How do I change the axis values of the rms r package! from horizontal to vertical like style. Below is a code extracted from the rms package! The axis values are in horizontal style. I want to change it to the vertical style and I can seem to figure it out, I appreciate any help! Thanks
n <- 1000 # define sample size
set.seed(17) # so can reproduce the results
d <- data.frame(age = rnorm(n, 50, 10),
blood.pressure = rnorm(n, 120, 15),
cholesterol = rnorm(n, 200, 25),
sex = factor(sample(c('female','male'), n,TRUE)))
# Specify population model for log odds that Y=1
# Simulate binary y to have Prob(y=1) = 1/[1+exp(-L)]
d <- upData(d,
L = .4*(sex=='male') + .045*(age-50) +
(log(cholesterol - 10)-5.2)*(-2*(sex=='female') + 2*(sex=='male')),
y = ifelse(runif(n) < plogis(L), 1, 0))
ddist <- datadist(d); options(datadist='ddist')
f <- lrm(y ~ lsp(age,50) + sex * rcs(cholesterol, 4) + blood.pressure,
data=d)
nom <- nomogram(f, fun=function(x)1/(1+exp(-x)), # or fun=plogis
fun.at=c(.001,.01,.05,seq(.1,.9,by=.1),.95,.99,.999),
funlabel="Risk of Death")
#Instead of fun.at, could have specified fun.lp.at=logit of
#sequence above - faster and slightly more accurate
plot(nom, xfrac=.45)
print(nom)
nom <- nomogram(f, age=seq(10,90,by=10))
plot(nom, xfrac=.45)
g <- lrm(y ~ sex + rcs(age, 3) * rcs(cholesterol, 3), data=d)
nom <- nomogram(g, interact=list(age=c(20,40,60)),
conf.int=c(.7,.9,.95))
plot(nom, col.conf=c(1,.5,.2), naxes=7)
w <- upData(d,
cens = 15 * runif(n),
h = .02 * exp(.04 * (age - 50) + .8 * (sex == 'Female')),
d.time = -log(runif(n)) / h,
death = ifelse(d.time <= cens, 1, 0),
d.time = pmin(d.time, cens))
f <- psm(Surv(d.time,death) ~ sex * age, data=w, dist='lognormal')
med <- Quantile(f)
surv <- Survival(f) # This would also work if f was from cph
plot(nomogram(f, fun=function(x) med(lp=x), funlabel="Median Survival Time"))
nom <- nomogram(f, fun=list(function(x) surv(3, x),
function(x) surv(6, x)),
funlabel=c("3-Month Survival Probability",
"6-month Survival Probability"))
plot(nom, xfrac=.7)
[![This is the nomogram result][1]][1]
Related
If I have a GLM, is there any way I can efficiently find the maximum output by changing one covariate and holding the others?
Using my simulated data:
# FUNCTIONS ====================================================================
logit <- function(p){
x = log(p/(1-p))
x
}
sigmoid <- function(x){
p = 1/(1 + exp(-x))
p
}
beta_duration <- function(D, select){
logit(
switch(select,
0.05 + 0.9 / (1 + exp(-2*D + 25)),
0.9 * exp(-exp(-0.5 * (D - 11))),
0.9 * exp(-exp(-(D - 11))),
0.9 * exp(-2 * exp(-(D - 9))),
sigmoid(0.847 + 0.210 * (D - 10)),
0.7 + 0.0015 * (D - 10) ^ 2,
0.7 - 0.0015 * (D - 10) ^ 2 + 0.03 * (D - 10)
)
)
}
beta_sex <- function(sex, OR = 1){
ifelse(sex == "Female", -0.5 * log(OR), 0.5 * log(OR))
}
plot_beta_duration <- function(select){
x <- seq(10, 20, by = 0.01)
y <- beta_duration(x, select)
data.frame(x = x,
y = y) %>%
ggplot(aes(x = x, y = y)) +
geom_line() +
ylim(0, 1)
}
# DATA SIMULATION ==============================================================
duration <- c(10, 12, 14, 18, 20)
sex <- factor(c("Female", "Male"))
eta <- function(duration, sex, duration_select, sex_OR, noise_sd){
beta_sex(sex, sex_OR) + beta_duration(duration, duration_select) + rnorm(length(duration), 0, noise_sd)
}
sim_data <- function(durations_type, sex_OR, noise_sd, p_female, n, seed){
set.seed(seed)
data.frame(
duration = sample(duration, n, TRUE),
sex = sample(sex, n, TRUE, c(p_female, 1 - p_female))
) %>%
rowwise() %>%
mutate(eta = eta(duration, sex, durations_type, sex_OR, noise_sd),
p = sigmoid(eta),
