Related
I want to achieve the exact same thing asked in this question:
How to plot the survival curve generated by survreg (package survival of R)?
Except for the fact that I don't want the data to be stratified by a variable (in the question above it was stratified by sex).
I just want the progression free survival for the whole group of treated patients.
So when I copy the code from the other question, here is where I get stuck:
library(survminer)
library(tidyr)
s <- with(lung,Surv(time,status))
fKM <- survfit(s ~ sex,data=lung)
sWei <- survreg(s ~ as.factor(sex),dist='weibull',data=lung) # in my case here I would replace as.factor(sex) by 1
pred.sex1 = predict(sWei, newdata=list(sex=1),type="quantile",p=seq(.01,.99,by=.01)) #Since I don't want to stratify, what do I do with these 2 lines of code?
pred.sex2 = predict(sWei, newdata=list(sex=2),type="quantile",p=seq(.01,.99,by=.01))
df = data.frame(y=seq(.99,.01,by=-.01), sex1=pred.sex1, sex2=pred.sex2)
df_long = gather(df, key= "sex", value="time", -y)
p = ggsurvplot(fKM, data = lung, risk.table = T)
p$plot = p$plot + geom_line(data=df_long, aes(x=time, y=y, group=sex))
I tried replacing as.factor(sex) by 1 and then the rest of the code just does not make sense, can someone help me with this?
Many thanks in advance!
If you just want to plot the overall empirical survival curve, you might do something like this:
library(survival)
library(survminer)
library(tidyr)
s <- with(lung, Surv(time, status))
fKM <- survfit(s ~ 1, data = survival::lung)
ggsurvplot(fKM, ggtheme = theme_bw())
However, if you want to fit a Weibull model with no predictors, then your formula is fine.
sWei <- survreg(s ~ 1, dist = 'weibull', data = lung)
probs <- seq(0.01, 1, by = 0.01)
time <- predict(sWei, type = "quantile", se = TRUE, p = probs)
The only problem is that time is now a named list of two matrices: fit and se.fit. Both have the same number of rows as lung, but all rows are identical, so we just take one from each and calculate the confidence interval in a data frame which we can then use to create a ggplot:
ggplot(data = data.frame(p = 1 - probs,
time = time$fit[1,],
upper = time$fit[1,] + 1.96 * time$se.fit[1,],
lower = time$fit[1,] - 1.96 * time$se.fit[1,])) +
geom_step(aes(p, time, colour = "All"), size = 1) +
geom_ribbon(aes(p, ymin = lower, ymax = upper, fill = "All"), alpha = 0.2) +
coord_flip(ylim = c(0, 1000)) +
scale_fill_discrete(name = "Strata") +
scale_color_discrete(name = "Strata") +
theme_bw() +
theme(legend.position = "top")
Which we can see looks like a pretty good fit.
If you want both in the same plot you can do something like:
df <- data.frame(p = 1 - probs,
time = time$fit[1,],
upper = time$fit[1,] + 1.96 * time$se.fit[1,],
lower = time$fit[1,] - 1.96 * time$se.fit[1,])
ggsurvplot(fKM, ggtheme = theme_bw())$plot +
geom_line(data = df, aes(time, p), linetype = 2, size = 1) +
geom_line(data = df, aes(upper, p), linetype = 2, size = 1) +
geom_line(data = df, aes(lower, p), linetype = 2, size = 1)
Created on 2020-08-18 by the reprex package (v0.3.0)
I am trying to replicate a lattice graph using ggplot2 for a mixed model. My ggplot graph looks very similar but I am not sure about about loess line model fitted.
