search for: 2.98

I have a dataset that I'm trying to rearrange for a repeated measures analysis: It looks like: patient basefev1 fev11h fev12h fev13h fev14h fev15h fev16h fev17h fev18h drug 201 2.46 2.68 2.76 2.50 2.30 2.14 2.40 2.33 2.20 a 202 3.50 3.95 3.65 2.93 2.53 3.04 3.37 3.14 2.62 a 203 1.96 2.28 2.34 2.29 2.43 2.06 2.18 2.28 2.29 a

How to add trendline (i.e. straight line passing through maximum points) in graph. I have worked on the data given below. Please tell me how to add trendline in the graph. The script is as follows =================================== start ==================================================== # The data is as follows data <- c( 0.01, 0.02, 0.04, 0.13, 0.17 , 0.19 , 0.21 , 0.27 , 0.27 ,

Hi, I am beginning to learn R and have a data table that I would like to produce a microarray-like plot. The table looks like this: 3 0 0 3 -377.61 1.94 3 0 0 3 -444.80 2.36 2 1 0 3 -519.60 2.39 1 1 1 3 -54.88 2.49 2 1 1 4 -536.55 2.53 1 0 1 2 108.29 2.62 2 0 0 2 39.56 2.62 3 0 1 4 108.32 2.63 2 0 0 2 -455.23 2.84 1 0 0 1 -432.30 2.98 ... I would like to assign colors to the first three columns

Dear list, I'm still trying to calculate the sd for V2 for each group in V1 if V3 is '0': > x V1 V2 V3 1 A01 2.40 0 2 A01 3.40 1 3 A01 2.80 0 4 A02 3.20 0 5 A02 4.20 0 6 A03 2.98 1 7 A03 2.31 0 8 A04 4.20 0 # Work x$vmean <- ave(x$V2, x$V1, x$V3 == 0, FUN = mean) # Work x$vsd2 <- ave(x$V2, x$V1, FUN = sd) # Doesn't work x$vsd <- ave(x$V2, x$V1, x$V3

Hi, I am a total beginner with this whole thing so please have patience! I am trying to run an S-plus program with a certain line: spline(1:nrow(y), y[,1],n=100); This crashes with: Error: NAs in foreign function call (arg 8) Apparently, this is caused by the last command of spline: u <- seq(xmin, xmax, length.out = n) .C("spline_eval", z$method, length(u), x = u, y =

Dear All. I have some problem with combined two data frame. .... I have first data frame .. GPAX THAI MATH SCINCE SOCIAL HEALT ART CAREER LANGUAGE 1227 2.99 3.32 2.50 2.64 3.05 3.60 3.72 3.57 2.62 1704 2.81 2.56 2.48 2.86 3.22 3.19 3.55 3.20 2.51 617 2.18 1.90 1.97 2.06 2.38 3.50 3.54 2.33 1.70 876 2.82 3.14 2.73 2.46 2.71 3.11 3.04 3.24 2.90

I want to do "moran's I test" in R language. I try to use "gearymoran" in Package "ade4","moran" in Package "spdep", and Moran.I in Package "ape". But I do not know how to do it because data format is different. My data: x y dbh 111.03 10.7 7 118.11 0.28 1.2 165.36 0.36 8.4

I would like to run a t-test within a "by" group function. My dataset, "error", is organized as the following (I have 133 Sites): Site week Dataset Region lat_map long_map mean_tsim diff20 diff40 diff80 ALFI 15 USACE UC 48.15625 -117.0938 8.87 1.34 1.90 2.98 ALFI 16 USACE UC 48.15625 -117.0938 10.28 0.57 1.08 2.27 ALFI 17

Hii, I will create boxplots from matrices. I have the following data sets: 5.0 1.78 2.99 2.019 0 10.0 1.79 3.00 1.744 0 15.0 1.78 2.98 1.936 0 20.0 1.78 2.99 1.975 0 25.0 1.73 2.91 3.591 0 30.0 1.79 3.00 1.966 0 35.0 1.79 3.00 2.451 0 40.0 1.79 3.00

