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 would like to use the following dataset to fit glm and visualize the predict().
y=c(-18.948,-19.007,-18.899,-19.022,-20.599,-19.778,-17.344,-20.265,-20.258,-19.886,-18.05,-19.824,-20.1,-20.508,-20.455,-16.573,-20.249,-20.205,-20.436,-16.358,-17.717,-19.794,-20.372,-19.944,-20.072,-19.889,-20.139,-19.132,-20.275,-19.953,-19.769,-20.2,-19.638,-17.419,-19.086,-18.347,-18.73,-18.872,-18.956,-19.28,-18.176,-19.036,-18.084,-20.11,-19.641,-19.656,-19.25,-18.68,-19.089,-18.969,-18.161,-17.603,-20.37,-19.233,-18.961,-19.083,-20.118,-19.795,-17.154,-16.75)
x1=c(9.698,9.583,9.356,9.326,9.438,9.733,8.803,8.973,9.141,9.044,8.788,9.377,9.26,10.186,9.035,9.569,9.431,9.09,8.776,9.117,9.393,9.408,9.307,8.868,8.398,8.407,9.364,9.074,8.444,9.122,10.11,7.81,9.777,6.472,9.521,8.92,9.341,9.446,9.08,8.071,8.047,8.019,7.419,9.022,9.981,9.337,9.989,10.013,9.31,10.843,8.337,9.103,6.438,9.372,9.071,8.749,9.016,8.181,9.284,8.44)
x2=c('S03','S03','S03','S03','S03','S03','S03','S03','S03','S03','S03','S03','S03','S03','S03','S03','S03','S03','S03','S03','S04','S04','S04','S04','S04','S04','S06','S06','S06','S06','S06','S06','S06','S06','S07','S07','S07','S07','S07','S07','S07','S07','S07','S08','S08','S09','S09','S09','S09','S09','S09','S09','S10','S03','S03','S03','S04','S04','S07','S07')
x3=c('A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','A1','P1','P1','P1','P1','P1','P1','P1')
mydata <- data.frame(y,x1,x2,x3)
Fit glm Model:
myglm <- glm(y ~ x1+x2+x3+x1:x2, family="gaussian", data= mydata)
Prediction:
1). Extract the range of x1
min <- min(mydata$x1)
max <- max(mydata$x1)
2). Create a new data frame.x
Here comes the question:
How should I include x2 and x3 in the new.x?
new.x <- data.frame(
x1=seq(min, max, length=60),
x2= ???
x3= ???)
Then predict new.y with myglm:
new.y = predict(myglm, newdata=new.x, se.fit=TRUE)
Combine new.x and new.y:
addThese <- data.frame(new.x, new.y)
interval
addThese <- mutate(addThese,
d15N=exp(fit),
lwr=exp(fit-1.96*se.fit),
upr=exp(fit+1.96*se.fit))
3). Visualization of the original data points and the glm prediction smooth line added:
ggplot(addThese, aes(x1, fit))+
geom_point(shape=21, size=3)+
geom_smooth(data=addThese,
aes(ymin=lwr, ymax=upr),
stat='identity')
I'm still wondering if this is a right way to create new.data, but I'll give it a try. So with your data, slightly modifying your code:
myglm <- glm(y ~ x1 + x2 + x3 + x1:x2, family = gaussian, data = mydata)
minx <- min(mydata$x1)
maxx <- max(mydata$x1)
# create data with all combinations of x1, x2, x3
new.data <- expand.grid(x1 = seq(minx, maxx, length.out = 60),
x2 = unique(mydata$x2),
x3 = unique(mydata$x3)
)
# visualize data
data.frame(predict(myglm, newdata = new.data, se.fit = T)[1:2]) %>%
bind_cols(new.data) %>%
mutate(d15N = exp(fit), lwr = fit - 1.96 * se.fit, upr = fit + 1.96 * se.fit) %>%
ggplot(aes(x = x1, y = fit, colour = interaction(x2, x3))) +
geom_point(size = 1, alpha = .75, pch = 19, position = "jitter") +
geom_smooth(aes(ymin = lwr, ymax = upr), stat = "identity", alpha = .5) +
facet_wrap(~interaction(x2, x3, sep = " : "), nrow = 5) +
ggthemes::theme_few() +
labs(y = "Predicted value", x = bquote(x[1])) +
theme(legend.position = "none")
I'm new with R and I have fit 3 models for my data as follows:
Model 1: y = a(x) + b
lm1 = lm(data$CBI ~ data$dNDVI)
Model 2: y = a(x)2 + b(x) + c
lm2 <- lm(CBI ~ dNDVI + I(dNDVI^2), data=data)
Model 3: y = x(a|x| + b)–1
lm3 = nls(CBI ~ dNDVI*(a*abs(dNDVI) + b) - 1, start = c(a = 1.5, b = 2.7), data = data)
Now I would like to plot all these three models in R but I could not find the way to do it, can you please help me? I have tried with the first two models as follow and it work but I don't know how to add the Model 3 on it:
ggplot(data = data, aes(x = dNDVI, y = CBI)) +
geom_point() +
geom_smooth(method = lm, formula = y ~ x, size = 1, se = FALSE) +
geom_smooth(method = lm, formula = y ~ x + I(x^2), size = 1, se = FALSE ) +
theme_bw()
I also would like to add a legend which show 3 different colours or types of lines/curves for the 3 models as well. Can you please guide me how to make it in the figure?