My goal is to add a loess line from the mixed model using ggplot2. Below is an example of my commands :
library(nlme)
library(ggplot2)
library(lattice)
library(lme4)
data(MathAchieve)
attach(MathAchieve)
mses <- tapply(SES, School, mean)
mses[as.character(MathAchSchool$School[1:10])]
Bryk <- as.data.frame(MathAchieve[, c("School", "SES", "MathAch")])
names(Bryk) <- c("school", "ses", "mathach")
sample20 <- sort(sample(7185, 20)) # 20 randomly sampled students
Bryk$meanses <- mses[as.character(Bryk$school)]
Bryk$cses <- Bryk$ses - Bryk$meanses
sector <- MathAchSchool$Sector
names(sector) <- row.names(MathAchSchool)
Bryk$sector <- sector[as.character(Bryk$school)]
attach(Bryk)
cat <- sample(unique(school[sector=="Catholic"]), 20)
Cat.20 <- groupedData(mathach ~ ses | school, data=Bryk[is.element(school, cat),])
Graph with Lattice:
trellis.device(color=T)
xyplot(mathach ~ ses | school, data=Cat.20, main="Catholic",
panel=function(x, y) {
panel.loess(x, y, span=1)
panel.xyplot(x, y)
panel.lmline(x, y, lty=2)
})
Graph with ggplot:
ggplot(Cat.20, aes(x = ses, y =mathach )) +
geom_point(size=1, shape=1) +
stat_smooth(method="lm",se=F)+
stat_smooth(, colour="Red",se=F)+
facet_wrap(school~., scale = "free_y")
Please any advice will be appreciated.
Preamble
Before going into the explanation, allow me to refer you to this question: Why is it not advisable to use attach() in R, and what should I use instead?
While it's recommendable that you made your question reproducible, the code you used can do with some clean-up. For example:
Don't include packages that aren't used in the code (I didn't see a need for the lme4 package);
There's no need to use data(...) to load MathAchieve. See the "Good Practices" section from ?data for more details.
As mentioned above, don't use attach().
For complete reproducibility, use set.seed() before any random sampling.
For a minimal example, don't plot 20 schools when a smaller number would do.
Since you are using one of the tidyverse packages for plotting, I recommend another from its collection for data manipulation:
library(nlme)
library(ggplot2)
library(lattice)
library(dplyr)
Bryk <- MathAchieve %>%
select(School, SES, MathAch) %>%
group_by(School) %>%
mutate(meanses = mean(SES),
cses = SES - meanses) %>%
ungroup() %>%
left_join(MathAchSchool %>% select(School, Sector),
by = "School")
colnames(Bryk) <- tolower(colnames(Bryk))
set.seed(123)
cat <- sample(unique(Bryk$school[Bryk$sector == "Catholic"]), 2)
Cat.2 <- groupedData(mathach ~ ses | school,
data = Bryk %>% filter(school %in% cat))
Explanation
Now that that's out of the way, let's look at the relevant functions for loess:
from ?panel.loess:
panel.loess(x, y, span = 2/3, degree = 1,
family = c("symmetric", "gaussian"),
... # omitted for space
)
from ?stat_smooth:
stat_smooth(mapping = NULL, data = NULL, geom = "smooth",
method = "auto", formula = y ~ x, span = 0.75, method.args = list(),
... # omitted for space
)
where method = "auto" defaults to loess from the stats package for <1000 observations.
from ?loess:
loess(formula, data, span = 0.75, degree = 2,
family = c("gaussian", "symmetric"),
... #omitted for space
)
In short, a loess plot's default parameters are span = 2/3, degree = 1, family = "symmetric" for the lattice package, and span = 0.75, degree = 2, family = "gaussian" for the ggplot2 package. You have to specify matching parameters if you want the resulting plots to match:
xyplot(mathach ~ ses | school, data = Cat.2, main = "Catholic",
panel=function(x, y) {
panel.loess(x, y, span=1, col = "red") # match ggplot's colours
panel.xyplot(x, y, col = "black") # to facilitate comparison
panel.lmline(x, y, lty=2, col = "blue")
})
ggplot(Cat.2, aes(x = ses, y = mathach)) +
geom_point(size = 2, shape = 1) +
stat_smooth(method = "lm", se = F)+
stat_smooth(span = 1,
method.args = list(degree = 1, family = "symmetric"),
colour = "red", se = F)+
facet_wrap(school ~ .) +
theme_classic() # less cluttered background to facilitate comparison
So I have 2 models for the data set that I am using:
> Bears1Fit1 <- lm(Weight ~ Neck.G)
>
> Bears2Fit2 <- lm(Weight ~ Neck.G + I(Neck.G)^2)
I want to plot these two models on the same scatterplot. I have this so far:
> plot(Neck.G, Weight, pch = c(1), main = "Black Bears Data: Weight Vs Neck Girth", xlab = "Neck Girth (inches) ", ylab = "Weight (pounds)")
> abline(Bears1Fit1)
However, I am unsure of how I should put the quadratic model on the same graph as well. I want to be able to have both lines on the same graph.
