search for: preval

Hi, I am wondering if there is a way to specify the prevalence of events in logistic regression using lrm() from Design package? Linear Discriminant Analysis using lda() from MASS library has an argument "prior=" that we can use to specify the prevalent of events when the actual dataset being analyzed does not have a representative prevalence. Ho...

Thank you Jim, I read the data as you suggested but I could not find K1 in col1. rbind(preval,mydat) Col1 Col2 col3 1 <NA> <NA> <NA> 2 X1 <NA> <NA> 3 Y1 <NA> <NA> 4 K2 <NA> <NA> 5 W1 <NA> <NA> 6 Z1 K1 K2 7 Z2 <NA> <NA> 8 Z3 X1 <NA> 9 Z4 Y1 W1 On Sat, Feb 24, 2018 at 6:18 PM, Jim...

...ld be first in the result produced by "unique": mydat <- read.table(textConnection("Col1 Col2 col3 Z1 K1 K2 Z2 NA NA Z3 X1 NA Z4 Y1 W1"),header = TRUE,stringsAsFactors=FALSE) val23<-unique(unlist(mydat[,c("Col2","col3")])) napos<-which(is.na(val23)) preval<-data.frame(Col1=val23[-napos], Col2=NA,col3=NA) mydat<-rbind(preval,mydat) mydat[is.na(mydat)]<-"0" mydat Jim On Sun, Feb 25, 2018 at 11:27 AM, Val <valkremk at gmail.com> wrote: > Thank you Jim, > > I read the data as you suggested but I could not find K1 in...

Hello Everyone, I was analysing big survey data using survey packages on RStudio. Survey package allows survey data analysis with the design effect.The survey package included functions for all other statistical analysis except two-proportion z tests. I was trying to calculate the difference in prevalence of Diabetes and Prediabetes between the year 2011 and 2017 (with 95%CI). I was able to calculate the weighted prevalence of diabetes and prediabetes in the Year 2011 and 2017 and just subtracted the prevalence of 2011 from the prevalence of 2017 to get the difference in prevalence. But I could...

Sorry , I hit the send key accidentally here is my complete message. Thank you Jim and all, I got it. I have one more question on the original question What does this "[-1] " do? preval<-data.frame(Col1=unique(unlist(mydat[,c("Col2","col3")]))[-1], Col2=NA,col3=NA) mydat <- read.table(textConnection("Col1 Col2 col3 Z1 K1 K2 Z2 NA NA Z3 X1 NA Z4 Y1 W1"),header = TRUE) preval<-data.frame(Col1=unique(unlist(mydat[,c("C...

...unique": > > mydat <- read.table(textConnection("Col1 Col2 col3 > Z1 K1 K2 > Z2 NA NA > Z3 X1 NA > Z4 Y1 W1"),header = TRUE,stringsAsFactors=FALSE) > val23<-unique(unlist(mydat[,c("Col2","col3")])) > napos<-which(is.na(val23)) > preval<-data.frame(Col1=val23[-napos], > Col2=NA,col3=NA) > mydat<-rbind(preval,mydat) > mydat[is.na(mydat)]<-"0" > mydat > > Jim > > On Sun, Feb 25, 2018 at 11:27 AM, Val <valkremk at gmail.com> wrote: > > Thank you Jim, > > > > I read t...

...analysing big survey data using survey packages on RStudio. Survey > package allows survey data analysis with the design effect.The survey > package included functions for all other statistical analysis except > two-proportion z tests. > > I was trying to calculate the difference in prevalence of Diabetes and > Prediabetes between the year 2011 and 2017 (with 95%CI). I was able to > calculate the weighted prevalence of diabetes and prediabetes in the Year > 2011 and 2017 and just subtracted the prevalence of 2011 from the > prevalence of 2017 to get the difference in prev...

hi Val, Your problem seems to be that the data are read in as a factor. The simplest way I can think of to get around this is: mydat <- read.table(textConnection("Col1 Col2 col3 Z1 K1 K2 Z2 NA NA Z3 X1 NA Z4 Y1 W1"),header = TRUE,stringsAsFactors=FALSE) preval<-data.frame(Col1=unique(unlist(mydat[,c("Col2","col3")]))[-1], Col2=NA,col3=NA) rbind(preval,mydat) mydat[is.na(mydat)]<-"0" Jiim On Sun, Feb 25, 2018 at 11:05 AM, Val <valkremk at gmail.com> wrote: > Sorry , I hit the send key accidentally here is...

...t <- read.table(textConnection("Col1 Col2 col3 >> Z1 K1 K2 >> Z2 NA NA >> Z3 X1 NA >> Z4 Y1 W1"),header = TRUE,stringsAsFactors=FALSE) >> val23<-unique(unlist(mydat[,c("Col2","col3")])) >> napos<-which(is.na(val23)) >> preval<-data.frame(Col1=val23[-napos], >> Col2=NA,col3=NA) >> mydat<-rbind(preval,mydat) >> mydat[is.na(mydat)]<-"0" >> mydat >> >> Jim >> >> On Sun, Feb 25, 2018 at 11:27 AM, Val <valkremk at gmail.com> wrote: >> > Thank y...

