similar to: Removing rows if certain elements are found in character string

I adjusted an exponential regression to the following data and wish to plot confidence bands as well. Is this possible? Any help greatly appreciated. Claudia x <- c(1989,1990,1991,1992,1993,1994,1995,1996,1997,1998,1999,2000,2001, 2002,2003,2004,2005,2006,2007,2008,2009) y <- c(987,937,810,749,1087,807,1050,1294,1455,1022,927,403,698,1191,1078, 1176,1125,936,1263,647,868) dat <-

Dear All, I would like to apply two different "for loops" to each set of four columns of a matrix (the loops here are simplifications of the actual loops I will be running which involve multiple if/else statements). I don't know how to "alternate" between the loops depending on which column is "running through the loop" at the time. ## Set up matrix J <- 10 N

I ran this code (several times) from the Quick-R web page ( http://www.statmethods.net/advstats/cart.html) but my cross-validation errors increase instead of decrease (same thing happens with an unrelated data set). Why does this happen? Am I doing something wrong? # Classification Tree with rpart library(rpart) # grow tree fit <- rpart(Kyphosis ~ Age + Number + Start,

(corrected version of previous posting) I fit a GAM to turtle growth data following methods in Limpus & Chaloupka 1997 (http://www.int-res.com/articles/meps/149/m149p023.pdf). I want to obtain figures similar to Fig 3 c & f in Limpus & Chaloupka (1997), which according to the figure legend are "expected size-at-age functions" obtained by numerically integrating

I am trying to apply methods used by Chaloupka & Limpus (1997) ( http://www.int-res.com/articles/meps/146/m146p001.pdf) to my own turtle growth data. I am having trouble with two things... 1) After the GAM is fit, the residuals are skewed. >m1 <- gam(growth~s(mean.size, bs="cr")+s(year,bs="cr",k=7)+s(cohort,bs="cr")+s(age,bs="cr"), data=grow,

Hi, set.seed(45) test1<-data.frame(columnA=rnorm(7,45),columnB=rnorm(7,10)) #used an example probably similar to your actual data apply(test1,2,function(x) sprintf("%.1f",median(x))) #columnA columnB # "44.5"? "10.2" par(mfrow=c(1,2)) lapply(test1,function(x) {b<-

Hello everybody, how can I reduce e.g. 30 days from a date? When I do the following "2011-05-01 CEST" -"2011-04-01 CEST" I get: "Time difference of 30 days" an thats fine. But when I try "2011-05-01 CEST" - 30 I get nonsense. So how can I subtract some days, month or years from a date? thanking you in anticipation Claudia Paladini

Hello all, I have a .csv file like below. Tool,Step_Number,Data1,Data2... etc up to 100 columns. A,1,0,1 A,2,3,1 A,3,2,1 . . B,1,3,2 B,2,1,2 B,3,3,2 . . ...... so on upto 50 rows where the column "*Tool*" has distinct steps in second column "*Step_Number*",but both have same entries in Step_Number column. I want the output like below.

Hello R users, This is more of a convenience question that I hope others might find useful if there is a better answer. I work with large datasets that requires multiple parsing stages for different analysis. For example, compare group 3 vs. group 4. A more complicated comparison would be time B in group 3 of group L with B in group 4 of group L. I normally subset each group with the

Hi all, may i know is it possible to plot a graph by first letter? for example: Name: Age: Angel 20 Amelia 20 Bernard 19 Stephanie 20 Vanessa 22 Angeline 23 Camel 21 If I want to plot the name started with letter 'A' and their Angel,

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 <- data.table(xx , yr1, yr2, yr3) dat2[, !names(dat2)

Hi, Assuming that you provided the sample data from the file. temp <- readLines(textConnection("#Hogd/met, Temp, 005[M], Value #Hogd/met, Difftemp, 051[M], Value BA0+ 1 MTEMP005 1 [deg.C] 2 MDTMP051 1 [deg.C] EOH 891231, 2400, -1.5, -0.21, 900101, 0100, -1.4, -0.25, 900101, 0200, -1.6, -0.28, 900101, 0300, -1.7, -0.25, 900101, 0400, -2.1, -0.0999999, 900101, 0500, -2.3, -0.0899999,

Let me start by saying I am rather new to R and generally consider myself to be a novice programmer...so don't assume I know what I'm doing :) I have a large matrix, approximately 300,000 x 14. It's essentially a 20-year dataset of 15-minute data. However, I only need the rows where the column I've named REC.TYPE contains the string "SAO " or "FL-15". My

Steven, Just want to add a few things to what people wrote. In base R, the methods mentioned will let you make a copy of your original DF that is missing the items you are selecting that match your pattern. That is fine. For some purposes, you want to keep the original data.frame and remove a column within it. You can do that in several ways but the simplest is something where you sat the

I want to write code that says "If you find an element equal to 4 in this vector for each person in the data set tested separately, then put in 1 for 2 and 2 for 4, else leave the variable as is" u.ppl <- (unique(init.dat1$grid)) l.ppl <- length(u.ppl) for (i in 1:l.ppl) { if (grep("4",init.dat1$Slide1_RESP)) {2 == 1, 4 == 2}; else

Hi, It would be better if you provided the output of dput(dataset).? I am not sure about the structure of your dataset. Just from reading the data as is shown. dat1<- read.table(text=" separator,tissID >,>,2 ,2,1 ,6,5 ,11,13 >,>,4 ,4,9 ,6,2 ,7,3 ,21,1 ,23,58 ,25,9 ,26,4 >,>,11 ,1,12 >,>,21 ,4,1 ,11,3

Hi Friends I am new here and have a problem Year Market Winner BID 1 1990 ABC Apple 0.1260 2 1990 ABC Apple 0.1395 3 1990 EFG Pear 0.1350 4 1991 EFG Apple 0.1113 5 1991 EFG Orange 0.1094 For each year and separately for the two

Hey R users, suppose we have data: [1] 2010.12.26 00:00:52 688,88 11,69 43,00 [2] 11,69 43,00 [3] 11,69 43,00 [4] 11,69 43,00 [5] 11,69 43,00 [6] 11,69 43,00 [7] 11,69 43,00 [8] 11,69 43,00 [9] 11,69 43,00 [10] 11,69 43,00 [11] 11,69 43,00 [12] 11,69 43,00 [13] 11,69 43,00 [14] 11,69 43,00 [15] 11,69

Hello, What would be the best set of R functions to parse and transform a file? My file looks as shown below. I would like to plot this data and I need to parse it into a single data frame that sorts of "transposes the data" with the following structure: > df <- data.frame(n=c(1,1,2,2),iter=c(1,2,1,2),step=as.factor(c('Step 1', 'Step2', 'Step 1',

Hi, I haven't tried the code yet. Is there a way to parse this data using R and create bar plots so that each program's 'RAM used' figures are grouped together. So 'uuidd' bars will be together. The data will have about 50 sets. So if there are 100 processes each will have about 50 bars. What is the recommended way to graph these big barplots ? I am looking