[R] colmeans not working

[arun]

Hi Eliza,

I tried with the example you gave.  Couldn't reproduce the error.  


res1<-list(read.table(text=" 
        2005      2006      2008      2009 
  1.7360776 0.8095275 1.6369044 0.8195241 
  0.6962079 3.8510720 0.4319758 2.3304495 
  1.0423625 2.7687266 0.2904245 0.7015527 
  2.4158326 1.2315324 1.4287387 1.5701019 
",sep="",header=TRUE),read.table(text=" 
         2008      2009      2010 
   1.4737028  2.314878  2.672661 
   1.6700918  2.609722  2.112421 
   3.2387775  7.305766  6.939536 
   6.7063592 18.745256 13.278218 
",sep="",header=TRUE))
 names(res1)<-c("EE","WW")
res1<-lapply(res1,function(x) {names(x)<-gsub("X","",names(x));x})
res1
#$EE
#       2005      2006      2008      2009
#1 1.7360776 0.8095275 1.6369044 0.8195241
#2 0.6962079 3.8510720 0.4319758 2.3304495
#3 1.0423625 2.7687266 0.2904245 0.7015527
#4 2.4158326 1.2315324 1.4287387 1.5701019
#
#$WW
#      2008      2009      2010
#1 1.473703  2.314878  2.672661
#2 1.670092  2.609722  2.112421
#3 3.238777  7.305766  6.939536
#4 6.706359 18.745256 13.278218
lapply(res1,colMeans)
#$EE
#     2005      2006      2008      2009 
#1.4726202 2.1652146 0.9470108 1.3554070 
#
#$WW
#    2008     2009     2010 
#3.272233 7.743906 6.250709 

 lapply(res1,rowMeans)
#$EE
#[1] 1.250508 1.827426 1.200767 1.661551
#
#$WW
#[1]  2.153747  2.130745  5.828026 12.909944


A.K.



----- Original Message -----
From: eliza botto <eliza_botto at hotmail.com>
To: bbolker at gmail.com; r-help at stat.math.ethz.ch
Cc: 
Sent: Sunday, December 23, 2012 7:48 PM
Subject: Re: [R] colmeans not working


Dear Ben,Thanks for replying but its still not working.your code was>lapply(res,colMeans)but i want to use "res1" instead of "res". when i did use it, i got same error.eliza
> To: r-help at stat.math.ethz.ch
> From: bbolker at gmail.com
> Date: Mon, 24 Dec 2012 00:31:41 +0000
> Subject: Re: [R] colmeans not working
> 
> 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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