by(df[,-1], df$Day, colMeans) ddf
df$Day: Tue val1 val2 2 6 ------------------------------------------------------------ df$Day: Wed val1 val2 4.5 13.5
ddf$Day
NULL
ddf$val1
NULL
In real data, instead of "days", I have around 6000 items, so I need
them to be in one column called "Days" (or whatever). OK. So correct
me if I understand wrongly what is happening here:
by() divides df in data frame subsets and applies a function
(colMeans) to each of them. The result of colMeans ... manual says
that colMeans returns the following:
A numeric or complex array of suitable size, or a vector if the
result is one-dimensional. The 'dimnames' (or 'names' for a
vector result) are taken from the original array.
...which doesn't tell me much. typeof(colMeans(...)) tells me
"double" but I think it lies. OK, lets assume it is a vector (should
be, I assume the result is one-dimensional, as I can hardly imagine a
multidimensional result).
So in the end I have a list with as many columns as I have days, and
in each column I have a vector with N named dimensions, where N is the
numbers of variables in the original data frame bar one. But what I
would like to have is a data frame with exactly the same column names,
and rows being just a summary. And no clue how to convert one in the
other :-)
More generally (eg the approach would work for medians as well) by(df[,1], df$Day, function(today) apply(today, 2, mean))
Huh? why is it df[,1] now? I think I'm completly lost.
df[,1] and df$Day both refer to the same first column.
Finally, you could just use aggregate().
Probably, yes. As soon as I figure out how to use it, that is :-) (an
aggregate(df[,-1], df[,1,drop = FALSE], mean) or aggregate(df[,-1], list(Day = df$Day), mean) The second arg of aggregate must be a list which is why we used drop = FALSE in the first instance and an explicit list in the second. Another alternative is to use summaryBy from the doBy package found at http://genetics.agrsci.dk/~sorenh/misc/ : library(doBy) summaryBy(cbind(var1, var2) ~ Day, data = df)
hour later: OK, I got it! yuppie!) However what I really needed was
smth like this:
ddf <- by(df[,-1], df$Day, function(z) { return(cor(z$val1,z$val2)) ; } )
(but I still don't know how to convert it to a friendly data frame...)
do.call("rbind", ddf)
Thanks for the answers! January -- ------------ January Weiner 3 ---------------------+--------------- Division of Bioinformatics, University of Muenster | Schlo??platz 4 (+49)(251)8321634 | D48149 M??nster http://www.uni-muenster.de/Biologie.Botanik/ebb/ | Germany
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