search for: yr

Trying to use contrast to look at differences within an lme lme.fnl.REML <- lme(Max ~ S + Tr + Yr + Tr:Yr, random = ~1 |TID, method = "REML") I have three levels of Tr I'm trying to contrast among different years (R, T97, T98), years = 1997-1999, so I'm interested in contrasts of the interaction term. > anova(lme.fnl.REML) numDF denDF F-value p-value...

Hi All, I've been searching the help archives but haven't found a workable solution to this problem. I'm running an lme model with the following call: >lme.fnl <- lme(Max ~ S + Tr + Yr + Tr:Yr, random = ~1 |TID) > anova(lme.fnl) numDF denDF F-value p-value (Intercept) 1 168 19255.389 <.0001 S 1 168 5.912 0.0161 Tr 2 116 15.919 <.0001 Yr 1 168 77.837 <.0001 Tr:Yr 2 168 47.5...

...ds on the next line. Usually this is fine, but sometimes R appears confused by line wrappings (more likely of course it is me who is confused). I tried ';' to explictly break the line, but no dice. An example (partial function). inrange _ function(v,Y,s,z,mlv,met){ lower _ mean(v[data$yr==Y])-(var(v[data$yr==Y])^.5) upper _ mean(v[data$yr==Y])+(var(v[data$yr==Y])^.5) ML _ mlv[ML$yr==Y & ML$metcode==met] diff _ 0 if (ML<lower) {diff <- lower-ML; # tried with and without ; v <<- v-diff} } gives: > inrange _ function(v,Y,s,z,mlv,met){ + lower _ mean(v[da...

Hi expeRts, I would like to calculate weighted mean by two factors. My code is as follows: R> tmp <- by(re$meta.sales.lkm[, c("pc", "sales")], re$meta.sales.lkm[, c("size", "yr")], function(x) weighted.mean(x[,1], x[,2])) The result is as follows: R> tmp size: micro yr: 1994 [1] 1.090 ------------------------------------------------------------ size: small yr: 1994 [1] 1.135 ------------------------------------------------------------ size:...

...ow can I calculate the future growth increment of each tree for the next 3 years. The following Rscript was written, #---------- a0 <- data.frame(d0=seq(5,50,5) , sp=gl(2,5,10), grow=rep(0.5,10)) a2<- list() for( i in 1:3){ a1 <- a0$d0+a0$grow a2[[i]] <- cbind(sp=a0$sp,d0=a1+i,yr=i) } as.data.frame(do.call(cbind,a2)) > as.data.frame(do.call(cbind,a2)) sp d0 yr sp d0 yr sp d0 yr 1 1 6.5 1 1 7.5 2 1 8.5 3 2 1 11.5 1 1 12.5 2 1 13.5 3 3 1 16.5 1 1 17.5 2 1 18.5 3 4 1 21.5 1 1 22.5 2 1 23.5 3 5 1 26.5 1 1 27.5 2 1 28.5 3 6...

...24325 The dataset runs from 1986 to 2007. I want to be able to take subsets of my data based on date e.g. data between 2000 - 2005. As it stands, I can't work with the dates as they are not in correct format. I tried successfully converting the dates to just the year using: transform(data, Yr = format(as.Date(as.character(Date),format = '%d-%b-%y'), "%y"))) This gives the following format: Date Price Open.Int. Comm.Long Comm.Short net.comm Yr 1 15-Jan-86 673.25 175645 65910 28425 37485 86 2 31-Jan-86 677.00 167350 54060 27120 269...

...le names are not very long, and are accepted by write.csv. alldata is a list containing 4 dataframes, with each dataframe having the the same variable names, which are: > names(avg8302) [1] "ID" "cluster" "rec.unit" "int.hib" "yr.hib" "yr0309.hib" "int.hib.se" "yr.hib.se" " yr0309.hib.se" "int.cl" [11] "yr.cl" "yr0309.cl" "int.cl.se" "yr.cl.se" " yr0309.cl.se" "int.ru"...

Can anybody help me write a code on the following data example, which fills out all NA values by using a linear interpolation with the two closest values? Doy is day of year (%j). Code example: yr<-c(rep(2000,14)) doy<-c(16:29) dat<-c(3.2,NA,NA,NA,NA,NA,NA,5.1,NA,NA,NA,NA,NA,4.6) ta<-cbind(yr,doy,dat) ta yr doy dat [1,] 2000 16 3.2 [2,] 2000 17 NA [3,] 2000 18 NA [4,] 2000 19 NA [5,] 2000 20 NA [6,] 2000 21 NA [7,] 2000 22 NA [8,] 2000 23 5.1 [9,] 2...

Can GLMM take formula derived from another object? foo <- glm (OVEN ~ h + h2, poisson, dataset) # ok bar <- GLMM (OVEN ~ h + h2, poisson, dataset, random = list (yr = ~1)) #error bar <- GLMM (foo$formula, poisson, dataset, random = list (yr = ~1)) #Error in foo$("formula" + yr + 1) : invalid subscript type I am using R2.1.0, lme4 0.8-2, windows xp. Below is a dataset if you need (yr is a factor). > dataset OVEN h h2 yr 1 2 30 900...

