[R] polygon
Dear all I try once again. In order to clearly mark values wich are larger than a treshold value, I would like to color the surface below the line given by plot (yy~xx). To color is only the surface between abline (treshold) and yy if they are larger than the specific limit. I guess I can use the function polygon, but I can not find any valuable solution. I'm grateful to you for an advice, an example or a link. Mike xx <- c(1:100) yy <- rnorm(100) plot (yy~xx,type="l") abline (h=0.5,col="red") #?? how can I use polygon() #in order to color surface below yy value and > abline?? polygon(xx,yy,col="gray") --- --- --- --- __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: RE: [R] import Excel file ?
You can also find different examples in the manual R for Beginners by Emmanuel Paradis (that you can easily download from the internet) under http://zoonek2.free.fr/UNIX/48_R/all.html you find a lot of information about R in french, which is also very useful Mike --- --- --- --- __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] polygon
Dear all In order to clearly mark values wich are larger than a treshold value, I would like to color the surface below the line given by plot (yy~xx). To color is only the surface between abline (treshold) and yy if they are larger than the specific limit. I guess I can use the function polygon, but I can not find any valuable solution. I'm grateful to you for an advice or an example. Mike xx <- c(1:100) yy <- rnorm(100) plot (yy~xx,type="l") abline (h=0.5,col="red") #?? how can I use polygon() #in order to color surface below yy value and > abline?? polygon(xx,yy,col="gray") --- 60 Minuten pro Monat gratis telefonieren: So werden bei sunrise treue Kunden belohnt! http://internet.sunrise.ch/de/wireline/wir_60.asp --- Téléphoner gratuitement pendant 60 minutes: sunrise sait récompenser les clients fidèles. http://internet.sunrise.ch/fr/wireline/wir_60.asp --- Telefonate gratuitamente per 60 minuti: sunrise premia così la vostra fiducia. http://internet.sunrise.ch/it/wireline/wir_60.asp --- __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] outliers
Dear all I would like to represent the outliers in the plot. These few outliers are much larger than the limit of 50 in the ylim-argument. plot(daten$month~daten$no,ylim=c(0,50)) I know that it is possible to introduce the information about the presence of outliers without changing the range of the axis. Do you have an advice? I hope the question is understandable!! Thanks a lot for your explantion Mike --- Aufgrund grosser Nachfrage: sunrise ADSL Null-Franken-Angebot verlängert bis Ende April 2004! --- En raison de la forte demande: offre sunrise ADSL à CHF 0.- prolongée jusqu'à fin avril 2004! --- Considerando la grande domanda, l'offerta sunrise ADSL a zero franchi sarà prolungata sino a fine aprile 2004! --- __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] plot & easy question
Dear all I guess I have a quite simple question I do not have any problem to plot the data (daily averages)in different plots as monthly time series (for example with a for loop over the different months). My problem begin when I want to represent always as a time serie 1999 data (from june) and 2000 data (until june) in a single plot(always as daily averages). Actually I would like to see the winter data in the middle of the plot. Do you have an advice? Thank you for your answer Mike > data Datum Bod Loc Mag Chi Lug Mil Com day month year Smog 1 19990601 NA NA 27.6 33 30.8 74.1 53.3 16 19991 2 19990602 NA NA 21.8 49 48.9 81.9 70.2 26 19991 3 19990603 NA NA 38.9 31 26.9 53.3 49.4 36 19991 4 19990604 NA NA 16.8 11 11.0 23.4 19.5 46 19991 5 19990605 NA NA 9.0 18 13.6 36.4 32.5 56 19991 6 19990606 NA NA 12.9 37 23.2 33.8 39.0 66 19991 579 2530 NA 16 12.9 16 11.3 48.1 15.6 30 12 2000 2 580 2531 NA 18 21.0 22 12.1 66.3 39.0 31 12 2000 2 --- Aufgrund grosser Nachfrage: sunrise ADSL Null-Franken-Angebot verlängert bis Ende April 2004! --- En raison de la forte demande: offre sunrise ADSL à CHF 0.- prolongée jusqu'à fin avril 2004! --- Considerando la grande domanda, l'offerta sunrise ADSL a zero franchi sarà prolungata sino a fine aprile 2004! --- __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] tapply
Dear all
I have a dataframe containing hourly data of 3 parameters.
I would like to create a dataframe containg daily mean values of these
parameters. Additionally I want to keep information about time of
measurement ("year","month","day").
With the function tapply I can average over a column of the dataframe.
I can repeat the function 2 time and merge the vectors. In this way I
obtain my new dataframe (see below).If I want to add the column day,
month and year I can repeat tapply other three time. This system works.
Question: is there a function that average in a single step over the 3
columns?
Thanks a lot for your answer!
Regards
Mike Campana
read the data
setwd("c:/R")
data <- NULL
data <- as.data.frame(read.table(file="Montreal.txt",header=F,skip=15))
colnames(data)
<-c("year","month","day","hour","min","temp","press","ozone")
### create mean value
temp_daily <-
tapply(data$temp,data$year*1+data$month*100+data$day,FUN=mean)
press_daily <-
tapply(data$press,data$year*1+data$month*100+data$day,FUN=mean)
ozone_daily <-
tapply(data$ozone,data$year*1+data$month*100+data$day,FUN=mean)
### merge the data
newdata <- as.data.frame (cbind(temp_daily,temp_daily,temp_daily))
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[R] creating a ps. file
Dear all I wrote a routine. At the end of each cycle of the loop I would like to save the result (plot) in a postcriptfile. Of course if I just use dev.print in the following way: dev.print(device=postcript, 'c:/Rfigures/plot_1.ps") I overwrite my results with the second cycle of the loop. I suppose there is a way to define the file name so that several plots are created(plot_1,plot_2...). Could you give me an advice? Thanks a lot Best Regards Mike --- sunrise ADSL Gratis, einfach und schnell: sunrise ADSL, inkl. USB Modem und einmalige Aufschaltgebühr http://internet.sunrise.ch/de/internet/int_ads.asp --- sunrise ADSL Gratuit, simple et rapide: sunrise ADSL, Modem USB inclus, aucun frais de mise en service http://internet.sunrise.ch/fr/internet/int_ads.asp --- sunrise ADSL Gratis, pratico e veloce: sunrise ADSL, incl. Modem USB et tassa attivazione http://internet.sunrise.ch/it/internet/int_ads.asp --- sunrise ADSL free, simple and fast: ADSL from sunrise, incl. Modem USB and one- time activation fee http://internet.sunrise.ch/de/internet/int_ads.asp --- __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] failure in simple calculation!!
Hello
I have a dataframe, at least I think I created it by using:
DELTA <- as.data.frame(DELTA)
The dataframe has 2 columns made of numbers("data" and "delta")
If I use DELTA$delta + 12
I get a sequence of NA Warning message:
"+" not meaningful for factors in: Ops.factor(DELTA$deltatemp, 12)
What does this mean? Can you explain me please why I can not make this
simple calculation? And what can I do to the calculation?
further information:
mode(DELTA)
[1] "list"
DELTA$delta
.
[685] 6.04 7.84 8.01 8.43 6.6 6.33 8.42 8.86 10.25 8.2 9.32 9.77
[697] 7.9 6.42 5.95
474 Levels: -0.02 -0.13 -0.22 -0.46 -0.55 -1.26 -1.83 -1.86 -2.45 -3.2 ...
9.9
What does this last row mean?
Thanks a lot for the answer and on general for the help contaned in the
mailing list
Mike
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