similar to: recursively Merging a list a zoo objects

Dear All, I have several sets of data such as this: year jday avg_m3s 1 1960 1 4.262307 2 1960 2 4.242308 3 1960 3 4.216923 4 1960 4 4.185385 5 1960 5 4.151538 6 1960 6 4.133846 ... There is a value for each day of multiple years. In this particular data set it goes up to 1974. I am am looking to obtain the minimum and maximum values for each year, but also know on which

Hi All, I am unable to get the plot geomap in googleVis package. data is as follows > head(index.ret) country ytd 1 Argentina -10.18 2 Australia -3.42 3 Austria -2.70 4 Belgium 1.94 5 Brazil -7.16 6 Canada 0.56 > map1 = gvisGeoMap(index.ret,locationvar = 'country', numvar = 'ytd') > plot(map1) But it just displays a blank page, showing an

Thank you for providing the example code... for the request of running it multiple times it would have helped if you could have confirmed that the example ran through without errors... there were a lot of mistakes in it. Look into using the reprex package to check your example next time. I don't do this kind of analysis... I really don't know what to expect from the functions. The

I am using the following script to make .jpg files. jpeg('plotx.jpg') ddat <-read.table("file",header=T) attach(ddat) tdat<-read.table("file1") plot(xx1,yy1,type='p',pch=1,col="blue",cex=0.2,xlim=c(0,3.5),ylim=c(-75,75)) points(tdat,col="green",pch=1,cex=0.2) dev.off() The problem is that I want the points to be very small;

Dear all, I have a data similar to this: myframe<- data.frame (ID=c("Ernie", "Ernie","Ernie","Ernie"), Timestamp=c("24.09.2012 08:00", "24.09.2012 09:00", "24.09.2012 10:00", "25.09.2012 10:00"), Longitude=c("8.481","8.482","8.483","8.481"),

[answers inline] On 18 August 2017 at 20:08, Dagmar <Ramgad82 at gmx.net> wrote: > > myframe<- data.frame (ID=c("Ernie", "Ernie","Ernie","Ernie"), > Timestamp=c("24.09.2012 08:00", "24.09.2012 09:00", "24.09.2012 10:00", > "25.09.2012 10:00"),

Hi, with newest R.app (1.11) / R2.1.0a if a package is installed for the current user, it is not installed properly (does not get an installed version entry and is not loadable). If installed system wide it works. The package I installed is xgobi. MAC OS X 10.4.1 Volker - - - - - - - - - - - - - ------ ---- ---- --- -- - - - - - - - - - - - - - - -

I have date data as a numeric and hourly data in 0 to 2300 hours in a dataframe. d <- rep(20110101,24) h <- seq(from = 0, to = 2300, by = 100) df <- data.frame(LST_DATE = d, LST_TIME = h, data = rnorm(24, 0, 1)) S <- chron(dates. = as.character(df$LST_DATE), times. = paste(as.character(df$LST_TIME/100), ":0:0", sep = ""), format =

i want to plot the histogram and the curve in the same graph.if i have a set of data ,i plot the histogram and also want to see what distribution it was.So i want to plot the curve to know what distribution it like. -- View this message in context: http://www.nabble.com/how-to-plot-the-histogram-and-the-curve-in--the-same-graph-tp20082506p20082506.html Sent from the R help mailing list archive at

Hi, I am trying to assign "Season" values to dates depending on when they occur. For example, the following dates would be assigned the following "Season" numbers based on the "season" intervals detailed below in the code: ddate Season 29/12/1998 20:00:33 1 02/01/1999 05:20:44 2 02/01/1999 06:18:36 2 02/02/1999

Fellow R users, I'm using the BCE {BCE} function to run a Bayesian sediment mixing model. The aim is to find the optimum % contribution from each of the 4 source areas that can yield the target geochemistry. I have geochemistry for 4 source areas called Rat: Rat<-read.table(text="CaO MgO Na2O Al2O3 Topsoils 2.511250 0.7445500 0.7085500 14.10375 ChannelBanks

Given a long zoo matrix, the goal is to "sweep" out a statistic from the entire length of the sequences. longzoomatrix<-zoo(matrix(rnorm(720),ncol=6),as.yearmon(outer(1900,seq(0,length=120)/12,"+"))) cnames<-c(12345,23456,34567,45678,56789,67890) colnames(longzoomatrix)<-cnames longzoomatrix[1:24,] 12345 23456 34567 45678

A quick question x <- as.yearmon(2000 + seq(0, 23)/12) x [1] "Jan 2000" "Feb 2000" "Mar 2000" "Apr 2000" "May 2000" "Jun 2000" "Jul 2000" "Aug 2000" "Sep 2000" "Oct 2000" "Nov 2000" "Dec 2000" "Jan 2001" [14] "Feb 2001" "Mar 2001" "Apr

I have a matrix of zoo series. each series is in a column. x <- as.yearmon(2000 + seq(0, 23)/12) # 24 months of data, lets make 20 sets of random data testData <- matrix(rnorm(480),ncol=20) # make a zoo object and columns will hold the 20 series TestZoo <- zoo(testData,order.by=x) # now run lm for just one series. m <- lm(TestZoo[,1]~time(TestZoo))$coeff[2] m time(TestZoo)

I have a 43MB dataframe ( 5 variables) and I'm trying to summarize subsets of the data. I've RTFM ( not very clear) and looked at a variety of samples but cant seem to figure out how to make these functions work. A sample of what I want to do would be this: ids<-seq(1,50) years<-c(rep(5,10),rep(6,10),rep(7,10),rep(8,20))

Dear all, I have several text files looking like this: 9063032 19700201 22:00 174.067 9063032 19700201 23:00 174.076 9063032 19700202 00:00 174.085 9063032 19700202 01:00 174.091 9063032 19700202 02:00 174.094 9063032 19700202 03:00 174.091 9063032 19700202 04:00 174.082 9063032 19700202 05:00 174.079 And I run this loop: for (j in 1:nr.of.files) { #Import: DF <-

j<-matrix(nrow=10,ncol=10) k<-matrix(seq(1:50), ncol=10) row.names(k) <- seq(2,10,by=2) j [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,] NA NA NA NA NA NA NA NA NA NA [2,] NA NA NA NA NA NA NA NA NA NA [3,] NA NA NA NA NA NA NA NA NA NA [4,] NA NA NA NA NA NA NA NA NA NA [5,] NA

Hi, I'm really struggling with barplot I have a data.frame with 3 columns. The first column represents an "incident" type The second column represents a "month" The third column represents a "time" Code for a sample data.frame incidents <- rep(c('a','b','d','e'), each =25) months <- rep(c(1,2), each =10) times

Given a matrix of m*n, I want to reorder it as a vector, using a row major transpose. so: > m<-matrix(seq(1,48),nrow=6,byrow=T) > m [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [1,] 1 2 3 4 5 6 7 8 [2,] 9 10 11 12 13 14 15 16 [3,] 17 18 19 20 21 22 23 24 [4,] 25 26 27 28 29 30 31 32 [5,] 33 34 35 36 37

Assuming a data frame or matrix with two columns representing variable that you want to aggregate over. you want to calculate column means, by year, for each Id example<-data.frame(id=c(rep(12345,5),rep(54321,6),rep(45678,7)),Year=rep(seq(1900,1902,by=1),6), x=seq(1,18,by=1),y=seq(18,1,by=-1)) example id Year x y 1 12345 1900 1 18 2 12345 1901 2 17 3 12345 1902 3 16 4 12345