search for: smoke

...here the raw data is not available. Each study analyst runs a model that includes an interaction term for, say, between sex and smoking and age. Here is an illustrative example example for one study: set.seed(1066) status <- rbinom( 1000, 1, 0.2 ) males <- rbinom( 1000, 1, 0.6 ) smoke <- rbinom( 1000, 1, 0.3 ) age <- runif(1000, min=20, max=80) coef( summary( f1 <- glm( status ~ males*smoke + age, family="binomial" ) ) ) # Estimate Std. Error z value Pr(>|z|) # (Intercept) -1.520399871 0.28446...

Hi, I'm sure this is simple, but I haven't been able to find this in TFM, say I have some data in R like this (pasted here: http://pastebin.com/raw.php?i=sjS9Zkup): > head(df) gender age smokes disease Y 1 female 65 ever control 0.18 2 female 77 never control 0.12 3 male 40 state1 0.11 4 female 67 ever control 0.20 5 male 63 ever state1 0.16 6 female 26 never state1 0.13 where unique(disease) == c("control", "state1", "...

...está de vacaciones, igual me pueden ayudar. Quiero ajustar unos modelos: REG_LOG <- glm (low ~ X, family = "binomial", data = DATOS) Ejemplo: library(MASS) data(birthwt, package="MASS") birthwt$low <- factor(birthwt$low) birthwt$race <- factor(birthwt$smoke) REG_LOG <- glm (low ~ smoke, family = "binomial", data = birthwt) summary (REG_LOG) Se pueden colocar etiquetas en las variable de la fórmula de manera que en los resultados salgan las etiquetas y no el nombre de la variable? Seria una cosa como: REG_LOG <- glm (low = "B...

dear professor: I have a problem about the nomogram.I have got the result through analysing the dataset "exp2.sav" through multinominal logistic regression by SPSS 17.0. and I want to deveop the nomogram through R-Projject,just like this : > n<-100 > set.seed(10) > T.Grade<-factor(0:3,labels=c("G0", "G1", "G2","G3")) >

dear professor: thank you for your help,witn your help i develop the nomogram successfully. after that i want to do the internal validation to the model.i ues the bootpred to do it,and then i encounter problem again,just like that.(´íÎóÓÚerror to :complete.cases(x, y, wt) : ²»ÊÇËùÓеIJÎÊý¶¼Ò»Ñù³¤(the length of the augment was different)) i hope you tell me where is the mistake,and maybe i have

...ons where I had to add or modify something? What else can I do? Thank you for your help! Johanna Brandt (I'm using R 2.0.1 under Windows 2000) ## Example for geese() from the R-Help ##################### I took the example from the help: > data(ohio) > summary(geese(resp ~ age + smoke + age:smoke, id=id, data=ohio, + family=binomial(link="logit"), corstr="exch", scale.fix=TRUE)) Call: geese(formula = resp ~ age + smoke + age:smoke, id = id, data = ohio, family = binomial(link = "logit"), scale.fix = TRUE, corstr = "exch&q...

...0 = BWT>2500g # age Age of Mother Years # lwt Weight of Mother at Pounds # Last Menstrual Period # race Race 1 = White # 2 = Black # 3 = Other # smoke Smoking Status 0 = No,1 = Yes # During Pregnancy # ptd History of Premature Labor 0 = None,1 = Yes # ht History of Hypertension 0 = No, 1 = Yes # ui Presence of Uterine 0 = No, 1 = Yes # Irritability pair<-rep(1:56, each=2) low<-rep...

Hello there, I have two groups (men and women) and I know per group how many of them smoke or don't smoke (women 40 of 200; men 100 of 300). I would like to know how I can compare in R if men and women differ significantly in their smoking. However, because there are more men in the sample than women I cannot just compare the number of smokers and non-smokers in both groups, right?!...

...ive Gauss-Hermite approximation with only 1 point, while 'xtmelogit' uses 7 points. In order to compare them, I tried also to change the corresponding parameters. This is the code for R: rm(list=ls()) library(faraway); library(lme4); library(MASS) data <- ohio pql <- glmmPQL(resp~smoke+factor(age), random=~1|id, family=binomial,data) summary(pql)$tTable["smoke",1:2] lap <- glmer(resp~smoke+factor(age)+(1|id), family=binomial,data) attributes(summary(lap))$coefs["smoke",1:2] agq7 <- glmer(resp~smoke+factor(age)+(1|id),nAGQ=7,family=binomial,data) write.cs...

On Mon, Sep 21, 2015 at 11:10:06AM +0300, Pasi K?rkk?inen wrote: > Hi, > > On Wed, Sep 09, 2015 at 02:24:15PM +0100, George Dunlap wrote: > > Updated Xen 4.4.3 packages have passed my local smoke tests, and are > > now available. > > > > Full C7 release is still waiting on suitable testing for the libvirt packages. > > > > If someone wants to step up to give the libvirt packages a decent > > smoke test, that will speed the C7 release process significant...

