Displaying 5 results from an estimated 5 matches for "surv5".
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surv
2011 Nov 29
2
Nomogram with stratified cph in Design package-- failure probability
..., x=T, y=T, time.inc=5)
surv<- Survival(f1)
haz<- Hazard(f1)
Here is the Error in UseMethod("Hazard") :
no applicable method for 'Hazard' applied to an object of class "c('cph',
'Design', 'coxph')"
surv10 <- function(lp) surv(10,lp)
surv5 <- function(lp) surv(5,lp)
quant <- Quantile(f1)
at.surv <- c(0.1, 0.3, 0.5, 0.7, 0.9)
at.med1<-c(2,3,4, 5,6,7,8, 10,15,20,25, 30)
par(cex=0.8)
nom<- nomogram(f1, conf.int=F,
fun=list(1-surv5, 1-surv10), funlabel=c('5-Year Survival Probability',
'10-Year Survival Pro...
2011 Nov 30
1
Nomogram with stratified cph in rms package, how to get failure probability
...T, x=T, y=T, time.inc=5)
surv<- Survival(f1)
haz<- Hazard(f1)
Here is the Error in UseMethod("Hazard") :
no applicable method for 'Hazard' applied to an object of class "c('cph',
'Design', 'coxph')"
surv10 <- function(lp) surv(10,lp)
surv5 <- function(lp) surv(5,lp)
quant <- Quantile(f1)
at.surv <- c(0.1, 0.3, 0.5, 0.7, 0.9)
at.med1<-c(2,3,4, 5,6,7,8, 10,15,20,25, 30)
par(cex=0.8)
nom<- nomogram(f1, conf.int=F,
fun=list(surv5, surv10), funlabel=c('5-Year Survival Probability', '10-Year
Survival Probabi...
2005 Jul 11
1
validation, calibration and Design
...ariables A and B, cens and time
columns (months)
ddist1 <- datadist(data1)
options(datadist='ddist1')
s1 <- Surv(data1$time, data1$cens)
cph.nomo <- cph(s1 ~ A+B, surv=T, x=T, y=T, time.inc=60)
survcph <- Survival(cph.nomo, x=T, y=T, time.inc=60, surv=T)
surv5 <- function(lp) survcph(60, lp)
nomogram(cph.nomo, lp=T, conf.int=F, fun=list(surv5, surv7),
funlabel=c("5 yr DFS"))
# now have a useful nomogram model, with good discrimination and
#calibration when checked with validate and calibrate (not shown)
#....move on to validation co...
2011 Nov 30
0
formula for calculating the survival probability for nomogram
...survival probability for this nomogram. Then
I can use this formula to do validation by using other data set. *
f1 <- cph(Surv(retime,dfs) ~ age+her2+t_stage+n_stage+er+cytcyt+Cyt_PCDK2 ,
data=data11,
surv=T, x=T, y=T, time.inc=5)
surv<- Survival(f1)
surv10 <- function(lp) surv(10,lp)
surv5 <- function(lp) surv(5,lp)
quant <- Quantile(f1)
at.surv <- c(0.1, 0.3, 0.5, 0.7, 0.9)
nom<- nomogram(f1, conf.int=F,
fun=list(surv5, surv10), funlabel=c('5-Year Survival Probability', '10-Year
Survival Probability' ), lp=F,
fun.at=c(at.surv, at.surv))
############...
2017 Oct 26
1
nomogram function error
...g to fit a model to the nomogram. Here is the codeand corresponding error:
?
?
>nomogramCF = nomogram(cph_age6_40avp4,
+????????????????????lp.at= seq(-10,10,by =0.5),lp = TRUE,
+??????????????????????
+??????????????????????funlabel="5year survival",
+??????????????????????fun=surv5.CFdata5,fun.at= c(0.01,seq(0.1,0.9,by=0.2),0.99))
Error inapprox(fu[s], xseq[s], fat, ties = mean) :
??needat least two non-NA values to interpolate
?
?
I have fit similar nomograms based on cox ph models usingsimilar code, so I?m not sure what I?m doing wrong this time.
?
I heard there mig...