cured = sample(0:1, 1, prob = c(1 - p, p)))
}
# DATA SIM PARAMETERS
durations_type <- 4 # See beta_duration for functions
sex_OR <- 3 # Odds of cure for male vs female (ref)
noise_sd <- 1
p_female <- 0.7 # proportion of females in the sample
n <- 500
data <- sim_data(durations_type = 1, # See beta_duration for functions
sex_OR = 3, # Odds of cure for male vs female (ref)
noise_sd = 1,
p_female = 0.7, # proportion of females in the sample
n = 500,
seed = 21874564)
I am fitting a fractional polynomial GLM:
library(mfp)
model1 <- mfp(cured ~ fp(duration) + sex,
family = binomial(link = "logit"),
data = data)
summary(model1)
Given that I am holding sex as constant, is there any way to find the value of duration within a certain range that gives me the highest predicted value? Something less inefficient than:
range <- seq(10, 20, by = 1e-4)
range[which.max(predict(model, type = "response", newdata = data.frame(duration = range, sex = "Male")))]
You can use optimize here. Just create a function which returns a prediction based on the value of duration:
f <- function(x) predict(model1, list(sex = 'Male', duration = x))
And we can find the value of duration which produces the maximum log odds within the range 0-20 by doing:
optimise(f, c(0, 20), maximum = TRUE)$maximum
#> [1] 17.95679
Given a fractional polynomial GLM, I am looking to find the value of a covariate that gives me an output of a given probability.
My data is simulated using:
# FUNCTIONS ====================================================================
logit <- function(p){
x = log(p/(1-p))
x
}
sigmoid <- function(x){
p = 1/(1 + exp(-x))
p
}
beta_duration <- function(D, select){
logit(
switch(select,
0.05 + 0.9 / (1 + exp(-2*D + 25)),
0.9 * exp(-exp(-0.5 * (D - 11))),
0.9 * exp(-exp(-(D - 11))),
0.9 * exp(-2 * exp(-(D - 9))),
sigmoid(0.847 + 0.210 * (D - 10)),
0.7 + 0.0015 * (D - 10) ^ 2,
0.7 - 0.0015 * (D - 10) ^ 2 + 0.03 * (D - 10)
)
)
}
beta_sex <- function(sex, OR = 1){
ifelse(sex == "Female", -0.5 * log(OR), 0.5 * log(OR))
}
plot_beta_duration <- function(select){
x <- seq(10, 20, by = 0.01)
y <- beta_duration(x, select)
data.frame(x = x,
y = y) %>%
ggplot(aes(x = x, y = y)) +
geom_line() +
ylim(0, 1)
}
# DATA SIMULATION ==============================================================
duration <- c(10, 12, 14, 18, 20)
sex <- factor(c("Female", "Male"))
eta <- function(duration, sex, duration_select, sex_OR, noise_sd){
beta_sex(sex, sex_OR) + beta_duration(duration, duration_select) + rnorm(length(duration), 0, noise_sd)
}
sim_data <- function(durations_type, sex_OR, noise_sd, p_female, n, seed){
set.seed(seed)
data.frame(
duration = sample(duration, n, TRUE),
sex = sample(sex, n, TRUE, c(p_female, 1 - p_female))
) %>%
rowwise() %>%
mutate(eta = eta(duration, sex, durations_type, sex_OR, noise_sd),
p = sigmoid(eta),
cured = sample(0:1, 1, prob = c(1 - p, p)))
}
# DATA SIM PARAMETERS
durations_type <- 4 # See beta_duration for functions
sex_OR <- 3 # Odds of cure for male vs female (ref)
noise_sd <- 1
p_female <- 0.7 # proportion of females in the sample
n <- 500
data <- sim_data(durations_type = 1, # See beta_duration for functions
sex_OR = 3, # Odds of cure for male vs female (ref)
noise_sd = 1,
p_female = 0.7, # proportion of females in the sample
n = 500,
seed = 21874564)
And my model is fitted by:
library(mfp)
model1 <- mfp(cured ~ fp(duration) + sex,
family = binomial(link = "logit"),
data = data)
summary(model1)
For each level of sex (i.e. "Male" or "Female"), I want to find the value of duration that gives me a probability equal to some value frontier <- 0.8.