...> Documents/Bioinfo_workshop/" > setwd("C:/Users/Masoud/Desktop/r/") > > #### Read data #### > > > > PAD1 <- read.delim("C:/Users/Masoud/Desktop/r/PAD.txt", header=TRUE) > > attach(PAD1) > > head(PAD1) Code PAD Age High_ratio Smoke DM1_2 HT HxVasc PI_NonPI n 1 EE-001 0 46 0 1 0 0 0 1 1 2 SK-002 0 56 1 0 0 1 1 1 1 3 EP-003 0 52 0 0 0 0 0 1 1 4 TS-004 0 48 0 0 1 1 0 1 1 5 XS-005 1 65 0...

Hi, Does anyone know if: with R can you take a set of numbers and aggregate them like you can in SPSS? For example, could you calculate the percentage of people who smoke based on a dataset like the following: smoke = 1 non-smoke = 2 variable 1 1 1 2 2 1 1 1 2 2 2 2 2 2 When aggregated, SPSS can tell you what percentage of persons are smokers based on the frequency of 1's and 2's. Can R statistical package do a similar thing? Thanks, Nat

greetings. my problem is i can't print from windows 2000 or windows xp. my solaris box can print fine. on smoking (solaris 10, samba host) smbclient '//smoking/hp2605' -U jgs%password put /etc/motd i get the printout. on windows, this works: echo hiya^L > \\smoking\hp2605 on windows, from M$ word, the print SEEMS to go into the queue. i click on the printer, get the box that

Dear UseRs, Suppose I have data regarding smoking habits of a prospective cohort and wish to determine the risk ratio of colorectal cancer in the smokers compared to the non-smokers. What do I do at the end of the study with people who die of heart disease? Can I just censor them exactly the same as people who become uncontactable or who die in a plane crash? If not, why not? I'm thinking that heart disease isn't independent of smoking...

...t; <Pine.LNX.4.64.0607210716270.12611 at gannet.stats.ox.ac.uk> Apologies for a non-selfcontained example. Here is what I should have sent: pyears <- scan() 18793 52407 10673 43248 5710 28612 2585 12663 1462 5317 deaths <- scan() 2 32 12 104 28 206 28 186 31 102 l <- log(pyears) Smoke <- gl(2,1,10,labels=c("No","Yes")) Age <- gl(5,2,10,labels=c("35-44","45-54","55-64","65-74","75-84"), ordered=TRUE) mod1.glm <- glm(deaths ~ Age * Smoke + offset(l),family=poisson) summary(mod1.glm) age <...

...advance if this is posted all ready I have not been able to find any information about it. I have this data frame and I want to sort smoking by retlevel. Age Gender BMI Calories Fat Fiber Alc retlevel Smoking 1 64 Female 18.87834 1828.0 63.4 14.7 0.0 Normal Non-Smoker 2 25 Female 20.64102 1517.4 59.1 5.9 0.0 Normal Smoker 3 50 Female 20.40345 1902.9 72.9 35.4 7.3 Normal Non-Smoker 4 32 Female 35.97525 3328.4 163.3 20.0 4.1 High Smoker 5 43 Female 25.58279 2501.6 121.1 19.5 0.0 High Sm...

Consider the following example (based on an example in Pat Altham's GLM notes) pyears <- scan() 18793 52407 10673 43248 5710 28612 2585 12663 1462 5317 deaths <- scan() 2 32 12 104 28 206 28 186 31 102 Smoke <- gl(2,1,10,labels=c("No","Yes")) Age <- gl(5,2,10,labels=c("35-44","45-54","55-64","65-74","75-84"), ordered=TRUE) mod1.glm <- glm(deaths ~ Age * Smoke + offset(l),family=poisson) summary(mod1.glm) age <...

...ec, and documentation mandatory for a feature patch. Now, thanks to Shyam and Nigel, we have the patch ready to automate this process [2]. Feel free to review the patch, and comment on this. A heads up on how it looks like after this patch gets in. * A patch for a github reference won't pass smoke unless these labels are present on github issue. * Everyone, feel free to review and comment on the issue / patch regarding the document. But, the label is expected to be provided only by Project's general architects, and any industry experts we as community nominate for validating feature. Ini...

I've got a dataset which looks like this in the beginning: cbr dust smoking expo 1 0 0.20 1 5 2 0 0.25 1 4 3 0 0.25 1 8 4 0 0.25 1 4 5 0 0.25 1 4 (till no. 1240, anyway, a huge set) I have to analyse cbr and smoking, I know it works with chisq.test() for the whole set, but I only need cbr and smoking, and I

Hi, Suppose I have: # Fit a base model d1.ph <- coxph(Surv(start, stop, event)~ ejec + diavol + score + smoking + beta + surg.done, data = data.frame(foo)) summary(update(d1.ph, . ~ . + td1)) summary(update(d1.ph, . ~ . + td2)) As I have many columns in my data frame, foo, called td's. e.g. td1, td2, td3, .... And I'd like to