similar to: Using apply for two datasets

Dear users, I have a double for loop that does exactly what I want, but is quite slow. It is not so much with this simplified example, but IRL it is slow. Can anyone help me improve it? The data and code for foo_reg() are available at the end of the email; I preferred going directly into the problematic part. Here is the code (I tried to simplify it but I cannot do it too much or else it

R helpers I am getting to grips with R but came across a small problem today that I could not fix by myself. I have 3 text files, each with a single column of data. I read them in using: myData1<-scan("C:/Program Files/R/myData1.txt") myData2<-scan("C:/Program Files/R/myData2.txt") myData3<-scan("C:/Program Files/R/myData3.txt") I wanted to produce a

Hello I have dataframes. mydata1 <-data.frame(value=c(15,20,25,30,45,50),dates=c("2005-05-25 07:00:00 ","2005-05-25 19:00:00","2005-06-25 07:00:00","2005-06-25 19:00:00 ","2005-07-25 07:00:00","2005-8-25 19:00:00")) or mydata2 <-data.frame(value=c(15,20,25,30,45,50),dates=c("2005-05-25 00:00:00 ","2005-05-25

Hi, It works for me to import spss datasets via library(foreign) with read.spss or via library Hmisc by (spss.get). But no matter which way I do import the data, user-defined missings from Spss are always lost. (it makes no difference if there are a single value, a range, or any combination of them. They are always ignored). Is there any way in R to find out if any value was user-defined missing

I'm using ks.test (mydata, dnorm) on my data. I know some of my different variable samples (mydata1, mydata2, etc) must be normally distributed but the p value is always < 2.0^-16 (the 2.0 can change but not the exponent). I want to test mydata against a normal distribution. What could I be doing wrong? I tried instead using rnorm to create a normal distribution: y = rnorm

Dear Jim, Here is again exactly what I did and with the output of Rprof (with this reduced dataset and with a simpler function, it is here much faster than in real life). Thanks you again for your help! ## CODE ## mydata1<- structure(list(species = structure(1:8, .Label = c("alsen","gogor", "loalb", "mafas", "pacyn", "patro",

Dear useRs, How could I obtain the confidence intervals for the means of my treatments, when my data was fitted to a GLM? I need the CI's for the Poisson and Negative Binomial distributions. Here's what I have: mydata1 <- data.frame('treatments'=gl(4,20), 'value'=rpois(80, 1)) model1 <- glm(value ~ treatments, data=mydata1, family=poisson) means1 <-

you could try mapply mydata2<-mapply("[", mydata, lapply(mydata, function(x) !x %in% A)) mydata2[[1]]<-A #to replace the obviously deleted elements of "A" mydata2 mydata2[[1]] mydata2[[2]] mydata2[[3]] mydata2[[4]] Stefano -----Messaggio originale----- Da: r-help-bounces at stat.math.ethz.ch [mailto:r-help-bounces at stat.math.ethz.ch]Per conto di jastar

Thanks a lot, Ista! I really appreciate it. How about a slightly different case as the following: set.seed(1) (tmp <- data.frame(x = 1:10, R1 = sample(LETTERS[1:5], 10, replace = TRUE), R2 = sample(LETTERS[2:6], 10, replace = TRUE))) x R1 R2 1 C B 2 B B 3 C E 4 E C 5 E B 6 D E 7 E E 8 D F 9 C D 10 A E Notice that the factor levels between

Hello everyone, If I want to convert or combine a (large) character matrix into a data frame without having any of its columns convert into a factor class, is there a simple solution? I() says it will operate on 'an object' but it seems that unless the object is a vector, the results are not what I expect. For instance, if g is a 2x2 character matrix, as.data.frame(I(g)) will return an

A very simple question. With a data frame like this: > n = c(2, 3, 5) > s = c("aa", "bb", "cc") > df = data.frame(n, s) I want df$s[1] or df[1,2], but how can I get rid of the extra line in the output about the factor levels: > df$s[1] [1] aa Levels: aa bb cc Thanks, Gang

Suppose I create an R program called myTest.R with only one line like the following: type <- as.integer(readline("input type (1: type1; 2: type2)? ")) Then I'd like to run myTest.R in batch mode by constructing an input file called answers.R with the following: source("myTest.R") 1 When I ran the following at the terminal: R CMD BATCH answer.R output.Rout it failed

Hi all, I am trying to understand Writing R Extension... Section 1.1.5, data: I include two datasets in a package, one using 'save', the other using 'write.table': --- 8< ---- myData1 <- data.frame(x=1:10) write.table(myData1,file="myData1.txt") myData2 <- data.frame(x=2:10) save(myData2,file="myData2.Rdata") --- 8< ---- Then R CMD check aks me to

Anybody knows what functions can be used to calculate variance/covariance with complex numbers? var and cov don't seem to work: > a 1 V1 0.00810014+0.00169366i V2 0.00813054+0.00158251i V3 0.00805489+0.00163295i V4 0.00809141+0.00159533i V5 0.00813976+0.00161850i > var(a) 1 1 1.141556e-09 Warning message: In var(a) : imaginary parts discarded in

Suppose I have a dataframe 'd' defined as L3 <- LETTERS[1:3] d0 <- data.frame(cbind(x = 1, y = 1:10), fac = sample(L3, 10, replace = TRUE)) (d <- d0[d0$fac %in% c('A', 'B'),]) x y fac 2 1 2 B 3 1 3 A 4 1 4 A 5 1 5 A 6 1 6 B 8 1 8 A Even though factor 'fac' in 'd' only has 2 levels, but it seems to bear the birthmark

Hi I am really new using R, so this is really a beginner stuff! I created a very small data set on excel and then converted it to .csv file. I am able to open the data on R using the command "read.table ("mydata1.csv", sep=",", header=T)" and it just works fine. But when I want to work on the data (e.g. calculate the mean of variable "X") R says

Hello, I am attempting to train a random forest model using the randomForest package on 500,000 rows and 8 columns (7 predictors, 1 response). The data set is the first block of data from the UCI Machine Learning Repo dataset "Record Linkage Comparison Patterns" with the slight modification that I dropped two columns with lots of NA's and I used knn imputation to fill in other gaps.

Hi All, I'm a little stumped by the following problem. I've got a dataset with the following structure: idxy ix iy country (other variables) 1 1 1 c1 x1 2 1 2 c1 x2 3 1 3 c1 x3 . . . . . 3739 55 67 c7 x3739 3740 55 68 c7 x3740 where ix and

Untested, but I expect that setting the levels to be the same across the two factors levels(tmp$R1) <- levels(tmp$R2) <- LETTERS[1:6] and proceeding as before should be fine. Best, Ista On Jul 6, 2017 6:54 PM, "Gang Chen" <gangchen6 at gmail.com> wrote: Thanks a lot, Ista! I really appreciate it. How about a slightly different case as the following: set.seed(1) (tmp

In a classical meta analysis model y_i = X_i * beta_i + e_i, data {y_i} are assumed to be independent effect sizes. However, I'm encountering the following two scenarios: (1) Each source has multiple effect sizes, thus {y_i} are not fully independent with each other. (2) Each source has multiple effect sizes, each of the effect size from a source can be categorized as one of a factor levels