similar to: Variable descriptions of a built-in dataset

Hi, I am just wondering if there is an easy way to count in a numeric vector how many numbers don't have replicates. For example, a=c(1,1,2,2,3,4,5), how can I know there are three numbers (3, 4 and 5) without replicates? Thank you! Jun =====

Hi Everyone, I just don't know how to extract the information I want from the summary of a linear regression model fitting. For example, I fit the following simple linear regression model: results = lm(y_var ~ x_var) summary(results) gives me: Call: lm(formula = y_var ~ x_var) Residuals: Min 1Q Median 3Q Max -5.9859 -1.5849 0.4574 2.0163 4.6015 Coefficients:

Dear all, I'm having some problems with using the step() function inside another function. I think it is an environment problem but I do not know how to overcome it. Any suggestions are appreciated. I've prepared a simple example to illustrate my problem: > library(MASS) > data(Boston) > my.fun <- function(dataset) { + l <- lm(medv ~ .,data=dataset) + final.l <-

Dear All, I have data which contains 14 variables. And I have to regress one of variables on each variable (simple 13 linear regressions) I try to make a loop and store only R-squared colnames(boston) [1] "CRIM" "ZN" "INDUS" "CHAS" "NOX" "RM" "AGE" [8] "DIS" "RAD"

> >  Dear moderators, Please help me encode the program instructed by follows. Thank u! Apply the methods introduced in Sections 4.2.1 and 4.2.2, say the > rank-based variable selection and BIC criterions, to the Boston housing > data. >  The Boston housing data contains 506 observations, and is publicly available in the R package mlbench (dataset “BostonHousing”).  The

data(Boston, package='MASS') y <- Boston$nox x <- Boston$dis nls(y~ A + B * exp(C * x), start=list(A=1, B=1, C=1)) Error in nls(y ~ A + B * exp(C * x), start = list(A = 1, B = 1, C = 1), : step factor 0.000488281 reduced below 'minFactor' of 0.000976562 I don't know how to fix this error. I think my problem is that I set the wrong start. Could somebody help please?

Hi All, I have a data frame dat1: dat1 <-data.frame(N=seq(1, 12,1), B=c("29_log","29_log", "29_log", "27_cat", "27_cat", "1_log", "1_log", "1_log", "1_log", "1_log",

Hi Sarah, Thank you so much!! I got your good ideas. Ding -----Original Message----- From: Sarah Goslee [mailto:sarah.goslee at gmail.com] Sent: Friday, May 11, 2018 11:40 AM To: Ding, Yuan Chun Cc: r-help mailing list Subject: Re: [R] add one variable to a data frame [Attention: This email came from an external source. Do not open attachments or click on links from unknown senders or

Adjust almost all the license headers to point to the GNU website instead of the FSF postal address. 3rd party code copies are unchanged. Spotted by Rpmlint. --- v1 is: https://www.redhat.com/archives/libguestfs/2019-September/msg00079.html Makefile.am | 3 +-- autogen.sh | 3 +-- bash/Makefile.am | 3 +--

Hi, Here's one way to approach it, using the coercion of factor to numeric. Note that I changed your data.frame() statement to avoid coercing strings to factors, just to make it simpler to set the levels. dat1 <-data.frame(N=seq(1, 12,1), B=c("29_log","29_log", "29_log", "27_cat", "27_cat", "1_log", "1_log",

Sarah's solutions are good, and here's another, even more basic: tmp1 <- unique(dat1$B) tmp2 <- seq_along(tmp1) dat1$C <- tmp2[ match( dat1$B, tmp1) ] > dat1 N B C 1 1 29_log 1 2 2 29_log 1 3 3 29_log 1 4 4 27_cat 2 5 5 27_cat 2 6 6 1_log 3 7 7 1_log 3 8 8 1_log 3 9 9 1_log 3 10 10 1_log 3 11 11 3_cat 4 12 12 3_cat 4 As a single line

I first try these command, it works quite well. library(lattice) data(cuckoos) levnam <- strsplit(levels(cuckoos$species), "\\.") BUT, i want to try : levnam <- strsplit(levels(cuckoos$species), ".") to see the difference. They maybe I modified the data file, because when I try again, it says: > data(cuckoos) Warning message: data set 'cuckoos' not found in:

According to the man page for formula, "a formula object has an associated environment". However, update.default doesn't use this environment, which creates problems like the following: make.model <- function(x) { lm(medv~.,x) } library(MASS) data(Boston) fit = make.model(Boston) fit = update(fit,".~.-crim") # Object "x" not found Here is a

Hi Rui, Thank you very much! Yes, I verified using real data, it worked correctly as expected after adding tidyr:: to the pivot_longer function and dplyr:: to the group_by and summarize Function. I did not know how to assign the tidyr and dplyr to the three functions because I do not really understand well the three functions and just got the code from a google search. I also tried your

Dear William and Ben, Thank you for your replies and elegant solutions. I am having trouble with the fact that two of the previous locations do not appear in current locations (that is no one moved to OKC and Dallas from other cities), so these two cities are not being included in the output. I have provided a better sample of the data and the ideal output (wide form - a 10x10 bilateral matrix)

Hello, Inline. ?s 19:03 de 12/06/2024, Yuan Chun Ding via R-help escreveu: > I am sorry that I know I should provide a dataset that allows to replicate my problem. > > It is a research dataset and quite large, so I can not share. > > Both Bert and Tim guessed my problem correctly. I also thought about the conflicting issue between different packages and function masking. > I

> On Apr 19, 2018, at 1:22 PM, David Winsemius <dwinsemius at comcast.net> wrote: > > >> On Apr 19, 2018, at 11:20 AM, Ding, Yuan Chun <ycding at coh.org> wrote: >> >> Hi All, >> >> I want to create a categorical variable, cat.pfoa, in the file of pfas.pheno (a data frame) based on log2pfoa values. I can do it using the following code.

I am sorry that I know I should provide a dataset that allows to replicate my problem. It is a research dataset and quite large, so I can not share. Both Bert and Tim guessed my problem correctly. I also thought about the conflicting issue between different packages and function masking. I just hope to that someone has similar experience, so providing me suggestion. For conflicting issue,

Dear Bert and Huzefa, Apologies for the late reply, my account got hacked and I have just managed to recover it. Thank you very much for your replies and the solutions. Both work well. I was wondering if there was any way to ensure (force) that all possible combinations show up in the output. The full dataset has 25 cities but of course people have not moved from Boston to all the other 24

xtabs does this automatically if your cross classifying variables are factors with levels all the cities (sorted, if you like): > x <- sample(letters[1:5],8, rep=TRUE) > y <- sample(letters[1:5],8,rep=TRUE) > xtabs(~ x + y) y x c d e a 1 0 0 b 0 0 1 c 1 0 0 d 1 1 1 e 1 1 0 > lvls <- sort(union(x,y)) > x <- factor(x, levels = lvls) > y <-