Dear All, I have a database of 200 observations named myD. In the dataframe there are a column named code (with codes varying from 1 to 77), a column named "prevalence" with some quantitative measurements are given and an column named Pr_mean, with no values. I would like to set a cycle to compute the average of prevalence values for each different code and store the averages under the empty field Pr_mean. This is what I wrote: # Set a cycle for (i in...

...authors who recomends the use of Cox regression as a substitute for Logistic regression (<a href="http://www.biomedcentral.com/1471-2288/3/21.pdf"> Barros AJD, Hirakata VN. BMCMedical Research Methodology, 2003; 3:21 </a>. The use of Cox regression permit the estimation of the prevalence rates rather than Odds ratios obtained by logistic regression analysis. Cox regression is used for time-to-event data. To obtain prevalence rates the time has to be constant. One of the problems of Cox regression is that the confidence intervals are overestimated. To correct this Barros &...

On 24/02/2018 1:53 PM, William Dunlap via R-help wrote: > x1 = rbind(unique(preval),mydat) > x2 <- x1[is.na(x1)] <- 0 > x2 # gives 0 > > Why introduce the 'x2'? x1[...] <- 0 alters x1 in place and I think that > altered x1 is what you want. > > You asked why x2 was zero. The value of the expression > f(a) <- b > and a...

Thank you Jim and all, I got it. I have one more question on the original question What does this "[-1] " do? preval<-data.frame(Col1=unique(unlist(mydat[,c("Col2","col3")]))[-1], Col2=NA,col3=NA) mydat <- read.table(textConnection("Col1 Col2 col3 Z1 K1 K2 Z2 NA NA Z3 X1 NA Z4 Y1 W1"),header = TRUE) preval<-data.frame(Col1=unique(unlist(mydat[,c("C...

Hey everyone, I'm working on a contour plot depicting asymptomatic prevalence at varying durations of infectiousness and force of infection. I've been able to work everything out except for this one - my legend title keeps getting cut off. Here's what I have: filled.contour(x=seq(2,30,length.out=nrow(asym_matrix)), y=seq(1,2,length.out=ncol(asym_matrix)), asym...

Hi all! I'm new to this forum so please excuse me if I don't conform perfectly to the protocols on this board! I'm trying to get an estimate of true prevalence based upon results from an imperfect test. I have various estimates of se/sp which could inform my priors (at least upper and lower limits even if with a uniform distribution) and found the following code on this website.. http://www.lancs.ac.uk/staff/diggle/prevalence_estimation.R/ (the foll...

Thank you Jim I wanted a final data frame after replacing the NA's to "0" x1 = rbind(unique(preval),mydat) x2 <- x1[is.na(x1)] <- 0 x2 but I got this, [1] 0 why I am getting this? On Sat, Feb 24, 2018 at 12:17 AM, Jim Lemon <drjimlemon at gmail.com> wrote: > Hi Val, > Try this: > > preval<-data.frame(Col1=unique(unlist(mydat[,c("Col2","col3")]...

I'm resending this message because I did not include a subject line in my first posting. Apologies for the inconvenience! Tanja > Hello, > > I'm trying to fit a generalized linear mixed model to estimate diabetes prevalence at US county level. To do this I'm using the glmer() function in package lme4. I can fit relatively simple models (i.e. few covariates) but when expanding the number of covariates I usually encounter the following error message. > > gm8 <- > glmer(DIAB05F~AGE+as.factor(SEX)+p...

Hello R users, I am working with the GAM to inspect the effect of some factors (year, area) and continuous variables (length, depth, latitude and longitude) on the intensity and prevalence of the common parasite Anisakis. I would like introduce interaction in my models, both "continuous variables-continuous variables" and "continuous variables-factor". I have read some questions-answers regard to this subject but I still have doubts. The solution that I have...

...# What I want to do is calculate a true proportion, with the number found divided by the number tested. # I have managed to do this: # calculates true proportion by only including values in the denominator that have non-NAs in the numerator # x= vector of numbers # total = vector of numbers calc.prevalence <- function(x, total) { sum.x = sum(x, na.rm=T) # calculate numerator sum.y = sum(total[!is.na(x)]) # calculate denominator return(sum.x/sum.y) } correct.prevalence <- calc.prevalence(sp$drpla, sp$total) incorrect.prevalence <- sum(sp$drpla, na.rm=T) / sum(sp$total, na.rm=T) prin...

Hi all, I am trying to produce a series of plots showing the prevalence of a condition, which is subject to censoring. In most cases the condition is temporary and resolves with time. I would like to use the method of Pepe et al Stat Med 1991; 413-421 - essentially the prevalence is the Kaplan-Meier prob[having the condition at time t] - KM prob[recovery by time t]...