[R] Looking for a sort of tapply() to data frames
January Weiner
january at uni-muenster.de
Thu Dec 15 12:55:40 CET 2005
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?
(by the way, why does typeof(df) show "list"? I thought that
read.table() returns 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.
> Finally, you could just use aggregate().
Probably, yes. As soon as I figure out how to use it, that is :-) (an
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...)
Thanks for the answers!
January
--
------------ January Weiner 3 ---------------------+---------------
Division of Bioinformatics, University of Muenster | Schloßplatz 4
(+49)(251)8321634 | D48149 Münster
http://www.uni-muenster.de/Biologie.Botanik/ebb/ | Germany
More information about the R-help
mailing list