similar to: more efficient way to parallel

Buenas tardes, He construido la función “myfun” al objeto de considerar aquellas persones que a partir de una determinada fecha de Apertura tienen como mínimo 65 años. Se tiene su fecha de nacimiento, su fecha de inicio en la institución y su fecha de salida de la misma. Doy vueltas al script y no acabo se saber cómo poder aplicar de un modo eficiente las instrucciones “if” ó bien “ifelse”, y me

I need to apply a yearly inflation factor to some wages and supply some simple sums by work category. I have gone at it with a brute force "for" loop approach which seems okay as it is a small dataset. It looks a bit inelegant and given all the warnings in the Intro to R, etc, about using loops I wondered if anyone could suggest something a bit simpler or more efficent? Example:

I think that there is a bug in the as.POSIXct function on Windows. Here is what I get on Win2000, Pentium III machine in R 1.8.1. > dd1 <- ISOdatetime(2003, 10, 26, 0, 59, 59) > dd2 <- ISOdatetime(2003, 10, 26, 1, 0, 0) > dd2 - dd1 Time difference of 1.000278 hours Now, the 26th of October was the day that change to the standard time occurred, so I suspect that this has

Dear Rusers, I am now using R and SAS to fit the piecewise linear functions, and what surprised me is that they have a great differrent result. See below. #R code--Knots for distance are 16.13 and 24, respectively, and Knots for y are -0.4357 and -0.3202 m.glm<-glm(mark~x+poly(elevation,2)+bs(distance,degree=1,knots=c(16.13,24)) +bs(y,degree=1,knots=c(-0.4357,-0.3202

Hi there, Problem :: When one tries to change one or some of the columns of a data.frame, R makes a copy of the whole data.frame using the '*tmp*' mechanism (this does not happen for components of a list, tracemem( ) on R-2.6.2 says so). Suggested solution :: Store the columns of the data.frame as a list inside of an environment slot of an S4 class, and define the '[',

Dear all, I have an xts object , t.xts with 4 columns: "v1" "DD1" "v2" "DD2" and created a data frame : t <- as.data.frame(t.xts) I would like to extract data and create a new data frame for when the values in column DD1 falls between 0 and 30 and extract the corresponding v1 value. How can I do this? Thanks. -- View this message in context:

Dear all, I just found this issue: dd1 = c(rep(NaN,82), rep(-1, 144), rep(1, 74)) xx = runmed(dd1, 21) -> R crashes reproducibly in R 3.4.3, R3.4.4 (Ubuntu 14.04/Ubuntu 16.04) With GDB: Program received signal SIGSEGV, Segmentation fault. swap (l=53, r=86, window=window at entry=0xc59308, outlist=outlist at entry=0x12ea2e8, nrlist=nrlist at entry=0x114fdd8, print_level=print_level at

Something probably obivous but I don't see it. I needed to find the first 1 or two digits of some 5 and 6 digit numbers since they identified research sites while the rest of the number was the plot id. I converted the numbers to characters, got the first 1 or 2 characters as appropriate and went to add the new vector to the data.frame. For some reason R is insisting on turning the

salutations! i am doing some longitudinal modeling with sem and thought calculating some "simple variables" would make my model more readable. this is the smallest subset of my model that illustrates the resulting problem. i have 2 observed exogenous variables (c1, d2) and 4 observed endogenous variables (dc1, dd1, dc2, dd2). c1 is the observed state at time 1, dc1 is the change in c

Hello! I have a function: zywnoscCalosc<- function( jedzenie, n1, n2, n3, n4, d1, d2, d3, d4 ) { ndf <- data.frame(nn1=n1,nn2=n2,nn3=n3,nn4=n4) ddf <- data.frame(dd1=d1,dd2=d2,dd3=d3,dd4=d4) for (i in 1:length(n1)){ wekt_n = ndf[i,] wekt_n_ok = wekt_n[!is.na(wekt_n)] dl_n = length(wekt_n_ok) wynik = (1*wekt_n_ok)/(1*dl_n) } } and I get an error like this: Error in 1 * wekt_n_ok :

Hello, I want to know what do you think about my function. I know that it isn't briliant :/ but what do you think? What I should do that my function will be better? (now is very slow and not ideal, sometimes I also get a mistake!) ########## My function ############################################# dzieci<-transform(dzieci, zywnosc=0) zywnoscCalosc<- function( jedzenie, sklep, n1, n2,

I am lousy at simple regex and I have not found a solution to a simple problem. I have a vector with some character values that I want to split. Sample data dd1 <- c( "XXY (mat harry)","XXY (jim bob)", "CAMP (joe blow)", "ALP (max jack)") Desired result dd2 <- data.frame( xx = c("XXY", "XXY", "CAMP",

--- John Kane <jrkrideau at yahoo.ca> wrote: > Date: Thu, 19 Apr 2007 15:50:36 -0400 (EDT) > From: John Kane <jrkrideau at yahoo.ca> > Subject: RE: [R] Character coerced to factor and I > cannot get it back > To: Jorge Cornejo-Donoso <jorgecornejo at uach.cl> > > > --- Jorge Cornejo-Donoso <jorgecornejo at uach.cl> > wrote: > > >

> sessionInfo() R version 2.13.1 (2011-07-08) Platform: x86_64-apple-darwin9.8.0/x86_64 (64-bit) locale: [1] en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] Matrix_0.999375-50 lattice_0.19-30 loaded via a namespace (and not attached): [1] grid_2.13.1

--- John Kane <jrkrideau at yahoo.ca> wrote: > Date: Fri, 20 Apr 2007 10:47:45 -0400 (EDT) > From: John Kane <jrkrideau at yahoo.ca> > Subject: Re: [R] Character coerced to factor and I > cannot get it back > To: jim holtman <jholtman at gmail.com> > > Thanks Jim, > > I can live with it as a factor or I will do as you > suggest. What is bothering

We would like to load data from Statistics Canada (http://www.statcan.gc.ca/) using R, for example, Employment and unemployment rates. It seems to me that the tables are displayed in HTML. I was wondering if you know how to load these tables. Thanks, -james

Hi all, If I run LDA on the same data (2-class classification) with default(no priors specified in the lda function) vs. "prior=c(0.5, 0.5)", the results are different. The (0.5, 0.5) priors give better 1-classify-to-1 rate, and the proportional priors(default, no priors specified in the lda function) give better 0-classify-to-0 rate, for both training and testing data sets. However,

Hi, > dd <- data.frame(A=c("b","c","a"), B=3:1) > dd A B 1 b 3 2 c 2 3 a 1 > unlist(dd) A1 A2 A3 B1 B2 B3 2 3 1 3 2 1 Someone else might get something different. It all depends on the values of its 'stringsAsFactors' option: > dd2 <- data.frame(A=c("b","c","a"), B=3:1, stringsAsFactors=FALSE) > dd2

Dear R-helpers, I have run the code below which I expected to make an object called dd1, but that object does not exist. So, in summary, my problem is that my function is meant to make an object (dd1), and it does indeed make that object (I know that the last line of the function prints it out) but then, after the function has run, the object has disappeared. It's late on a Friday so I may

In barplot for degree distribution x-axis is not seen. See the example below > g = barabasi.game(500, 0.4) > dd1 = degree.distribution(g) > plot(dd1, xlab="degree", ylab = "frequency") whereas barplot doesnot have any x-axis > barplot(dd1, xlab = "degree", ylab = "frequency") Please see the figures attached.