search for: cassie

Dear all, I created a n*2 matrix and I want to replace the ith row with a vector. Can anyone suggest me what is the simple way to do it? Sorry for bothering you with such simple question. I appreciate any hints or advice. Thanks in advance. Cassie [[alternative HTML version deleted]]

Dear all, I am trying to simulate from truncated Pareto distribution. I know there is a package called PtProcess for Pareto distribution...but it is not for truncated one. Can anyone please help me with this? Thanks in advance. Cassie [[alternative HTML version deleted]]

Dear all, I am trying to simulate from discrete uniform distribution. But I could not find any in-built code in R. Could anyone help me please? Thanks in advance for the time and help. Cassie [[alternative HTML version deleted]]

...esponding to those locations. I am using image.plot command for that. But I want to overlay a map of US on that plot. I know we can use map('usa') to draw it, but I dont know how can I overlay it on the plot of the precipitation data. Can anyone please help me with it? Thanks in advance. Cassie. [[alternative HTML version deleted]]

...follows: a=array(NA,c(m,2,N,G)), Even though the numbers N and G are fixed, m is a vector. So, I can't define 'a' as above. For eg, N=3 G=6, but m=c(2,7,5) and length(m)=N. Can anyone suggest me how can I define 'a' when m is a vector? Thanks in advance for your help. Cassie [[alternative HTML version deleted]]

Does anyone know how to do a fry plot in R? I have 600-800 points per image, and I really don't want to attempt that manually. Thank you! Cassie [[alternative HTML version deleted]]

...gh and hard code the data to the specific stations. I tried stnID <- stnid[which(duplicated(stnid)==FALSE)] for(i in 1:length(stnID)) { ppt[i] <- ppt[which(stnid==[i])] } but it doesn't like to use the which function inside a for loop? Any idea's here is a sample of my code... thanks-Cassie ### label data ### stnid <- dat1[,1] #list of stations lat <- dat1[,2] #latitude of station lon <- dat1[,3] #longitude of station year <- dat1[,4] # data year month <- dat1[,5] #data month ppt <- dat2 #precipiation data #### station's missing zero years ### ppt2 <- matr...

..., I need to simulate values from a Poisson distribution which is truncated at certain value 'a'. Can anyone tell me if there is in-built package in R which can simulate from a truncated Poisson? If not, what should be the steps to write a function which would do that? Thanks in advance. Cassie [[alternative HTML version deleted]]

...]=3 and I want to replace x[3,2] by x[2,2]. Similarly, I want to replace x[5,2] with x[4,2], which means that for duplicate entries in the 1st column, I want to replace the corresponding entries in the 2nd column with its 1st entry of the duplicate row. Can anybody help me in this regard? Thanks, Cassie [[alternative HTML version deleted]]

...in 'seconds' unit to the 12 hour time format. For example, suppose the data set looks like x=c(36885,84000,20) #x in seconds I want to get the output as [1] " 11:14:45 AM" [2] " 11:20:00 PM" [3] "12:20:00 AM" Does anyone have any idea? Thanks in advance. Cassie [[alternative HTML version deleted]]

...ounter to be 2 and regenerate from i=2. But I guess I have not put the break command in the right place, so even though the 'if' condition becomes true and it sets i=2, it does not regenerate the values of 'a' since the beginning. Any help is really appreciated. Thanks in advance. Cassie [[alternative HTML version deleted]]

...run the code I get the following error: Error in image.default(values,x=sort(long),y=sort(lat), increasing 'x' and 'y' values expected Even though I sort the latitude and longitude in increasing order, I am getting this error. Any help is greatly appreciated. Thanks in advance, Cassie. [[alternative HTML version deleted]]

..., where we can break the x-axis into several parts putting a 'break' at the discontinuity..what I mean is if the x-axis start from 0 and break at say 3, then start at 19 and break at 21 and the last part starts at say 98. In this way, the histograms would be prominent. Thanks in advance. Cassie [[alternative HTML version deleted]]

...have approximately 10000 observations, so I can't go through all the observations individually in order to find which row in the array has this format. Can anyone suggest me how to handle this issue? Is there any way to convert %I %M %p format to %I %M %S %p? Thanks in advance for the help. Cassie [[alternative HTML version deleted]]

..., where i=1:n and j=1:m. Unlike the usual for loop, i should go in the following way: For j=1, i=1,2,3,....n For j=2, i=n,n-1,n-2,......,1 For j=3, i=1,2,3,.....n etc. which means i should go in either increasing or decreasing order alternatively. Can anyone please help me in doing this? Thanks, Cassie [[alternative HTML version deleted]]

...dev.off() RStudioGD 2 when I try colnames(data2) or names(data2) ,it shows as follows: > colnames(data2) [1] "X" "X1" "X0" "." "..1" I think this may be the problem. Could you please tell me how to treat it? Thank you! Cassie [[alternative HTML version deleted]]

Hi, i'm not english and i'm not very familiar to R, and i'm asking if you can help me. I'm wondering how to create a multivariate normal an then repeat this for a sample of T=1000, and the save this result. Thank you very much for your helping -- View this message in context: http://r.789695.n4.nabble.com/Multivariate-Normal-and-loops-tp4633504.html Sent from the R help mailing

...b.loc, envir = dataEnv) : data set 'bindat.txt' not found Warning in utils::data(list = f, package = package, lib.loc = lib.loc, envir = dataEnv) : data set 'binpar.txt' not found Warning in utils::data(list = f, package = package, lib.loc = lib.loc, envir = dataEnv) : data set 'cassie.txt' not found Warning in utils::data(list = f, package = package, lib.loc = lib.loc, envir = dataEnv) : data set 'expdat.txt' not found Warning in utils::data(list = f, package = package, lib.loc = lib.loc, envir = dataEnv) : data set 'fiftn80.txt' not found Warning in utils::d...

Hi everyone, i'm not very good with R, and i know that my question is quite simple. I generated this vector for a sample of 100 and i want to know how to save these as ez1, ez1, ez3...ez100. I guess i have to do it by "for", but i didn't succeed.. i put here my output ezt [1] -0.05470228 -1.76356668 -1.68843463 -0.77283767 0.25077826 -0.65236833 1.05230056 -0.48900254

Dear all, I am writing a program using for loop. The loop contains i which runs from 1:n. There is a feature I need to include here. That is, when i=n, i+1 would be 1, not n+1. Basically the loop should run in a circular fashion. That also means, if i=1, i-1=n. Can anyone help me with this? How can I define it? Thanks. [[alternative HTML version deleted]]