survival-analysis

问题 This problem has confounded me for more hours than I care to admit. I have isolated the problem so I can replicate it. library(survival) library(survminer) set.seed(123) test <- data.frame(rnorm(10000)+5, sample(0:1, 10000, replace = TRUE)) colnames(test)<- c("time", "event") #sum(test$event) = 4975 survfitted <- survfit(Surv(time = time, event = event) ~ 1, data = test) plot(survfitted, fun = "event") Why does this curve sum up to 100% when only 49.75% experience an event? What would be the

问题 I have the following risk table for a Kaplan Meier curve constructed with surv_miner in RStudio: Kaplan Meier curve I would like to string wrap the text in the risk table so that the number is on top, and the percentage (in parenthesis) is on another line below the number, so that they all fit in the table. I tried using the stringr package, but I don't quite know how to incorporate that into the theme. Any help would be appreciated. I've posted just my code, since it is a bit tricky to add

问题 I am trying to plot Kaplan-Meyer curve using ggsurvplot from survminer package. I'm unable to plot it when I pass a survfit object saved in a list. Let me use lung dataset as a example. Everything works below: library("survival") library("survminer") fit <- survfit(Surv(time, status) ~ sex, data = lung) ggsurvplot(fit, conf.int = TRUE, risk.table.col = "strata", palette = c("#E7B800", "#2E9FDF"), xlim = c(0, 600)) Now I do survfit on two variables and save the model result in a list. Then

问题 I have a table that looks like this: ID Survival Event Allele 2 5 1 WildType 2 0 1 WildType 3 3 1 WildType 4 38 0 Variant I want to do a kaplan meier plot, and tell me if wild type or variants tend to survive longer. I have this code: library(survival) Table <-read.table("Table1",header=T) fit=survfit(Surv(Table$Survival,Table$Event)~Table$Allele) plot(fit,lty=2:3,col=3:4) From the fit p value, I can see that the survival of these two groups have significantly different survival curves.

问题 I have a table that looks like this: ID Survival Event Allele 2 5 1 WildType 2 0 1 WildType 3 3 1 WildType 4 38 0 Variant I want to do a kaplan meier plot, and tell me if wild type or variants tend to survive longer. I have this code: library(survival) Table <-read.table("Table1",header=T) fit=survfit(Surv(Table$Survival,Table$Event)~Table$Allele) plot(fit,lty=2:3,col=3:4) From the fit p value, I can see that the survival of these two groups have significantly different survival curves.

问题 I have a table that looks like this: ID Survival Event Allele 2 5 1 WildType 2 0 1 WildType 3 3 1 WildType 4 38 0 Variant I want to do a kaplan meier plot, and tell me if wild type or variants tend to survive longer. I have this code: library(survival) Table <-read.table("Table1",header=T) fit=survfit(Surv(Table$Survival,Table$Event)~Table$Allele) plot(fit,lty=2:3,col=3:4) From the fit p value, I can see that the survival of these two groups have significantly different survival curves.

问题 I used coxph() from the survival package in multiply imputed dataset and encountered a warning when trying to pool the results. The warning message states: "In mice.df(m, lambda, dfcom, method) : Large sample assumed. A reproducible example is below (with publically available data, without worrying to much about the appropriateness of using both mice and coxph is these data): library(mice) library(survival) #load publically available data data(pbc) #select variables for the reproducable

问题 I used coxph() from the survival package in multiply imputed dataset and encountered a warning when trying to pool the results. The warning message states: "In mice.df(m, lambda, dfcom, method) : Large sample assumed. A reproducible example is below (with publically available data, without worrying to much about the appropriateness of using both mice and coxph is these data): library(mice) library(survival) #load publically available data data(pbc) #select variables for the reproducable

问题 The question I am posting here is closely linked to another question I posted two days ago about gompertz aging analysis. I am trying to construct a survival object, see ?Surv, in R. This will hopefully be used to perform Gompertz analysis to produce an output of two values (see original question for further details). I have survival data from an experiment in flies which examines rates of aging in various genotypes. The data is available to me in several layouts so the choice of which is up

问题 I am new to survival analysis and survfit in R. I want to extract survival probabilities for 4 groups (diseases) at specified time periods (0,10,20,30 years since diagnosis) in a table. Here is the setup: fit <- survfit((time=time,event=death)~group) surv.prob <- summary(fit,time=c(0,10,20,30))$surv surv.prob contains 16 probabilities, that is, survival probabilities for 4 groups estimated at 4 different time periods listed above. I want to create a table like this: Group time.period prob 1 0