So far, I can only think of using an approximation using a vector of possibilities:
pred_duration <- seq(10, 20, by = 0.1)
pred <- data.frame(expand.grid(duration = pred_duration,
sex = sex),
p = predict(model1,
newdata = expand.grid(duration = pred_duration,
sex = sex),
type = "response"))
pred[which(pred$p > 0.8), ] %>%
group_by(sex) %>%
summarize(min(duration))
But I am really after an exact solution.
The function uniroot allows you to detect the point at which the output of a function equals 0. If you create a function that takes duration as input, calculates the predicted probability from that duration, then subtracts the desired probability, then this function will have an output of 0 at the desired value of duration. uniroot will find this value for you. If you wrap this process in a little function, it makes it very easy to use:
find_prob <- function(p) {
f <- function(v) {
predict(model1, type = 'response',
newdata = data.frame(duration = v, sex = 'Male')) - p
}
uniroot(f, interval = range(data$duration), tol = 1e-9)$root
}
So, for example, to find the duration that gives an 80% probability, we just do:
find_prob(0.8)
#> [1] 12.86089
To prove that this is the correct value, we can feed it directly into predict to see what the predicted probability will be given sex = male and duration = 12.86089
predict(model1, type = 'response',
newdata = data.frame(sex = 'Male', duration = find_prob(0.8)))
#> 1
#> 0.8
I have the following script
Posdef <- function (n, ev = runif(n, 0, 10))
{
Z <- matrix(ncol=n, rnorm(n^2))
decomp <- qr(Z)
Q <- qr.Q(decomp)
R <- qr.R(decomp)
d <- diag(R)
ph <- d / abs(d)
O <- Q %*% diag(ph)
Z <- t(O) %*% diag(ev) %*% O
return(Z)
}
Sigma <- Posdef(n = 11)
mu <- runif(11,0,10)
data <- as.data.frame(mvrnorm(n=1000, mu, Sigma))
data[data < 0] <- 0 #setting a floor#
data[data > 10] <- 10 #setting a ceiling#
names(data) = c('criteria_1', 'criteria_2', 'criteria_3', 'criteria_4', 'criteria_5',
'criteria_6', 'criteria_7', 'criteria_8', 'criteria_9', 'criteria_10',
'outcome')
data$outcome <- ifelse(data$outcome > 5, 1, 0)
data <- data[, sapply(data, is.numeric)]
maxValue <- as.numeric(apply (data, 2, max))
minValue <- as.numeric(apply (data, 2, min))
data_scaled <- as.data.frame(scale(data, center = minValue,
scale = maxValue-minValue))
ind <- sample (1:nrow(data_scaled), 600)
train <- data_scaled[ind,]
test <- data_scaled[-ind,]
model <- glm (formula =
outcome ~ criteria_1 + criteria_2 + criteria_3 + criteria_4 + criteria_5 +
criteria_6 + criteria_7 + criteria_8 + criteria_9 + criteria_10,
family = "binomial",
data = train)
summary (model)
predicted_model <- predict(model, test)
neural_model <- neuralnet(formula =
outcome ~ criteria_1 + criteria_2 + criteria_3 + criteria_4 + criteria_5 +
criteria_6 + criteria_7 + criteria_8 + criteria_9 + criteria_10,
hidden = c(2,2) ,
threshold = 0.01,
stepmax = 1e+07,
startweights = NULL,
rep = 1,
learningrate = NULL,
algorithm = "rprop+",
linear.output=FALSE,
data= train)
plot (neural_model)
results <- compute (neural_model, test[1:10])
results <- results$net.result*(max(data$outcome)-
min(data$outcome))+ min(data$outcome)
Values <- (test$outcome)*(max(data$outcome)-
min(data$outcome)) + min(data$outcome)
MSE_nueral_model <- sum((results - Values)^2)/nrow(test)
MSE_model <- sum((predicted_model - test$outcome)^2)/nrow(test)
print(MSE_model - MSE_nueral_model)
R1 <- (MSE_model - MSE_nueral_model)
The purpose of this script is to generate some arbitrary multivariate distribution and then compare two methods. In this case its a neural net and logistic regression. The end result is a difference in mean square error.
