Hi Jacqueline,
(1)
x <- as.matrix(d[rownames(d) != 'nc', colnames(d) != 'nr'])
nc <- d['nc', ]
nr <- d[, 'nr']
e <- (x - nc %o% nr)^2 / (nc %o% nr / 2)
(2) if I understand correctly, ?max.col is what you need.
Wuming
On Tue, Aug 20, 2013 at 11:43 AM, Jacqueline Oehri <
jacqueline.oe...@gmail.com> wrote:
> Dear R users
>
>
> I have a question concerning applying a function to each element of a
> dataframe:
>
>
> 1)
> --> I have a dataframe like this: "d":
> (columnames: names of Landcovertypes, rownames: coordinates, nr:
> rowsums, nc:colummnsums)
> (look at the end of the mail for the structure of d, dput(d) )
> here, "d" has 14 rows and 6 colummns:
>
> > d
> PL_7_1_7.txt PL_7_1_8.txt PUEH_4_0.txt PUEH_7_1_2.txt UEH_7_2_2.txt
> nr
> 821194 0 0 0 0 0
> 29
> 821202 0 0 0 0 0
> 8
> 821206 1 0 0 0 0
> 2
> 827162 1 0 0 0 0
> 6
> 827166 0 1 1 1 1
> 17
> 827178 0 0 0 0 0
> 0
> 827182 1 0 0 0 0
> 4
> 827186 0 0 0 0 0
> 16
> 827190 0 0 0 0 0
> 16
> 827194 0 0 0 0 0
> 18
> 827198 0 0 0 0 0
> 19
> 827206 0 0 0 0 0
> 19
> 833166 0 0 0 0 0
> 8
> nc 86 120 905 300 309
> 18733
>
>
> -->And i want to apply the following function "f" to each element xij
> of the dataframe "d":
> (xij is the element of the dataframe "d" at row nr. "i" and colummn
> nr. "j", x11 is therefore the element in the first row & the first
> collumn, which in case of "d" is equal to "0".)
>
> f = (x[i][j] -((nr[i]*nc[j])/n))^2/((nr[i]*nc[j])/n)
>
>
> so that in the end I will have a new dataframe "e", which contains the
> results of the function "f" as its elements instead of the original
> values! (do you know what I mean?)
> Do you have any hints how to do that?
>
> 2) After this, I wanted to filter out for EACH ROW in "e" the maximum
> value in the row & assign or link the respective columname of this
> maxiumum value to the respective rowname;
> so that in the end I will know for each rowname, which columname "fits
> best to it" i.e. which columname had the biggest value for this
> respective row.
> For example, in dataframe "d", in the third row called "821206 ", the
> maximum-value lies in the first colummn, which is named "PL_7_1_7.txt
> ". In this example I would link the name "821206 " somehow to the name
> "PL_7_1_7.txt ".
>
> Do you have any suggestions for me, how to do this the best way? or
> where i should look up possible solutions? I m really lost...
>
> What i tried until now was this:
>
> >
> f.good <- function(x, nr, nc, n) {
> n <- d[14,6]
> nr <- d[,6]
> nc <- d[14,]
> z1 <- (x-((nr*nc)/n))^2/((nr*nc)/n)
> return(z1)
> }
>
> and then i wanted to use the "apply" function:
>
> >
> apply(d, c(1,2), f.good)
>
> but it never worked at all, and I think Im far away from a solution!
>
> Can somebody help me out and give me a hint what to do? does somebody
> know a clever way to achieve tasks 1) &2) ?
>
> Im very glad about every input!!!!!
>
> Thanks a lot already!!! Have a nice day!
>
> Best wishes,
> Jacqueline
>
>
> > dput(d)
> structure(list(PL_7_1_7.txt = c(0, 0, 1, 1, 0, 0, 1, 0, 0, 0,
> 0, 0, 0, 86), PL_7_1_8.txt = c(0, 0, 0, 0, 1, 0, 0, 0, 0, 0,
> 0, 0, 0, 120), PUEH_4_0.txt = c(0, 0, 0, 0, 1, 0, 0, 0, 0, 0,
> 0, 0, 0, 905), PUEH_7_1_2.txt = c(0, 0, 0, 0, 1, 0, 0, 0, 0,
> 0, 0, 0, 0, 300), UEH_7_2_2.txt = c(0, 0, 0, 0, 1, 0, 0, 0, 0,
> 0, 0, 0, 0, 309), nr = c(29, 8, 2, 6, 17, 0, 4, 16, 16, 18, 19,
> 19, 8, 18733)), .Names = c("PL_7_1_7.txt", "PL_7_1_8.txt", "PUEH_4_0.txt",
> "PUEH_7_1_2.txt", "UEH_7_2_2.txt", "nr"), row.names = c("821194",
> "821202", "821206", "827162", "827166", "827178", "827182", "827186",
> "827190", "827194", "827198", "827206", "833166", "nc"), class =
> "data.frame")
>
> ______________________________________________
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