-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf
Of Yogesh Tiwari
Sent: Wednesday, September 24, 2008 10:55 PM
To: r-help at stat.math.ethz.ch; sarah.goslee at gmail.com; ssefick at gmail.com;
milton.ruser at gmail.com
Subject: Re: [R] climatological standard deviation- (question re-posted)
Dear R Users,
Sorry for not explaining the problem clearly.
Here we go.......
I am using R on windows OS....
Here is an example data and script and then the problem what I want to
solve.
# (data period 1993-2002)
#example data
yr mo co2
1993 2 359.543
1993 2 358.359
1993 2 359.315
1993 2 359.293
1993 4 362.756
1993 4 363.699
1993 4 363.505
1993 4 363.034
1993 6 358.482
1993 6 359.283
1993 7 356.739
1993 7 357.203
1993 7 357.257
1993 7 357.062
1993 8 357.618
1993 8 356.965
1993 8 356.11
1993 8 356.389
1993 9 356.418
1993 9 356.274
1993 9 355.7
1993 9 357.389
1993 10 351.972
1993 10 352.132
1993 10 352.181
1993 10 352.015
1993 10 355.774
1993 10 348.42
1993 10 348.376
1993 10 348.599
1993 11 349.954
1993 11 350.575
1993 11 350.301
1993 12 354.179
1993 12 353.917
1993 12 356.546
1994 2 366.965
1994 2 367.716
1994 2 367.711
1994 2 368.595
1994 3 367.167
1994 3 367.732
1994 3 368.427
1994 3 368.547
1994 3 367.833
1994 3 368.423
1994 4 361.591
1994 4 361.682
1994 5 362.952
1994 5 362.629
1994 5 360.703
1994 5 360.813
1994 5 366.137
1994 5 364.946
1994 6 357.072
1994 6 357.351
1994 7 357.888
1994 7 361.456
1994 7 357.311
1994 7 357.503
1994 8 356.242
1994 8 356.184
1994 8 357.733
1994 8 356.729
1994 9 357.147
1994 9 356.34
1994 9 357.626
1994 10 354.157
1994 10 353.994
1994 10 365.504
1994 10 365.036
1994 11 355.671
1994 11 355.209
1994 11 352.027
1994 11 351.612
1994 12 357.318
1994 12 357.762
1994 12 352.377
1994 12 352.719
1994 12 355.844
1994 12 354.784
1995 1 363.532
1995 1 364.478
1995 1 358.964
1995 1 359.402
1995 2 363.664
1995 2 363.746
1995 2 365.041
1995 2 364.56
1995 3 368.469
1995 3 367.798
1995 3 362.802
1995 3 363.027
1995 4 363.698
1995 4 363.772
1995 4 364.125
1995 4 364.408
1995 5 362.638
1995 5 362.534
1995 5 362.701
1995 6 366.4
1995 6 362.5
1995 6 359.634
1995 6 360.256
....
....
#Script to compute monthly climatology (12 values, Jan-Dec),
file<-read.csv("CRI-run109-run110.csv")
names(file)
attach(file)
co2.month.mean <- tapply(co2,mo,mean)
x<- as.integer(row.names(co2.month.mean),co2.month.mean)
plot(x, co2.month.mean, xlim=c(1,12),
ylim=c(358,372),xlab=NA,ylab=NA,xaxs="i",yaxs="i",col="grey",type="o",axes=F)
#
axis(1,at=1:12,labels=c("J","F","M","A","M","J","J","A","S","O","N","D"))
axis(2)
axis(3, at=1:12, labels=FALSE)
axis(4, at=NULL, labels=FALSE)
#standard deviation calulation for each month
#....jan
time_span <- file[(file$mo==1),]
sub_co2 <- (time_span$co2)
var.jan<-var(sub_co2, na.rm=TRUE)
sd.jan<-sqrt(var.jan)
#....feb
time_span <- file[(file$mo==2),]
sub_co2 <- (time_span$co2)
var.feb<-var(sub_co2, na.rm=TRUE)
sd.feb<-sqrt(var.feb)
#....mar
...
#Problem:
Need to compute climatological standard deviation i.e the 12 values
calculated from the above data january values, february values, etc.
and need to plot as an error bar on the ploted curve.
Is this the best way for such a calculation ?
Kindly help,
Best Regards,
Yogesh