Here is an example with cars data set:
data(cars)
make models:
model_lm <- lm(speed ~ dist, data = cars)
model_lm2 <- lm(speed ~ dist + I(dist^2), data = cars)
make new data:
new.data <- data.frame(dist = seq(from = min(cars$dist),
to = max(cars$dist), length.out = 200))
predict:
pred_lm <- predict(model_lm, newdata = new.data)
pred_lm2 <- predict(model_lm2, newdata = new.data)
plot:
plot(speed ~ dist, data = cars)
lines(pred_lm ~ new.data$dist, col = "red")
lines(pred_lm2 ~ new.data$dist, col = "blue")
legend("topleft", c("linear", "quadratic"), col = c("red", "blue"), lty = 1)
with ggplot2
library(ggplot2)
put all data in one data frame and convert to long format using melt from reshape2
preds <- data.frame(new.data,
linear = pred_lm,
quadratic = pred_lm2)
preds <- reshape2::melt(preds,
id.vars = 1)
plot
ggplot(data = preds)+
geom_line(aes(x = dist, y = value, color = variable ))+
geom_point(data = cars, aes(x = dist, y = speed))+
theme_bw()
EDIT: another way using just ggplot2 using two geom_smooth layers, one with the default formula y ~ x (so it need not be specified) and one with a quadratic model formula = y ~ x + I(x^2). In order to get a legend we can specify color within the aes call naming the desired entry as we want it to show in the legend.
ggplot(cars,
aes(x = dist, y = speed)) +
geom_point() +
geom_smooth(method = "lm",
aes(color = "linear"),
se = FALSE) +
geom_smooth(method = "lm",
formula = y ~ x + I(x^2),
aes(color = "quadratic"),
se = FALSE) +
theme_bw()
I would like to fit a weibull curve to some event data and then include the fitted weibull curve in a survival plot plotted by survminer::ggsurvplot. Any ideas of how?
Here is an example to work on:
A function for simulating weibull data:
# N = sample size
# lambda = scale parameter in h0()
# rho = shape parameter in h0()
# beta = fixed effect parameter
# rateC = rate parameter of the exponential distribution of C
simulWeib <- function(N, lambda, rho, beta, rateC)
{
# covariate --> N Bernoulli trials
x <- sample(x=c(0, 1), size=N, replace=TRUE, prob=c(0.5, 0.5))
# Weibull latent event times
v <- runif(n=N)
Tlat <- (- log(v) / (lambda * exp(x * beta)))^(1 / rho)
# censoring times
C <- rexp(n=N, rate=rateC)
# follow-up times and event indicators
time <- pmin(Tlat, C)
status <- as.numeric(Tlat <= C)
# data set
data.frame(id=1:N,
time=time,
status=status,
x=x)
}
generate data
set.seed(1234)
betaHat <- rep(NA, 1e3)
for(k in 1:1e3)
{
dat <- simulWeib(N=100, lambda=0.01, rho=1, beta=-0.6, rateC=0.001)
fit <- coxph(Surv(time, status) ~ x, data=dat)
betaHat[k] <- fit$coef
}
#Estimate a survival function
survfit(Surv(as.numeric(time), x)~1, data=dat) -> out0
#plot
library(survminer)
ggsurvplot(out0, data = dat, risk.table = TRUE)
gg1 <- ggsurvplot(
out0, # survfit object with calculated statistics.
data = dat, # data used to fit survival curves.
risk.table = TRUE, # show risk table.
pval = TRUE, # show p-value of log-rank test.
