How can I find the maximum output of a function - r

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

Related

Find value of covariate given a probability in R

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

Using `cor.test()` on ranked data

I would like to do a Spearman correlation test using rank data. How can I do this with cor.test()? I don't want the function to rerank the data.
Additionally, what form does the data need to be in? From the help, it seems to be the raw data as compared to a correlation matrix.
Consider this example
## Hollander & Wolfe (1973), p. 187f.
## Assessment of tuna quality. We compare the Hunter L measure of
## lightness to the averages of consumer panel scores (recoded as
## integer values from 1 to 6 and averaged over 80 such values) in
## 9 lots of canned tuna.
library(tidyverse)
A <- tibble(
x = c(44.4, 45.9, 41.9, 53.3, 44.7, 44.1, 50.7, 45.2, 60.1),
y = c( 2.6, 3.1, 2.5, 5.0, 3.6, 4.0, 5.2, 2.8, 3.8)
) %>%
mutate(rank_x = rank(x),
rank_y = rank(y)
)
Spearman's correlation coefficient is defined as Pearson's correlation between ranked variables
cor(A$x, A$y, method = "spearman")
#[1] 0.6
cor(A$rank_x, A$rank_y, method = "pearson")
#[1] 0.6
what about cor.test()? Can I use the rank data as its input?
x1 <- cor.test(A$x, A$y, method = "spearman")
x1
# Spearman's rank correlation rho
#
# data: A$x and A$y
# S = 48, p-value = 0.1
# alternative hypothesis: true rho is not equal to 0
# sample estimates:
# rho
# 0.6
x2 <- cor.test(A$rank_x, A$rank_y, method = "pearson")
x2
# Pearson's product-moment correlation
# data: A$rank_x and A$rank_y
# t = 2, df = 7, p-value = 0.09
# alternative hypothesis: true correlation is not equal to 0
# 95 percent confidence interval:
# -0.11 0.90
# sample estimates:
# cor
# 0.6
x3 <- cor.test(A$rank_x, A$rank_y, method = "spearman")
# Spearman's rank correlation rho
#
# data: A$rank_x and A$rank_y
# S = 48, p-value = 0.1
# alternative hypothesis: true rho is not equal to 0
# sample estimates:
# rho
# 0.6
Yes, you should use method = Spearman for ranked or original data. If rank data is used, the data is not reranked in the function.
As the help file implies, using method=Pearson with rank data conducts a Pearson's correlation test on the ranks, which would follow a t-distribution. However, since the ranks are not continuous variables, this approach is not correct.
getAnywhere(cor.test.default)
A single object matching ‘cor.test.default’ was found
It was found in the following places
registered S3 method for cor.test from namespace stats
namespace:stats
with value
function (x, y, alternative = c("two.sided", "less",
"greater"), method = c("pearson", "kendall",
"spearman"), exact = NULL, conf.level = 0.95, continuity = FALSE,
...)
{
alternative <- match.arg(alternative)
method <- match.arg(method)
DNAME <- paste(deparse1(substitute(x)), "and", deparse1(substitute(y)))
if (!is.numeric(x))
stop("'x' must be a numeric vector")
if (!is.numeric(y))
stop("'y' must be a numeric vector")
if (length(x) != length(y))
stop("'x' and 'y' must have the same length")
OK <- complete.cases(x, y)
x <- x[OK]
y <- y[OK]
n <- length(x)
NVAL <- 0
conf.int <- FALSE
if (method == "pearson") {
if (n < 3L)
stop("not enough finite observations")
method <- "Pearson's product-moment correlation"
names(NVAL) <- "correlation"
r <- cor(x, y)
df <- n - 2L
ESTIMATE <- c(cor = r)
PARAMETER <- c(df = df)
STATISTIC <- c(t = sqrt(df) * r/sqrt(1 - r^2))
if (n > 3) {
if (!missing(conf.level) && (length(conf.level) !=
1 || !is.finite(conf.level) || conf.level < 0 ||
conf.level > 1))
stop("'conf.level' must be a single number between 0 and 1")
conf.int <- TRUE
z <- atanh(r)
sigma <- 1/sqrt(n - 3)
cint <- switch(alternative, less = c(-Inf, z + sigma *
qnorm(conf.level)), greater = c(z - sigma * qnorm(conf.level),
Inf), two.sided = z + c(-1, 1) * sigma * qnorm((1 +
conf.level)/2))
cint <- tanh(cint)
attr(cint, "conf.level") <- conf.level
}
PVAL <- switch(alternative, less = pt(STATISTIC, df),
greater = pt(STATISTIC, df, lower.tail = FALSE),
two.sided = 2 * min(pt(STATISTIC, df), pt(STATISTIC,
df, lower.tail = FALSE)))
}
else {
if (n < 2)
stop("not enough finite observations")
PARAMETER <- NULL
TIES <- (min(length(unique(x)), length(unique(y))) <
n)
if (method == "kendall") {
method <- "Kendall's rank correlation tau"
names(NVAL) <- "tau"
r <- cor(x, y, method = "kendall")
ESTIMATE <- c(tau = r)
if (!is.finite(ESTIMATE)) {
ESTIMATE[] <- NA
STATISTIC <- c(T = NA)
PVAL <- NA
}
else {
if (is.null(exact))
exact <- (n < 50)
if (exact && !TIES) {
q <- round((r + 1) * n * (n - 1)/4)
STATISTIC <- c(T = q)
pkendall <- function(q, n) .Call(C_pKendall,
q, n)
PVAL <- switch(alternative, two.sided = {
if (q > n * (n - 1)/4) p <- 1 - pkendall(q -
1, n) else p <- pkendall(q, n)
min(2 * p, 1)
}, greater = 1 - pkendall(q - 1, n), less = pkendall(q,
n))
}
else {
xties <- table(x[duplicated(x)]) + 1
yties <- table(y[duplicated(y)]) + 1
T0 <- n * (n - 1)/2
T1 <- sum(xties * (xties - 1))/2
T2 <- sum(yties * (yties - 1))/2
S <- r * sqrt((T0 - T1) * (T0 - T2))
v0 <- n * (n - 1) * (2 * n + 5)
vt <- sum(xties * (xties - 1) * (2 * xties +
5))
vu <- sum(yties * (yties - 1) * (2 * yties +
5))
v1 <- sum(xties * (xties - 1)) * sum(yties *
(yties - 1))
v2 <- sum(xties * (xties - 1) * (xties - 2)) *
sum(yties * (yties - 1) * (yties - 2))
var_S <- (v0 - vt - vu)/18 + v1/(2 * n * (n -
1)) + v2/(9 * n * (n - 1) * (n - 2))
if (exact && TIES)
warning("Cannot compute exact p-value with ties")
if (continuity)
S <- sign(S) * (abs(S) - 1)
STATISTIC <- c(z = S/sqrt(var_S))
PVAL <- switch(alternative, less = pnorm(STATISTIC),
greater = pnorm(STATISTIC, lower.tail = FALSE),
two.sided = 2 * min(pnorm(STATISTIC), pnorm(STATISTIC,
lower.tail = FALSE)))
}
}
}
else {
method <- "Spearman's rank correlation rho"
if (is.null(exact))
exact <- TRUE
names(NVAL) <- "rho"
r <- cor(rank(x), rank(y))
ESTIMATE <- c(rho = r)
if (!is.finite(ESTIMATE)) {
ESTIMATE[] <- NA
STATISTIC <- c(S = NA)
PVAL <- NA
}
else {
pspearman <- function(q, n, lower.tail = TRUE) {
if (n <= 1290 && exact)
.Call(C_pRho, round(q) + 2 * lower.tail,
n, lower.tail)
else {
den <- (n * (n^2 - 1))/6
if (continuity)
den <- den + 1
r <- 1 - q/den
pt(r/sqrt((1 - r^2)/(n - 2)), df = n - 2,
lower.tail = !lower.tail)
}
}
q <- (n^3 - n) * (1 - r)/6
STATISTIC <- c(S = q)
if (TIES && exact) {
exact <- FALSE
warning("Cannot compute exact p-value with ties")
}
PVAL <- switch(alternative, two.sided = {
p <- if (q > (n^3 - n)/6) pspearman(q, n, lower.tail = FALSE) else pspearman(q,
n, lower.tail = TRUE)
min(2 * p, 1)
}, greater = pspearman(q, n, lower.tail = TRUE),
less = pspearman(q, n, lower.tail = FALSE))
}
}
}
RVAL <- list(statistic = STATISTIC, parameter = PARAMETER,
p.value = as.numeric(PVAL), estimate = ESTIMATE, null.value = NVAL,
alternative = alternative, method = method, data.name = DNAME)
if (conf.int)
RVAL <- c(RVAL, list(conf.int = cint))
class(RVAL) <- "htest"
RVAL
}
<bytecode: 0x0000018603fa9418>
<environment: namespace:stats>