Using iris as a dummy set to represent the three models:
new.dat <- data.frame(Sepal.Length=seq(min(iris$Sepal.Length),
max(iris$Sepal.Length), length.out=50)) #new data.frame to predict the fitted values for each model
m1 <- lm(Petal.Length ~ Sepal.Length, iris)
m2 <- lm(Petal.Length ~ Sepal.Length + I(Sepal.Length^2), data=iris)
m3 <- nls(Petal.Length ~ Sepal.Length*(a*abs(Sepal.Length) + b) - 1,
start = c(a = 1.5, b = 2.7), data = iris)
new.dat$m1.fitted <- predict(m1, new.dat)
new.dat$m2.fitted <- predict(m2, new.dat)
new.dat$m3.fitted <- predict(m3, new.dat)
new.dat <- new.dat %>% gather(var, val, m1.fitted:m3.fitted) #stacked format of fitted data of three models (to automatically generate the legend in ggplot)
ggplot(new.dat, aes(Sepal.Length, val, colour=var)) +
geom_line()
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 am trying to visualize the results of an nlme object without success. When I do so with an lmer object, the correct plot is created. My goal is to use nlme and visualize a fitted growth curve for each individual with ggplot2. The predict() function seems to work differently with nlme and lmer objects.
model:
#AR1 with REML
autoregressive <- lme(NPI ~ time,
data = data,
random = ~time|patient,
method = "REML",
na.action = "na.omit",
control = list(maxlter=5000, opt="optim"),
correlation = corAR1())
nlme visualization attempt:
data <- na.omit(data)
data$patient <- factor(data$patient,
levels = 1:23)
ggplot(data, aes(x=time, y=NPI, colour=factor(patient))) +
geom_point(size=1) +
#facet_wrap(~patient) +
geom_line(aes(y = predict(autoregressive,
level = 1)), size = 1)
when I use:
data$fit<-fitted(autoregressive, level = 1)
geom_line(aes(y = fitted(autoregressive), group = patient))
it returns the same fitted values for each individual and so ggplot produces the same growth curve for each. Running test <-data.frame(ranef(autoregressive, level=1)) returns varying intercepts and slopes by patient id. Interestingly, when I fit the model with lmer and run the below code it returns the correct plot. Why does predict() work differently with nlme and lmer objects?
timeREML <- lmer(NPI ~ time + (time | patient),
data = data,
REML=T, na.action=na.omit)
ggplot(data, aes(x = time, y = NPI, colour = factor(patient))) +
geom_point(size=3) +
#facet_wrap(~patient) +
geom_line(aes(y = predict(timeREML)))
In creating a reproducible example, I found that the error was not occurring in predict() nor in ggplot() but instead in the lme model.
Data:
###libraries
library(nlme)
library(tidyr)
library(ggplot2)
###example data
df <- data.frame(replicate(78, sample(seq(from = 0,
to = 100, by = 2), size = 25,
replace = F)))
##add id
df$id <- 1:nrow(df)
##rearrange cols
df <- df[c(79, 1:78)]
##sort columns
df[,2:79] <- lapply(df[,2:79], sort)
##long format
df <- gather(df, time, value, 2:79)
##convert time to numeric
df$time <- factor(df$time)
df$time <- as.numeric(df$time)
##order by id, time, value
df <- df[order(df$id, df$time),]
##order value
df$value <- sort(df$value)
Model 1 with no NA values fits successfully.
###model1
model1 <- lme(value ~ time,
data = df,
random = ~time|id,
method = "ML",
na.action = "na.omit",
control = list(maxlter=5000, opt="optim"),
correlation = corAR1(0, form=~time|id,
fixed=F))
Introducing NA's causes invertible coefficient matrix error in model 1.
###model 1 with one NA value
df[3,3] <- NA
model1 <- lme(value ~ time,
data = df,
random = ~time|id,
method = "ML",
na.action = "na.omit",
control = list(maxlter=2000, opt="optim"),
correlation = corAR1(0, form=~time|id,
fixed=F))
But not in model 2, which has a more simplistic within-group AR(1) correlation structure.
###but not in model2
model2 <- lme(value ~ time,
data = df,
random = ~time|id,
method = "ML",
na.action = "na.omit",
control = list(maxlter=2000, opt="optim"),
correlation = corAR1(0, form = ~1 | id))
However, changing opt="optim" to opt="nlminb" fits model 1 successfully.
###however changing the opt to "nlminb", model 1 runs
model3 <- lme(value ~ time,
data = df,
random = ~time|id,
method = "ML",
na.action = "na.omit",
control = list(maxlter=2000, opt="nlminb"),
correlation = corAR1(0, form=~time|id,
fixed=F))
The code below visualizes model 3 (formerly model 1) successfully.
df <- na.omit(df)
ggplot(df, aes(x=time, y=value)) +
geom_point(aes(colour = factor(id))) +
#facet_wrap(~id) +
geom_line(aes(y = predict(model3, level = 0)), size = 1.3, colour = "black") +
geom_line(aes(y = predict(model3, level=1, group=id), colour = factor(id)), size = 1)
Note that I am not exactly sure what changing the optimizer from "optim" to "nlminb" does and why it works.