[R] Looking for a sort of tapply() to data frames
Gabor Grothendieck
ggrothendieck at gmail.com
Fri Dec 16 20:11:26 CET 2005
One other point. The cor example could be done using tapply like
this:
tapply(rownames(df), df$Day, function(r) cor(df[r,"val1"], df[r, "val2"]))
On 12/16/05, Gabor Grothendieck <ggrothendieck at gmail.com> wrote:
> On 12/16/05, January Weiner <january at uni-muenster.de> wrote:
> > Hi,
> >
> > On 12/15/05, Gabor Grothendieck <ggrothendieck at gmail.com> wrote:
> > > You don't get them as a column but you get them as the
> > > component labels.
> > >
> > > by(df, df$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$Day, function(x) colMeans(x[,-1])))
> >
> > Thanks! that helps a lot. But I still run into problems with this.
> > Sorry for bothering you with newbie questions, if my problems are
> > trivial, point me to a suitable guide (I did read the introductory
> > materials on R).
> >
> > First: it works for colMeans, but it does not work for a function like this:
> >
> > do.call("rbind", by(df, df$Day, function(x) cor(df$val1, df$val2))
>
> There are a number of problems:
>
> 1. the function does not depend on x and therefore will return the
> same result for each day group.
>
> 2. although ?by says it returns a list, it apparently simplifies the result,
> contrary to the documentation, in certain cases. Try this:
>
> do.call("rbind", as.list(by(df, df$Day, function(x) cor(x$val1, x$val2))))
>
> or this:
>
> do.call("rbind", by(df, df$Day, function(x) list(cor = cor(x$val1, x$val2))))
>
>
> 3. In your sample data val1 is constant for Wed so you won't be able
> to get a correlation. That's the source of the warning that you get
> when running the line in #2.
>
> >
> > it says "Error in do.call(....) : second argument must be a list". I
> > do not understand this, as the second argument is "b" of the class
> > "by", as it was in the case of colMeans, so it did not change...?
> >
> > Second: in case of colMeans (where it works) it returns a matrix, and
> > I have troubles getting it back to the data.frame, so I can access
> > blah$Day. Instead, I have smth like that:
>
> Try blah[,"Day"] which works with both matrices and data frames.
>
> >
> > > do.call("rbind",b)
> > V2 V3 V4 V5 V7
> > Tue 19 15 2 0 1.538462
> > Wed 5 3 6 1 1.285714
>
>
> Another possibility is to coerce it to a data frame:
>
> as.data.frame(do.call("rbind", b))
>
> or change your function to return a list.
>
> >
> > ...and I do not know how to acces, for example, values for "Tue",
> > except with [1,] -- which is somewhat problematic. For example, I
> > would like to display the 3 days for which V7 is highest. How can I
> > do that?
> >
> > > I think you want class(df) which shows its a data frame.
> >
> > Ops. Sorry, I didn't guess it from the manual :-)
> >
> > > aggregate(df[,-1], df[,1,drop = FALSE], mean)
> >
> > But why is df[,1,drop=FALSE] a list? I don't get it...
>
> Because df is a one column data frame and data frames are lists.
> Had we not specified drop, it would have automatically dropped it
> since it has only one dimension simplifying it to a non-list.
> We do not want that simplification here.
>
> >
> > > aggregate(df[,-1], list(Day = df$Day), mean)
> >
> > Yeah, I figured out that one.
> >
> > > 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)
> >
> > I think I am not confident enough with the basic data types in R, I
> > need to understand them before I go over to specialized packages :-)
> > Again, thanks a lot,
> > 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
> >
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>
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