Now my issue with creating a loop has been with generating the 1000 observations.
I am able to create a loop without the data simulation portion of the script, putting that into the loop seems to make things go haywire. I tried creating a column vector filled with NA's but all I ended up getting was a single value returned rather than a vector of length n populated by the MSE reductions for each iteration of the loop.
Any help would be greatly appreciated.
I am trying to fit a crossed non-linear random effect model as the linear random effect models as mentioned in this question and in this mailing list post using the nlme package. Though, I get an error regardless of what I try. Here is an example
library(nlme)
#####
# simulate data
set.seed(18112003)
na <- 30
nb <- 30
sigma_a <- 1
sigma_b <- .5
sigma_res <- .33
n <- na*nb
a <- gl(na,1,n)
b <- gl(nb,na,n)
u <- gl(1,1,n)
x <- runif(n, -3, 3)
y_no_noise <- x + sin(2 * x)
y <-
x + sin(2 * x) +
rnorm(na, sd = sigma_a)[as.integer(a)] +
rnorm(nb, sd = sigma_b)[as.integer(b)] +
rnorm(n, sd = sigma_res)
#####
# works in the linear model where we know the true parameter
fit <- lme(
# somehow we found the right values
y ~ x + sin(2 * x),
random = list(u = pdBlocked(list(pdIdent(~ a - 1), pdIdent(~ b - 1)))))
vv <- VarCorr(fit)
vv2 <- vv[c("a1", "b1"), ]
storage.mode(vv2) <- "numeric"
print(vv2,digits=4)
#R Variance StdDev
#R a1 1.016 1.0082
#R b1 0.221 0.4701
#####
# now try to do the same with `nlme`
fit <- nlme(
y ~ c0 + sin(c1),
fixed = list(c0 ~ x, c1 ~ x - 1),
random = list(u = pdBlocked(list(pdIdent(~ a - 1), pdIdent(~ b - 1)))),
start = c(0, 0.5, 1))
#R Error in nlme.formula(y ~ a * x + sin(b * x), fixed = list(a ~ 1, b ~ :
#R 'random' must be a formula or list of formulae
The lme example is similar to the one page 163-166 of "Mixed-effects Models in S and S-PLUS" with only 2 random effects instead of 3.
I should haved used a two-sided formula as written in help("nlme")
fit <- nlme(
y ~ c0 + c1 + sin(c2),
fixed = list(c0 ~ 1, c1 ~ x - 1, c2 ~ x - 1),
random = list(u = pdBlocked(list(pdIdent(c0 ~ a - 1), pdIdent(c1 ~ b - 1)))),
start = c(0, 0.5, 1))
# fixed effects estimates
fixef(fit)
#R c0.(Intercept) c1.x c2.x
#R -0.1788218 0.9956076 2.0022338
# covariance estimates
vv <- VarCorr(fit)
vv2 <- vv[c("c0.a1", "c1.b1"), ]
storage.mode(vv2) <- "numeric"
print(vv2,digits=4)
#R Variance StdDev
#R c0.a1 0.9884 0.9942
#R c1.b1 0.2197 0.4688
I am using sjPlot, the sjp.int function, to plot an interaction of an lme.
The options for the moderator values are means +/- sd, quartiles, all, max/min. Is there a way to plot the mean +/- 2sd?
