similar to: lm and loop over variables

Hello, I have got a dataframe which looks like: y <- c(1,5,6,2,5,10) # response x <- c(2,12,8,1,16,17) # predictor group <- factor(c(1,2,2,3,4,4)) # group df <- data.frame(y,x,group) Now I'd like to resample that dataset. I want to get dataset (row) per group. So per total sample I get 4 rows into a new data frame. How can I do that? Is there any simple approach using an

Hello, I am doing a simple regression using lm(Y~X). As my response and my predictor seemed to be skewed and I can't meet the model assumptions. Therefore I need to transform my variables. I wanted to ask what is the preferred way to find out if predictor and/or response needs to be transformed and if yes how (log-transform?). I found a procedure in "A modern approach to Regressoin in

Just to inform: I posted that before in R-sig-ecology but as it might be interesting also for other useRs, I post it also to the general r-user list: Hello Alexandre, thank you very much. I also found another way to extract summarizing information from lm results over e.g. 1000 repeated model runs: results2 <- t(as.data.frame(results)) summary(results2) Although some questions popped up in

Hi, everyone. I need to run lm with the same response vector but with varying predictor vectors. (i.e. 1 response vector on each individual 6,000 predictor vectors) After looking through the R archive, I found roughly 3 methods that has been suggested. Unfortunately, I need to run this task multiple times(~ 5,000 times) and would like to find a faster way than the existing methods. All three

Hi, I just started with writing functions in R and so some questions popped up. I provide some values as argument to my function such as: function(a,b,c){} Now i want that the function first checks if the arguments are valid for the function. E.g argument "a" has to be a number in the range 0-1. How can that easily done? So far I have: a <- as.numeric(a) if(0 <= a &&

Hello, I want to use a for loop for repeadely calculating a maxent model (package dismo, function maxent()) which creates an object of the class maxent (S4). I want to collect all the resulting object in a list. I tried to simplify my for loop to explain what I want. There are two problems/questions: 1) How can I create the new variables in the loop (using paste) and assign the objects 2) How

Dear R-devel list members, I've discovered an inconsistency in how lm() and similar functions handle logical predictors as opposed to factor or character predictors. An "lm" object for a model that includes factor or character predictors includes the levels of a factor or unique values of a character predictor in the $xlevels component of the object, but not the FALSE/TRUE values

Dear Abby, > On Aug 30, 2019, at 8:20 PM, Abby Spurdle <spurdle.a at gmail.com> wrote: > >> I think that it would be better to handle factors, character predictors, and logical predictors consistently. > > "logical predictors" can be regarded as categorical or continuous (i.e. 0 or 1). > And the model matrix should be the same, either way. I think that

Hi, I have a dataframe and want to remove columns from it that are populated with a similar value (for the total column) (the variation of that column is 0). Is there an easier way than to calculate the statistics and then remove them by hand? A <- runif(100) B <- rep(1,100) C <- rep(2.42,100) D <- runif(100) df <- data.frame(A,B,C,D) # if want to conditionally remove column B and

Hi all, I have one array of predictors, one observation per row, and one array of responses, also arranged one observation per row. I arrange these into a data.frame and call lm() with a pasted-together formula. I would like to call lm() with a number of responses in excess of 100, but for some reason, 39 seems to be a limit. Why do I get an "invalid variable names" error from

Hi, I have a 1-column dataframe with rownames and I want to sort it based on the single column. The typical procedure that is recommended in diverse posts is to use order in the index. But that "destroys" my dataframe structure. Probabaly it is a very simple solution. Here is a short reproducable example: x <- c(1,3,51,2,34,44,12,33,2,8) df <- data.frame(x) rownames(df) <-

Dear R-devel list, I realized that removing a predictor in lm through the "-"'s operator in formula() does not affect the complete cases that are considered. A minimal example is: summary(lm(Wind ~ ., data = airquality)) # 42 observations deleted due to missingness summary(lm(Wind ~ . - Ozone, data = airquality)) # still 42 observations deleted due to missingness, even if only 7

Hello, just to follow up a question from last week. Here what I've done so far (here an example): library(MCMCpack) Y=c(15,14,23,18,19,9,19,13) X1=c(0.2,0.6,0.45,0.27,0.6,0.14,0.1,0.52) X2a=c(17,22,21,18,19,25,8,19) X2b=c(22,22,29,34,19,26,17,22) X2 <- function()runif(length(X2a), X2a, X2b) model1 <- MCMCregress(Y~X1+X2()) summary(model1) but I am not sure if my X2-function is

Hi, I'd like to know if it is possible to use wildcards * for indexing... E.g. I have a vector of strings. Now I'd like to select all elements which start with A_*? I'd also need to combine that with logical operators: "Select all elements of a vector that start with A (A*) OR that start with B (B*)" Probably that is quite easy. I looked into grep() which I think might

Good Morning I want to perform a synthetic control method with R. For this purpose, I created the following code: # Re-load packages library(Synth) library(readxl) # Pfadeinstellung Excel-Blatt excel_file_path <- ("C:\\Users\\xxxxx\\Desktop\\DATA_INVESTMENTVOLUMEN_FOR_R_WITHOUT_NA.xlsx") # Load the Excel file INVESTMENTVOLUME <- read_excel(excel_file_path) #

Hello, assume we have following dataframe: group <-c(rep("A",5),rep("B",6),rep("C",4)) x <- c(runif(5,1,5),runif(6,1,10),runif(4,2,15)) df <- data.frame(group,x) Now I want to select all cases (rows) for those groups which have more or equal 5 cases (so I want to select all cases of group A and B). How can I use the indexing for such questions? df[??]...

I'm fitting a lme growth curve model with two predictors and their interaction as predictors. The multilevel model is nested so that level 1 is time within the individual, and level 2 is the individual. I would like to plot the mean group-level trajectories at plus and minus 1 SD from the mean of the main effects composing the interaction term. Thus, the plot should have 4 lines (mean

Hi, I would like to plot some extra text in my plot. This should be a two line text including a special character (sigma). I tried so far a to use expression in combination with paste and "\n"... but I can't get the line break... Here what I've done so far: plot(1,type="n", xaxt='n', yaxt='n', ann=FALSE) text(1,1,labels=expression(paste(sigma,"\n

Hello R-people! I have a general statistical question about regressions. I just want to describe my case: I have got a dataset of around 150 observations and 1 dependent and 2 independent variables. The dependent variable is of metric nature (in my case meters in a range from around 0.5-10000 m). The first independent is also metric (in mm ranging from 50-700 mm) and it is assumed to be

Hello, I am trying to automate linear regression for many different datasets, each with the same rough format (the last variable is the response). I've been doing something like this: lm=lm(x[,dim(x)[2]] ~ ., data=x) where the dot denotes all variables. However, this means that the response is included as a predictor, which is obviously what I don't want. How do I request that all