conf.int = TRUE, # show confidence intervals for
# point estimaes of survival curves.
xlim = c(0,2000), # present narrower X axis, but not affect
# survival estimates.
break.time.by = 500, # break X axis in time intervals by 500.
ggtheme = theme_minimal(), # customize plot and risk table with a theme.
risk.table.y.text.col = T, # colour risk table text annotations.
risk.table.y.text = FALSE,
surv.median.line = "hv",
color = "darkgreen",
conf.int.fill = "lightblue",
title = "Survival probability",# show bars instead of names in text annotations
# in legend of risk table
)
gg1
As far as I see this, it is not possible do it with ggsurvplot at this moment.
I created an issue requesting this feature: https://github.com/kassambara/survminer/issues/276
You can plot survivor curves of a weibull model with ggplot2 like this:
library("survival")
wbmod <- survreg(Surv(time, status) ~ x, data = dat)
s <- seq(.01, .99, by = .01)
t_0 <- predict(wbmod, newdata = data.frame(x = 0),
type = "quantile", p = s)
t_1 <- predict(wbmod, newdata = data.frame(x = 1),
type = "quantile", p = s)
smod <- data.frame(time = c(t_0, t_1),
surv = rep(1 - s, times = 2),
strata = rep(c(0, 1), each = length(s)),
upper = NA, lower = NA)
head(surv_summary(cm))
library("ggplot2")
ggplot() +
geom_line(data = smod, aes(x = time, y = surv, color = factor(strata))) +
theme_classic()
However to my knowledge you cannot use survminer (yet):
library("survminer")
# wrong:
ggsurvplot(smod)
# does not work:
gg1$plot + geom_line(data = smod, aes(x = time, y = surv, color = factor(strata)))
The following works for me. Probably the credit goes to Heidi filling a feature request.
Hope, someone finds this useful.
library(survminer)
library(tidyr)
s <- with(lung,Surv(time,status))
sWei <- survreg(s ~ as.factor(sex),dist='weibull',data=lung)
fKM <- survfit(s ~ sex,data=lung)
pred.sex1 = predict(sWei, newdata=list(sex=1),type="quantile",p=seq(.01,.99,by=.01))
pred.sex2 = predict(sWei, newdata=list(sex=2),type="quantile",p=seq(.01,.99,by=.01))
df = data.frame(y=seq(.99,.01,by=-.01), sex1=pred.sex1, sex2=pred.sex2)
df_long = gather(df, key= "sex", value="time", -y)
p = ggsurvplot(fKM, data = lung, risk.table = T)
p$plot = p$plot + geom_line(data=df_long, aes(x=time, y=y, group=sex))
I have created an multiple linear regression model and would now like to plot it. But I can't seem to figure it out. Any help would be greatly appreciated! I used baruto to find the feature attributes and then used train() to get the model. When I try to plot model_lm I get the error:
There are no tuning parameters with more than 1 value.