Typically it would be like this:
model <- lme(outcome ~ var1+var2*time, random=~1|ID, data=mydata, na.action="na.omit")
sjp.int(model, show.ci=T, mdrt.values="meansd")
Many thanks
Reproducible example:
#create data
mydata <- data.frame( SID=sample(1:150,400,replace=TRUE),age=sample(50:70,400,replace=TRUE), sex=sample(c("Male","Female"),200, replace=TRUE),time= seq(0.7, 6.2, length.out=400), Vol =rnorm(400),HCD =rnorm(400))
mydata$time <- as.numeric(mydata$time)
#insert random NAs
NAins <- NAinsert <- function(df, prop = .1){
n <- nrow(df)
m <- ncol(df)
num.to.na <- ceiling(prop*n*m)
id <- sample(0:(m*n-1), num.to.na, replace = FALSE)
rows <- id %/% m + 1
cols <- id %% m + 1
sapply(seq(num.to.na), function(x){
df[rows[x], cols[x]] <<- NA
}
)
return(df)
}
mydata2 <- NAins(mydata,0.1)
#run the lme which gives error message
model = lme(Vol ~ age+sex*time+time* HCD, random=~time|SID,na.action="na.omit",data=mydata2);summary(model)
mydf <- ggpredict(model, terms=c("time","HCD [-2.5, -0.5, 2.0]"))
#lmer works
model2 = lmer(Vol ~ age+sex*time+time* HCD+(time|SID),control=lmerControl(check.nobs.vs.nlev = "ignore",check.nobs.vs.rankZ = "ignore", check.nobs.vs.nRE="ignore"), na.action="na.omit",data=mydata2);summary(model)
mydf <- ggpredict(model2, terms=c("time","HCD [-2.5, -0.5, 2.0]"))
#plotting gives problems (jittered lines)
plot(mydf)
With sjPlot, it's currently not possible. However, I have written a package especially dedicated to compute and plot marginal effects: ggeffects. This package is a bit more flexible (for marginal effects plots).
In the ggeffects-package, there's a ggpredict()-function, where you can compute marginal effects at specific values. Once you know the sd of your model term in question, you can specify these values in the function call to plot your interaction:
library(ggeffects)
# plot interaction for time and var2, for values
# 10, 30 and 50 of var2
mydf <- ggpredict(model, terms = c("time", "var2 [10,30,50]"))
plot(mydf)
There are some examples in the package-vignette, see especially this section.
Edit
Here are the results, based on your reproducible example (note that GitHub-Version is currently required!):
# requires at least the GitHub-Versiob 0.1.0.9000!
library(ggeffects)
library(nlme)
library(lme4)
library(glmmTMB)
#create data
mydata <-
data.frame(
SID = sample(1:150, 400, replace = TRUE),
age = sample(50:70, 400, replace = TRUE),
sex = sample(c("Male", "Female"), 200, replace = TRUE),
time = seq(0.7, 6.2, length.out = 400),
Vol = rnorm(400),
HCD = rnorm(400)
)
mydata$time <- as.numeric(mydata$time)
#insert random NAs
NAins <- NAinsert <- function(df, prop = .1) {
n <- nrow(df)
m <- ncol(df)
num.to.na <- ceiling(prop * n * m)
id <- sample(0:(m * n - 1), num.to.na, replace = FALSE)
rows <- id %/% m + 1
cols <- id %% m + 1
sapply(seq(num.to.na), function(x) {
df[rows[x], cols[x]] <<- NA
})
return(df)
}
mydata2 <- NAins(mydata, 0.1)
# run the lme, works now
model = lme(
Vol ~ age + sex * time + time * HCD,
random = ~ time |
SID,
na.action = "na.omit",
data = mydata2
)
summary(model)
mydf <- ggpredict(model, terms = c("time", "HCD [-2.5, -0.5, 2.0]"))
plot(mydf)
lme-plot
# lmer also works
model2 <- lmer(
Vol ~ age + sex * time + time * HCD + (time |
SID),
control = lmerControl(
check.nobs.vs.nlev = "ignore",
check.nobs.vs.rankZ = "ignore",
check.nobs.vs.nRE = "ignore"
),
na.action = "na.omit",
data = mydata2
)
summary(model)
mydf <- ggpredict(model2, terms = c("time", "HCD [-2.5, -0.5, 2.0]"), ci.lvl = NA)
# plotting works, but only w/o CI
plot(mydf)
lmer-plot
# lmer also works
model3 <- glmmTMB(
Vol ~ age + sex * time + time * HCD + (time | SID),
data = mydata2
)
summary(model)
mydf <- ggpredict(model3, terms = c("time", "HCD [-2.5, -0.5, 2.0]"))
plot(mydf)
plot(mydf, facets = T)
glmmTMB-plots