Hi, Few days ago I have asked about spatial join on the minimum distance between 2 sets of points with coordinates and attributes in 2 different data frames. Simon Knapp sent code to do it when calculating distance on a sphere using lat, long coordinates and I've change his code to use Euclidian distances since my data had UTM coordinates. Typically one data frame has around 30 000 points

Interestingly, as.list(substitute(...())) also works. On Sun, Aug 12, 2018 at 1:16 PM, Duncan Murdoch <murdoch.duncan at gmail.com> wrote: > On 12/08/2018 4:00 PM, Henrik Bengtsson wrote: >> >> Hi. For any number of *known* arguments, we can do: >> >> one <- function(a) list(a = substitute(a)) >> two <- function(a, b) list(a = substitute(a), b =

First, here's the problem I'm working on so you understand the context. I have a data frame of travel activity characteristics with 70,000+ records. These activities are identified by unique chain numbers. (Activities are part of trip chains.) There are 17,500 chains. I use the chain numbers as factors to split various data fields into lists of chain characteristics with each element of

Since you're already using bang-bang ;) library(rlang) dots1 <- function(...) as.list(substitute(list(...)))[-1L] dots2 <- function(...) as.list(substitute(...())) dots3 <- function(...) match.call(expand.dots = FALSE)[["..."]] dots4 <- function(...) exprs(...) bench::mark( dots1(1+2, "a", rnorm(3), stop("bang!")), dots2(1+2, "a",

Hello, I have 4 web servers with lighttpd to serve one web site with DNS load sharing. On the 2 SMP-enable web servers, there will be many php-cgi frozen in 'sbwait' state every day. It means the php-cgi stay in 'sbwait' state, and never be back to 'accept' or other state. If I restart them, there will be frozen php-cgi appear some hours later. There is no problem on the

Dear R-help, A colleague of mine was running some code on two of our boxes, and noticed a rather large difference in running time. We've so far isolated the problem to the difference between R 1.7.1 and 1.8.0, but not more than that. The exact same code took 933.5 seconds in 1.7.1, and 3594.4 seconds in 1.8.1, on the same box. Basically, the code calls boot() to bootstrap fitting mixture

On Sat, 2011-10-29 at 15:16 -0500, Hal Finkel wrote: > On Sat, 2011-10-29 at 14:02 -0500, Hal Finkel wrote: > > On Sat, 2011-10-29 at 12:30 -0500, Hal Finkel wrote: > > > Ralf, et al., > > > > > > Attached is the latest version of my autovectorization patch. llvmdev > > > has been CC'd (as had been suggested to me); this e-mail contains > >

There is a problem when importing an spss-file containing explicitly declared missing values in R using the read.spss function from the foreign package. I'm not sure these problems are the same in every version of spss, I am using the latest version 16.0.2. I included http://www.nabble.com/file/p18776776/missingdata.sav missingdata.sav and

Dear friend Duncan, Thank you so much for your kind reply. Yes, that is exactly what is happening, there are a lot of NA values at the start, so R assumes that the field is of type boolean. The challenge that I am facing is that I want to read into R an Excel file that has many sheets (46 in this case) but I wanted to combine all 46 sheets into a single dataframe (since the columns are exactly

Hi, I am trying to find m breakpoints of a linear regression model. I used the segmented package. It works fine for small number of predicators and breakpoints.(3 r.v. 3 points). However, my model has 14 variables it even would not work even for just one breakpoints!. The error message is always estimated breakpoints are out of range. Since my problem is time related problem. So I

"It seems therefore that there is no other way than read in individually > 100 weather tables using read.tables., right? Using file.choose() doesn't change the work." Yes. With that many files, file.choose() does not make sense. However, I still do not understand what is the problem with using lapply() on the character vector of file names with read.table() as you did in your