HI Eliza,
Just a doubt:
YOu mentioned that
"i then eliminated the first column as i was interested in knowing the row
wise mean of each sublist, by using" and
"afterwards when i applied the "colMeans" command i was not able to calculate
the mean as i got the following error"
Do you need colMeans or rowMeans?
>From the subset of data from your earlier email, I was able to get both
>rowMeans or colMeans.
dat1<-read.table("Eliza.txt",sep="",header=TRUE,stringsAsFactors=FALSE)
library(reshape2)
res<-lapply(split(dat1,dat1$st.),function(x)
dcast(x,month~year,mean,value.var="population_in_million"))
lapply(res,function(x) colMeans(x[,-1]))
#$Sa
#1955 1956 1957 1958 #didn't remove the NaN's
# NaN NaN NaN NaN
#$Ta
# 1966 1967 1968 1969
#2.355556 2.400000 2.355556 2.400000
lapply(res,function(x) rowMeans(x[,-1]))
#$Sa
# 1 2 3 4 5 6 7 8 9 10 11 12
#NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN
#
#$Ta
# 1 2 3
#2.400000 2.400000 2.333333
A.K.
----- Original Message -----
From: eliza botto <eliza_bo...@hotmail.com>
To: "r-help@r-project.org" <r-help@r-project.org>
Cc:
Sent: Sunday, December 23, 2012 7:17 PM
Subject: [R] colmeans not working
[text file is also attached in case you find the format of email difficult to
understand]
Dear useRs,You must all the planning for the christmas, but i am stucked in my
office on the following issue
i had a file containg information about station name, year, month, day, and
discharge information. i opened it by using
following command
> dat1<-read.table("EL.csv",header=TRUE, sep=",",na.strings="NA")
then by using following codes suggested by arun and rui i managed to obtain an
output
library(reshape2)
res<-lapply(split(dat1,dat1$st),function(x)
dcast(x,month~year,mean,value.var="discharge"))
> res
$EE month 2005 2006 2008 20091 1 1.7360776 0.8095275
1.6369044 0.81952412 2 0.6962079 3.8510720 0.4319758 2.33044953 3
1.0423625 2.7687266 0.2904245 0.70155274 4 2.4158326 1.2315324 1.4287387
1.5701019
$WW month 2008 2009 20101 1 1.4737028 2.314878
2.6726612 2 1.6700918 2.609722 2.1124213 3 3.2387775 7.305766
6.9395364 4 6.7063592 18.745256 13.278218
i then eliminated the first column as i was interested in knowing the row wise
mean of each sublist, by using
res1 <- lapply(res, function(x)x[,-c(1) ])
$EE 2005 2006 2008 20091 1.7360776 0.8095275 1.6369044
0.81952412 0.6962079 3.8510720 0.4319758 2.33044953 1.0423625 2.7687266
0.2904245 0.70155274 2.4158326 1.2315324 1.4287387 1.5701019
$WW 2008 2009 20101 1.4737028 2.314878 2.6726612
1.6700918 2.609722 2.1124213 3.2387775 7.305766 6.9395364 6.7063592
18.745256 13.278218
afterwards when i applied the "colMeans" command i was not able to calculate
the mean as i got the following error
>'x' must be an array of at least two dimensions
i cant spot the mistake in the last 5 hours.
you help is neededthanks in advance
eliza
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______________________________________________
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
