similar to: Splitting Large Data Frame into Two

Hi, I have a list of mutations, called "mutList", of the form: > head(mutList) Alu 1 AluJ 2 AluJ/F(R)AM 3 AluJ/FLAM 4 AluJ/FRAM 5 AluJ/monomer 6 AluJb It contains about 500 elements and not all of them contain the sequence "Alu". I tried using this code: Alu<-mutList[which(grep("Alu",mutList)==1)] But that simply returned

Hi All, I generated 5000 samples using the following script test<- rnorm(5000,1000,100) test1 <- subset(test, subset=(test > 1100)) d <- density(test) plot(d, main="Density of production") abline(v=mean(test1) I wanted to do the following but faced difficulties 1. to shade or color (blue) the curve using the criterion that any

Dear R-user, I'm trying to compare two sets of results and wanted to find out which element in the two data frame/matrix are different. I wrote the following function and it works ok, and gives me a long list of "good" as outcomes. CHECK<- function (x = "file1", y = "file2") { for (i in 1:nrow(x)) { for (j in 1:ncol(x)) { if (x[i, j]

Dear R Gurus, I am currently working on the two dataset ( A and B), they both have the same fields: ID , REGION, OFFICE, CSTART, CEND, NCYCLE, STATUS and CB. I want to merge the two data set by ID. The problem I have is that the in data A, the ID's are unique. However in the data set B, the ID's are not unique, thus some repeat themselves. How do I the merge or retrieve the

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",

Dear Rers, is there a way to color counties on a full US map based on a criterion one establishes (i.e., all counties I assign the same number should be the same color)? I explored a bit and looks like the package "maps" might be of help. library(maps) One could get a map of the US: map('usa') One could get countries within a US state: map('county', 'iowa', fill

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",

? data=read.table("http://statcourse.com/research/boston.csv", , sep=",", header = TRUE) ? library(rpart) ? fit=rpart (MV~ CRIM+ZN+INDUS+CHAS+NOX+RM+AGE+DIS+RAD+TAX+ PT+B+LSTAT) Please: Show me the tree. Mark -------- Original Message -------- Subject: Re: [R] help with rpart From: "Stephen Milborrow" <[1]milbo at sonic.net>

1) how to modify the the tickment of x-axis or y-axis. boxplot(data[,1:5]) the tickment in x-axis in V1 V2 V3 V4 V5 ,I want to be some name for example name<-c("1day","2day","3day","4day","5day") 2) how to overlap two plot into one figure? plot(data[1:5]) boxplot(newdata[,1:5]) ? -- TANG Jie [[alternative HTML version deleted]]

I thought some of you might be amused by this. In my non-work time, I'm an avid weaver and teacher of weaving. I'm working on a project involving creating many detailed weaving patterns, so I wrote R code to automate it. Details here: http://stringpage.com/blog/?p=822 If the overlap between R users and avid tablet weavers turns out to be >> 1, I'll polish it up and turn it

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

Hi Sarah, Thank you for your help. I tried using CR1<-as.matrix(CR1) but gives error Error in corrplot(CR1, method = "circle") : The matrix is not in [-1, 1]!. I am using a corrplot library. Please find the reproducible example: dput(head(CR1,10)) structure(c(26L, 46L, 39L, 38L, 47L, 59L, 56L, 61L, 43L, 60L, 78L, 63L, 2L, 58L, 8L, 1L, 1L, 9L, 11L, 2L, 1037500L, 46747L, 346300L,

Sarah et. al.: As a matter of aesthetics (i.e. my personal ocd-ness) I prefer using the public API of an object, i.e. *not* to makes use of the representation of a factor as essentially an integer vector with labels, but rather to use its documented behavior. (Feel free to ignore this remark!) Anyway, >cumsum(!duplicated(dat1$B)) [1] 1 1 1 2 2 3 3 3 3 3 4 4 will do it. This is very

Hi all, I'm working with a bunch of large graphs, and stumbled across something useful. Probably many of you know this, but I didn't and so others might benefit. Using pch="." speeds up plotting considerably over using symbols. > x <- runif(1000000) > y <- runif(1000000) > system.time(plot(x, y, pch=".")) user system elapsed 1.042 0.030 1.077

I am trying to replicate the SAS proc univariate in R. I got most of the stats I needed for a by grouping in a data frame using: all1 <- ddply(all,"ACT_NAME", summarise, mean=mean(COUNTS), sd=sd(COUNTS), q25=quantile(COUNTS,.25),median=quantile(COUNTS,.50), q75=quantile(COUNTS,.75), q90=quantile(COUNTS,.90), q95=quantile(COUNTS,.95), q99=quantile(COUNTS,.99) )

Hi, I have a set of x, y points where x represents dates and y actual values. I am trying to plot a line graph of the data with points on top, but R is connecting the wrong points with lines. Does anyone know how I can rectify this. Please see sample below: x= 24/09/2009 09:13 16/10/2009 11:17 24/10/2009 21:43 11/09/2009 18:34 22/08/2009 15:45 10/08/2009 00:30 14/08/2009 14:52 24/09/2009

Hi just a quick question in which I must use the "function and return" part of R. If I was asked to define a function called "median" where the function has a single argument x and returns the median value of a vector. How would I go about saying the median = element x in terms of n (say (n + 1)/2 or whatever it would be) if there is an even amount of elements, or x in terms

I'm trying to test if a correlation matrix is positive semidefinite. My understanding is that a matrix is positive semidefinite if it is Hermitian and all its eigenvalues are positive. The values in my correlation matrix are real and the layout means that it is symmetric. This seems to satisfy the Hermitian criterion so I figure that my real challenge is to check if the eigenvalues are all

That does clarify for me that you're missing a step: I didn't clearly follow your description at first. corrplot expects a correlation matrix, not your original data. You need to use cor() first. That's pretty clear in the documentation. See for instance the examples: data(mtcars) M <- cor(mtcars) corrplot(M) Sarah On Sat, Mar 17, 2018 at 12:00 PM Shivi Bhatia <shivipmp82 at

Hi, Can R read in .xyz files? If so, what is the package, thanks -- Shane [[alternative HTML version deleted]]