Dear all, given I have data in a data.frame which indicate the number of people in a specific year at a specific age: n <- 10 mydf <- data.frame(yr=sample(1:10, size=n, replace=FALSE), age=sample(1:12, size=n, replace=FALSE), no=sample(1:10, size=n, replace=FALSE)) Now I would like to make a matrix with (in this simple example) 10 columns (for the years) and 12 rows (for the ages). In each cell, I would l...

...gmail.com> wrote: > > ?You'll want to use grep() or grepl(). By default, grep() uses extended > regular expressions to find matches, but you can also use perl regular > expressions and globbing (after converting to a regular expression). > For example: > > grepl("^yr", colnames(mydata)) > > will tell you which 'colnames' start with "yr". If you'd rather you > use globbing: > > grepl(glob2rx("yr*"), colnames(mydata)) > > Then you might write something like this to remove the columns starting with yr: &...

...r would try should help me with my present task as well as future problems. Below I've included some lines that will generate a sample data set similar to the one I'm working with: # #Example data: # case <- sample(seq(1,10000,1),5000,replace=FALSE) hsgpa <- rbeta(5000,7,1.5)*4.25 yr <- sample(seq(1993,2005,1),5000,replace=TRUE) conf <- sample(letters[1:5],5000,replace=TRUE) data <- data.frame(case=case,hsgpa=hsgpa,yr=yr,conf=conf) data$conf <- as.character(data$conf) s1 <- sample(seq(1,5000,1),500,replace=FALSE) k <- data$hsgpa k[row.names(data) %in% s1] <...

I have a, no doubt, simple question. I wish to write a function such that a <- 9 b <- 10 changer _ function(x,y) { if (y>x){ x <<- Y+1}} Of course there are easier ways to accomplish the task above, but I am more interested in how to have the "x <<- Y+1" part of the function to change x in place for purposes of a much larger function. I have been wrestling with

You'll want to use grep() or grepl(). By default, grep() uses extended regular expressions to find matches, but you can also use perl regular expressions and globbing (after converting to a regular expression). For example: grepl("^yr", colnames(mydata)) will tell you which 'colnames' start with "yr". If you'd rather you use globbing: grepl(glob2rx("yr*"), colnames(mydata)) Then you might write something like this to remove the columns starting with yr: mydata <- mydata[, !grepl("...

...edule calendar to display in the view. I''m just starting with this and am simply trying to display the parameters I''m trying to pass in. Here''s what I''ve got so far: Controller: class SchedulesController < ApplicationController def index show_cal(yr, mo) render :action => ''show_cal'' end #...other functions not listed def show_cal(yr, mo) # set up dates based on yr, and mo end end View: show_cal.rhtml <h2>Calendar</h2> <%= yr %> <%= mo %> <%= link_to ''previous month&...

I'd like to sum a matrix only where the matrix meets a specific condition. The matrix has 365 rows and about 50,000 columns. str(cdem.mat.yr) num [1:365, 1:41772] -43.5 -48.4 -45.9 -38.4 -32 ... I'm having trouble replicating this because it is not working out for me, so I'm unsure I can provide an solid working example (apologies). I would like to subset my matrix where the values are greater than zero. I can do this easily...

Hi All. I have the commands below to create a stripchart/plot. I was hoping to color the points in the plot by yr, and use a symbol that varied with resp. However, the outcome makes it appear as though the point by point col and pch data is not being passed properly. Any suggestions? And truthfully, I?d rather be doing this with Lattice, but I?ve tried several variations of stripplot and can?t even get somet...

...le to do it in R, but when i use sweave, I can only print the last page. any help would be greatly appreciated Thanks Cameron #R code library(fImport) IBM <- yahooSeries("IBM", from="2000-01-01") IBM.Close <- IBM[,"IBM.Close"] rng=range(time(IBM.Close)) Syr <- as.numeric(format(rng[1],"%Y")) Eyr <- as.numeric(format(rng[2],"%Y")) Smth <- as.numeric(format(rng[1],"%m")) for( yr in Syr:Eyr){ par(mfrow=c(4,3)) Temp1 <- IBM.Close[which(format(time(IBM.Close),"%Y")==yr),] Temp3 <- tapply(Temp1[...

Dear List, I need to add a subscript and a superscript to some of the ions in the labels on some plots. I have got to here but now I'm stuck: plot(1:10, xlab = expression(paste("nm SO"[4], " ", mu, "eq cm"^{-2}, " yr"^{-1}))) Which gives almost what I require. No matter what I tried, however, I could not get bot a sub script *and* a superscript attached to the SO in the label. In LaTeX I would just do $SO_4^{2-}$ but taking that to R produces a syntax error: plot(1:10, xlab = expression(paste("...

You rang sir? library(tidyverse) xx = 1:10 yr1 = yr2 = yr3 = rnorm(10) dat1 <- data.frame(xx , yr1, yr2, y3) dat1 %>% select(!starts_with("yr")) or for something a bit more exotic as I have been trying to learn a bit about the "data.table package library(data.table) xx = 1:10 yr1 = yr2 = yr3 = rnorm(10) dat2 <- d...