Message-ID: <1356313215.53070.YahooMailNeo@web142606.mail.bf1.yahoo.com>
Date: 2012-12-24T01:40:15Z
From: arun
Subject: colmeans not working
In-Reply-To: <BLU170-W114BD0A95FE69330F74D6C1893B0@phx.gbl>
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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and provide commented, minimal, self-contained, reproducible code.