search for: prevalent

Displaying 20 results from an estimated 295 matches for "prevalent".

2009 Jul 10
1
prevalence in logistic regression lrm()
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. How can we incorporate this information in lrm()? The concern I have is if my dataset does not have a representative prevalence, then the probability generated by lrm() will not be meaningful. Thanks John
2018 Feb 25
3
include
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
2018 Feb 25
0
include
Hi Val, My fault - I assumed that the NA would 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],
2024 Jan 17
1
Is there any design based two proportions z test?
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
2018 Feb 25
2
include
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
2018 Feb 25
2
include
HI Jim and all, I want to put one more condition. Include col2 and col3 if they are not in col1. Here is the data mydat <- read.table(textConnection("Col1 Col2 col3 K2 X1 NA Z1 K1 K2 Z2 NA NA Z3 X1 NA Z4 Y1 W1"),header = TRUE,stringsAsFactors=FALSE) The desired out put would be Col1 Col2 col3 1 X1 0 0 2 K1 0 0 3 Y1 0 0 4 W1 0 0 6 K2 X1
2024 Jan 17
1
Is there any design based two proportions z test?
Dear Md Kamruzzaman, To answer your second question first, you could just use the svychisq() function. The difference-of-proportion test is equivalent to a chisquare test for the 2-by-2 table. You don't say how you computed the confidence intervals for the two separate proportions, but if you have their standard errors (and if not, you should be able to infer them from the confidence
2018 Feb 25
0
include
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)
2018 Feb 25
0
include
Jim has been exceedingly patient (and may well continue to be so), but this smells like "failure to launch". At what point will you start showing your (failed) attempts at solving your own problems so we can help you work on your specific weaknesses and become self-sufficient? -- Sent from my phone. Please excuse my brevity. On February 25, 2018 7:55:55 AM PST, Val <valkremk at
2008 Jul 08
6
Question: Beginner stuck in a R cycle
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
2004 Sep 06
4
Cox regression for prevalence estimates
Hello, I'm an MD working in an eye clinic. I'm learning by myself to use R for use in my research works and for implementation in a software project. There are some 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;
2018 Feb 24
3
include
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 assignments
2018 Feb 24
0
include
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)
2012 Oct 03
2
Legend Truncated Using filled.contour
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_matrix, color =
2012 Jan 05
2
Bayesian estimate of prevalence with an imperfect test
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..
2018 Feb 24
2
include
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: > >
2009 Feb 26
1
error message and convergence issues in fitting glmer in package lme4
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
2010 Sep 03
2
Interactions in GAM
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
2007 Jun 26
2
Aggregation of data frame with calculations of proportions
Dear all, I have been stuck on this problem, am rather struggling and would appreciate some advice if anyone can help. I apologise if this is a bit long-winded, but I've tried to limit it to the bare essentials, but don't know how to make it more generic! I have some slightly odd real world data that I'm looking at representing number of positive diagnoses for different diseases,
2006 Jan 29
0
actuarial prevalence plots
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] (also divided by 1-KM[death by