similar to: plotting axes

Dear friends. As promised ver 1.01 solved the former problem and now dates plot OK in windows. But. In the following code, "cex.axis"=0.5 works OK but "xaxp"=c(14620,14680,4) has seemingly no effect. The limits for xaxp was found from the default plot and inspecting par(). I need to put the date marks at specific points. Could I suppress them altogether and make a new xaxis

Dear friends. I'm in windows version R1.0.0. The code below produced a very beautiful and efficient graph in R Version 0.99.0 Patched (February 9, 2000) but in the newest, the x-axis is unreadable with dates written unformatted. I tried to set the numbers using par(xaxp) or xaxs but to no avail. Since it was OK just in the last version I guess there might perhaps be something wrong

At 22:31 03/03/00 +0100, Troels Ring wrote: >Dear friends. A very humble admirer of the progress of R with a very simple problem. I have the data shown below > >month day year dosis vanco >12 31 1999 1 3.8 >1 3 2000 1 2.4 >1 7 2000 1 3.4 >1 12 2000 2 7.1 >1 14 2000 2 5.3 >1 17 2000 2 5.7 >1 19 2000 3 7.7 >1 21 2000 3 8.3 >1 24 2000 4 8.7 >1 26 2000 4

Dear friends. Can anyone explain what is happening here ? I am told that no data entry editor is available in this version of R ? The example is not far from that shown in help(data.entry) ? Besides, could an empty spreadsheet entry be an option to start inputting data ? R : Copyright 2000, The R Development Core Team Version 1.0.0 (February 29, 2000) R is free software and comes with

Hello I've stumbled upon a problem for inversion of a matrix with large values, and I haven't found a solution yet... I wondered if someone could give a hand. (It is about automatic optimisation of a calibration process, which involves the inverse of the information matrix) code: ********************* > macht=0.8698965 > coeff=1.106836*10^(-8) >

Dear R-experts, I guess I have a problem with my fast function (fast tau estimator) here below. Indeed, zero errors look highly suspicious. I guess there is a bug in my R code. How could I correct my R code ? # install.packages( "robustbase" ) # install.packages( "MASS" ) # install.packages( "quantreg" ) # install.packages( "RobPer" ) #

Dear All, I faced the following problem. With the same data.frame the results are different under Linux and Windows. Could you help on this topic? Thanks in advance, Ildiko Linux: > d = read.csv("CRP.csv") > d$drugCode = as.numeric(d$drug) > cor(d, use="pairwise.complete.obs") PATIENT BL.CRP X24HR.CRP X48HR.CRP drug drugCode PATIENT NA

H ola Argel, Muchisimas gracias por los datos y el codigo! ;-) Una forma es hacer bootstrap. En http://www.mayin.org/ajayshah/KB/R/documents/boot.html hay una corta introduccion. Seria algo similar a lo siguiente: d <- LD50.1 - LD50.2 # diferencia real d #p = 0.5: #37.31248 # ------------ # bootstrap # ------------ B <- 1000 # numero de muestras bootstrap d.sample <-

Hi, ? I am trying to solve a problem related to Poincare circles ( for more info http://www.ms.uky.edu/~droyster/courses/spring08/math6118/Classnotes/Chapter09.pdf). In a nutshell i am trying to replicate the method in the above pdf section 9.2.1. that explains in broad terms how to draw the arc inside a circle that goes through 2 previously set points on the first circle. ? I think i came up

Hola compañeros de la lista. Tengo el análisis que muestro abajo, para la estimación de la Dosis Letal Media (DL50) por medio de regresión logística, de dos compuestos similares (denominados 1 y 2) aplicados a grupos de individuos similares, evaluando el número de individuos vivos y muertos al final del ensayo. Mi duda es qué forma me recomiendan para determinar si la diferencia es

I forgot, you should also set.seed() before calling boot() to make the results reproducible. Rui Barradas On 5/22/2018 10:00 AM, Rui Barradas wrote: > Hello, > > If you want to bootstrap a statistic, I suggest you use base package boot. > You would need the data in a data.frame, see how you could do it. > > > library(boot) > > bootMedianSE <- function(data,

Hello, If you want to bootstrap a statistic, I suggest you use base package boot. You would need the data in a data.frame, see how you could do it. library(boot) bootMedianSE <- function(data, indices){ d <- data[indices, ] fit <- rq(crp ~ bmi + glucose, tau = 0.5, data = d) ypred <- predict(fit) y <- d$crp median(y - ypred)^2 } dat <-

Hello, Right! I copied from the OP's question without thinking about it. Corrected would be bootMedianSE <- function(data, indices){ d <- data[indices, ] fit <- rq(crp ~ bmi + glucose, tau = 0.5, data = d) ypred <- predict(fit) y <- d$crp median((y - ypred)^2) } Sorry, rui Barradas On 5/22/2018 11:32 AM, Daniel Nordlund wrote: > On 5/22/2018

I tried to draw a point on all boxplots for their means, I did: boxplot( Daten$weight~interaction(Daten$Dosis,Daten$sex, drop=TRUE)) means<-tapply( Daten$weight, Daten$Dosis, mean) points(means, pch=5, col="red", lwd=5) but only the boxplots for male got that point on them, its really weird because I don't think that I separated the sex in the codes above [[alternative HTML

I'm wondering if hdlc can be the culprit (not sure what it is and what it does). Should I set hdlc to yes in misdn.conf (I'm asking before testing because this is a production system)? misdn.conf: [general] misdn_init=/etc/misdn-init.conf debug=0 ntdebugflags=0 ntdebugfile=/var/log/misdn-nt.log ntkeepcalls=no bridging=no stop_tone_after_first_digit=yes append_digits2exten=yes

Dear R-experts, I am trying to bootstrap (and average) the median squared error evaluation metric for a robust regression. I can't get it. What is going wrong ? Here is the reproducible example. ############################# install.packages( "quantreg" ) library(quantreg) crp <-c(12,14,13,24,25,34,45,56,25,34,47,44,35,24,53,44,55,46,36,67) bmi

Hola a todos . ¿cómo puedo ordenar los niveles de un factor? Ejemplo > x<-c("0","0-9","10-49","50-99","100-199",">200") > dosis<-factor(rep(x,10)) > dosis [1] 0 0-9 10-49 50-99 100-199 >200 0 0-9 10-49 [10] 50-99 100-199 >200 0 0-9 10-49 50-99 100-199 >200

On 5/22/2018 2:32 AM, Rui Barradas wrote: > bootMedianSE <- function(data, indices){ > ???? d <- data[indices, ] > ???? fit <- rq(crp ~ bmi + glucose, tau = 0.5, data = d) > ???? ypred <- predict(fit) > ???? y <- d$crp > ???? median(y - ypred)^2 > } since the OP is looking for the "median squared error", shouldn't the final line of the

Hi Rosa something like plot(1,1, sub=expression(lambda^"2")) So with your example, do you want something like plot(c(1:5), CRP7raw[1,], type = "n", xlim=c(1,5), ylim=c(-10,5) , xlab="Day in ICU", ylab="CRP (mg/dL)", sub = mtext(expression(lambda^2))) CRP7graph <- apply(CRP7, 1, lines, col="gray") Cheers Petr >

I try to use ColorRamp as ColorRampPalette (i.e. with the same gradient), but it seems there is a nuance that I've missed. pal.crp<-colorRampPalette( c("blue", "white", "red"), space = "rgb") plot(rep(0,40),pch=16,col=pal.crp(40)) # is great But, using the same gradient with colorRamp is giving erratic colors. pal.cr<-colorRamp(