Here is my code at what I have attempted so far:
rt_train <- rttotal2
rt_train$year <- NULL
#rt_train$box_office <- NULL
#impute na and address multicoliniearity
preproc <- preProcess(rt_train, method = c("knnImpute","center",
"scale"))
rt_proc <- predict(preproc, rt_train)
rt_proc$box_office <- rt_train$box_office
sum(is.na(rt_proc))
titles <- rt_proc$titles
rt_proc$titles <- NULL
#rt_train$interval <- as.factor(rt_train$interval)
dmy <- dummyVars(" ~ .", data = rt_proc,fullRank = T)
rt_transform <- data.frame(predict(dmy, newdata = rt_proc))
index <- createDataPartition(rt_transform$interval, p =.75, list = FALSE)
train_m <- rt_transform[index, ]
rt_test <- rt_transform[-index, ]
str(rt_train)
y_train <- train_m$box_office
y_test <-rt_test$box_office
train_m$box_office <- NULL
rt_test$box_office <- NULL
#selected feature attributes
boruta.train <- Boruta(interval~., train_m, doTrace =1)
#graph to see most important var to interval
lz<-lapply(1:ncol(boruta.train$ImpHistory),function(i)
boruta.train$ImpHistory[is.finite(boruta.train$ImpHistory[,i]),i])
names(lz) <- colnames(boruta.train$ImpHistory)
plot(boruta.train, xlab = "", xaxt = "n")
Labels <- sort(sapply(lz,median))
axis(side = 1,las=2,labels = names(Labels),
at = 1:ncol(boruta.train$ImpHistory), cex.axis = 0.7)
#get most important attributes
final.boruta <- TentativeRoughFix(boruta.train)
print(final.boruta)
getSelectedAttributes(final.boruta, withTentative = F)
boruta.rt_df <- attStats(final.boruta)
boruta.rt_df
boruta.rt_df <- setDT(boruta.rt_df, keep.rownames = TRUE)[]
predictors <- boruta.rt_df %>%
filter(., decision =="Confirmed") %>%
select(., rn)
predictors <- unlist(predictors)
control <- trainControl(method="repeatedcv",
number=10,
repeats=6)
#look at residuals
#p-value is very small so reject H0 that predictors have no effect so
#we can use rotten tomatoes to predict box_office ranges
train_m$interval <- NULL
model_lm <- train(train_m[,predictors],
y_train, method='lm',
trControl = control, tuneLength = 10)
model_lm #.568
#
plot(model_lm)
plot(model_lm)
z <- varImp(object=model_lm)
z <- setDT(z, keep.rownames = TRUE)
z$model <- NULL
z$calledFrom <- NULL
row.names(z)
plot(varImp(object=model_lm),main="Linear Model Variable Importance")
predictions<-predict.train(object=model_lm,rt_test[,predictors],type="raw")
table(predictions)
#get coeff
interc <- coef(model_lm$finalModel)
slope <- coef(model_lm$finalModel)
ggplot(data = rt_train, aes(y = box_office)) +
geom_point() +
geom_abline(slope = slope, intercept = interc, color = 'red')
This is what some of my input looks like. Thank you!!
Here is an example using the inbuilt data set cars:
data(cars, package = "datasets")
library(caret)
build the model
control <- trainControl(method = "repeatedcv",
number = 10,
repeats = 6)
model_lm <- train(dist ~ speed, data = cars, method='lm',
trControl = control, tuneLength = 10)
I will assume you would like to plot the final model.
You can use the caret predict.train function to get the predictions from the model and plot them:
pred <- predict(model_lm, cars)
pred <- data.frame(pred = pred, speed = cars$speed)
additionally you can provide the cars data set to geom point and plot the observations:
library(ggplot2)
ggplot(data = pred)+
geom_line(aes(x = speed, y = pred))+
geom_point(data = cars, aes(x=speed, y = dist))
if you would like to obtain the confidence or prediction interval you can use the predict.lm function on model_lm$finalModel:
Here is an example for the prediction interval:
pred <- predict(model_lm$finalModel, cars, se.fit = TRUE, interval = "prediction")
pred <- data.frame(pred = pred$fit[,1], speed = cars$speed, lwr = pred$fit[,2], upr = pred$fit[,3])
pred_int <- ggplot(data = pred)+
geom_line(aes(x = speed, y = pred))+
geom_point(data = cars, aes(x = speed, y = dist)) +
geom_ribbon(aes(ymin = lwr, ymax = upr, x = speed), alpha = 0.2)
or the confidence interval:
pred <- predict(model_lm$finalModel, cars, se.fit = TRUE, interval = "confidence")
pred <- data.frame(pred = pred$fit[,1], speed = cars$speed, lwr = pred$fit[,2], upr = pred$fit[,3])
pred_conf <- ggplot(data = pred)+
geom_line(aes(x = speed, y = pred))+
geom_point(data = cars, aes(x = speed, y = dist)) +
geom_ribbon(aes(ymin = lwr, ymax = upr, x = speed), alpha = 0.2)
plotting them side by side:
library(cowplot)
plot_grid(pred_int, pred_conf)
to plot the linear dependence on two variables you can use a 3D plot, for more than 3 it will be a problem.