I want to find the survival rate at 5 years.

[smilehappysmile]

library(randomForestSRC)
na.omit(pbc)
data(pbc, package = "randomForestSRC")
pbc<-na.omit(pbc)
set.seed(42)
n <- nrow(pbc)
n_train <- 200

train_data <- pbc[1:n_train, ]
test_data <- pbc[-(1:n_train), ]

rf_model <- rfsrc(Surv(days, status) ~ ., data = train_data)

predictions <- predict(rf_model, test_data)

predictions$survival

In this part, there are 104 columns, but I don't understand why a matrix appears, and I want to know what the columns mean. I want to find the survival rate at 5 years and draw a roc curve with the survival rate and status. If you know how, please let me know.

For example, it doesn't seem to be selecting column 5. But I chose it and drew it.

predicted_survival_5_years <- as.vector(predictions$survival[, 5])

library(pROC)
roc_obj <- roc(test_data$status, predicted_survival_5_years)

plot(roc_obj, main = "ROC Curve for Survival at 5 Years")

[ishwaran]

The rows represent the data points, so row i is the estimated survival for patient i. The columns are the estimated survival for the time points rf_model$time.interest. Column 5 is the estimated survival at time rf_model$time.interest[5] and note that for the PBC data, time is given in days so you need to divide by 365 to convert to years. According to my calculations, 5 years is approximately coordinate 73 for the full data set, but you will need to figure out what the coordinate is for your training data.

Therefore in your code, change to the following:

j <- sum(rf_model$time.interest/365<=5)
predicted_survival_5_years <- as.vector(predictions$survival[, j])
library(pROC)
roc_obj <- roc(test_data$status, predicted_survival_5_years)
plot(roc_obj, main = "ROC Curve for Survival at 5 Years")

[smilehappysmile]

Thank you so much for your reply. One more thing I'm curious about... Speaking of 5 years, I wonder why rf_model$time.interest/365<=5
is less than 5 in the current code.
Because in the rf_model$time.interest code, point 1827 is greater than 5 years, but it is the 68th.
However, the point you answered is the 67th.

And in the past,
t <- 1825
surv.prob28 <- NULL # strategy for survival probability
for (i in 1:nrow(data.ts28)) { # loop start
temp <- survfit(fit28, newdata = data.ts28[i,]) # fit survival curve at the ith obs in test set
id <- max(which(temp$time <= t)) # get id corresponding to t = 1825(in this case)
surv.prob28[i] <- temp$surv[id] # compute survival probability
}

Even in this code, id <- max(which(temp$time <= t), I found the smaller point in time... I wonder if I shouldn't find the smallest point in time after 5 years.

In summary, I'm curious to find the part that is less than or equal to <=5. This seems to be correct... but I'm wondering if min(>=5) doesn't work like this. I am curious as to whether the timing of not exceeding 5 years is more accurate than the timing of exceeding 5 years.

I wonder if it's because if you find the smallest point over 5 years, it could be 6 or 7 years. Could it be that it would be more accurate to find a point less than 5 years than to go beyond that period? So it could be 4 years... Is that more accurate than overdue in medical statistics?