similar to: Linear Model Prediction

Dear R Users, Could you please help me on the following issue? I have a real large yearly data set. For each year I have 365 flow values. Some of the flow values are not known and that’s why you will see NA written in those slots. I wanted to know, is there a way that I can estimate those values? I tried approx command but it seems least helpful for the kind of issue I am up against.

Hello I have dataframe 101 2008-07 0.2898966 102 2008-08 0.3101667 103 2008-09 0.3730476 104 2008-10 0.2717037 105 2008-11 0.1344286 106 2008-12 0.1375000 107 2009-01 0.1781000 108 2009-02 0.2146667 109 2009-03 0.2808235 110 2009-04 0.4326250 111 2009-05 0.3420741 112 2009-06 0.2675238 113 2009-07 0.2478667 114 2009-08 0.3147000 115 2009-09 0.3437826 116 2009-10 0.2057391 117 2009-11 0.1824737 118

Dear all, I think this is an easy task, but I don't know how to do it. Specifically, I have 69 columns with 300.000 rows. In each cell there is a code like "2E3", "4RR", etc. I now have a list that replaces this with values, e.g., old new 2E3 5 4RR 3 etc. The list ist about 1600 rows long, so also to extensive for normal solutions Do you how to do that? It would

HI, You may try this: dat1<- read.table(text=" CustID TripDate Store Bread Butter Milk Eggs 1 2-Jan-12 a 2 0 2 1 1 6-Jan-12 c 0 3 3 0 1 9-Jan-12 a 3 3 0 0 1 31-Mar-13 a 3 0 0 0 2 31-Aug-12 a 0 3 3 0 2 24-Sep-12 a 3 3 0 0 2 25-Sep-12 b 3 0 0 0 ",sep="",header=TRUE,stringsAsFactors=FALSE) dat2<- dat1[,-c(1:3)] res<- lapply(seq_len(ncol(dat2)),function(i)

HI Elisa, You could use ?cut() vec1<-c(33,18,13,47,30,10,6,21,39,25,40,29,14,16,44,1,41,4,15,20,46,32,38,5,31,12,48,27,36,24,34,2,35,11,42,9,8,7,26,22,43,17,19,28,23,3,49,37,50,45) label1<-unlist(lapply(mapply(c,lapply(seq(0,45,5),function(x) x),lapply(seq(5,50,5),function(x) x),SIMPLIFY=FALSE),function(i) paste(i[1],"<x<=",i[2],sep="")))

Hello, I'm working with a very large dataset (250,000+ lines in its' current form) that includes presence only data on various species (which is nested within different sites and sampling dates). I need to convert this into a dataset with presence/absence for each species. For example, I would like to expand "My current data" to "Desired data": My current data

[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,

I need to write an if/then statement saying something along the lines of: if (VALUE is in list) {... How do I write that in R's language. Thanks! -- View this message in context: http://r.789695.n4.nabble.com/If-then-statement-if-in-a-list-then-tp4638346.html Sent from the R help mailing list archive at Nabble.com.

Any reason I'd get an error like this? Error in if (round(pos[o + 1]) == (pos[o + 1] - 0.4)) { : missing value where TRUE/FALSE needed but when i do it individually, out of the for loop, > (round(pos[o+1])==(pos[o+1]-.4) ) 65 TRUE -- View this message in context: http://r.789695.n4.nabble.com/TRUE-FALSE-tp4636748.html Sent from the R help mailing list archive at Nabble.com.

What i have is a for loop to name files. I want it to skip over the file if it doesn't exist. Is there a way to do that? s=seq(from = chron("03/15/2012"), to = chron("06/15/2012")) day=format(as.Date(s), "%Y%m%d") for (k in 1:length(day)){ B1=read.csv(paste("S:/file_", day[k], ".csv", sep="")) Date=strptime(B1[,1],

How does one coerce predict.gls to incorporate the fitted correlation structure from the gls object into predictions? In the example below the AR(1) process with phi=0.545 is not used with predict.gls. Is there another function that does this? I'm going to want to fit a few dozen models varying in order from AR(1) to AR(3) and would like to look at the fits with the correlation structure

Hi All, I have tried every fix on my try or tryCatch that I have found on the internet, but so far have not been able to get my R code to continue with the "for loop" after the lmer model results in an error. Here is two attemps of my code, the input is a 3D array file, but really any function would do.... metatrialstry<-function(mydata){ a<-matrix(data=NA, nrow=dim(mydata)[3],

Hello, I've written a simple (although probably overly roundabout) function to test whether two regression slope coefficients from two linear models on independent data sets are significantly different. I'm a bit concerned, because when I test it on simulated data with different sample sizes and variances, the function seems to be extremely sensitive both of these. I am wondering if

Excuse the simple question... I'm not sure what I'm doing wrong with predict, but let me use this example: Suppose I do: dat<-matrix(c(0,0,10,20),2,byrow=T) lm1<-lm(dat[,2]~dat[,1]) Suppose I want to generate the linearly-interpolated y-values between the point (0,0) and (0,20) at every unit interval. I thought I just do: predict(lm1, data.frame(seq(0,10,1))) to get

Dear list members, I'm interested in showing that within-group statistical dependence is negligible, so I can use ordinary linear models without including random effects. However, I can find no mention of testing a model with vs. without random effects in either Venable & Ripley (2002) or Pinheiro and Bates (2000). Our in-house statisticians are not familiar with this, either,

Dear R-help list, I would like to perform multiple comparisons for lme. Can you report to me if my way to is correct or not? Please, note that I am not nor a statistician nor a mathematician, so, some understandings are sometimes quite hard for me. According to the previous helps on the topic in R-help list May 2003 (please, see Torsten Hothorn advices) and books such as Venables &

Hello, I am work with a linear regression model: y=ax+b with the function of lm. y= observed migration distance of butterflies x= predicted migration distance of butterflies Usually the result will show if the linear term a is significantly different from zero based on the p-value. Now I would like to test if the linear term is significantly different from one. (because I want to know

Hi All: the package "MuMIn" can be used to select the model based on AIC or AICc. The code is as follows: data(Cement) lm1 <- lm(y ~ ., data = Cement) dd <- dredge(lm1,rank="AIC") print(dd) If I want to select the model by BIC, what code do I need to use? And when to select the best model based on AIC, what the differences between the function "dredge" in

I found myself wanting to average a vector [vec] within each level of a factor [Fac], returning a vector of the same length as vec. After a while I realised that lm1 <- lm(vec ~ Fac) fitted(lm1) did what I want. But there must be another way to do this, and it would be good to be able to apply other functions than mean() in this way. Cheers, Murray -- Dr Murray Jorgensen

This continues the message "data.frame within a function (PR#9294)" that was posted on 2006/10/12. Duncan Murdoch kindly replied. I'm using the current version R 2.4.0, but the same issue exists. Just copy and paste the following code under R, and compare the output of f1() and f2() and the output of f3() and f4(). Does anybody have any idea? Thanks.