eliza botto <eliza_botto <at> hotmail.com> writes:
> Dear useRs,You must all the planning for the christmas, but i am
> stucked in my office on the following issue i had a file containg
> information about station name, year, month, day, and discharge
> information. i opened it by using following command
> > dat1<-read.table("EL.csv",header=TRUE, sep=",",na.strings="NA")
You can probably use
dat1 <- read.csv("EL.csv")
(although you may have to double-check some of the other
default differences between read.csv and read.table, e.g.
quote and comment.char arguments)
> then by using following codes suggested by arun and rui i managed to obtain an
output
library(reshape2)
res <- lapply(split(dat1,dat1$st),
function(x) dcast(x,month~year,mean,value.var="discharge"))
[snip]
res1 <- lapply(res, function(x)x[,-1])
(c() is redundant here)
> $EE
> 2005 2006 2008 2009
> 1 1.7360776 0.8095275 1.6369044 0.8195241
> 2 0.6962079 3.8510720 0.4319758 2.3304495
> 3 1.0423625 2.7687266 0.2904245 0.7015527
> 4 2.4158326 1.2315324 1.4287387 1.5701019
>
> $WW
> 2008 2009 2010
> 1 1.4737028 2.314878 2.672661
> 2 1.6700918 2.609722 2.112421
> 3 3.2387775 7.305766 6.939536
> 4 6.7063592 18.745256 13.278218
>
>
Now you just need
lapply(res,colMeans)
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