KNN outlier detection in R

I am trying to run a script I was given to perform outlier detection using a weighted KNN outlier score, but keep getting the following error:
Error in apply(kNNdist(x = dat, k = k), 1, mean) :
dim(X) must have a positive length
The script I am trying to run is as below. It is a single block of script, but I have added a comment directly above the section of the script that is causing the error, which is the function:
WKNN_Outlier <- apply(kNNdist(x=dat, k = k), 1, mean)
If anyone has any better or simpler ideas for unsupervised outlier detection, I am all ears (so to speak...)
library(dbscan)
library(ggplot2)
set.seed(0)
x11 <- rnorm(n = 100, mean = 10, sd = 1) # Cluster 1 (x1 coordinate)
x21 <- rnorm(n = 100, mean = 10, sd = 1) # Cluster 1 (x2 coordinate)
x12 <- rnorm(n = 100, mean = 20, sd = 1) # Cluster 2 (x1 coordinate)
x22 <- rnorm(n = 100, mean = 10, sd = 1) # Cluster 2 (x2 coordinate)
x13 <- rnorm(n = 100, mean = 15, sd = 3) # Cluster 3 (x1 coordinate)
x23 <- rnorm(n = 100, mean = 25, sd = 3) # Cluster 3 (x2 coordinate)
x14 <- rnorm(n = 50, mean = 25, sd = 1) # Cluster 4 (x1 coordinate)
x24 <- rnorm(n = 50, mean = 25, sd = 1) # Cluster 4 (x2 coordinate)
dat <- data.frame(x1 = c(x11,x12,x13,x14), x2 = c(x21,x22,x23,x24))
( g0a <- ggplot() + geom_point(data=dat, mapping=aes(x=x1, y=x2), shape = 19) )
k <- 4 # KNN parameter
top_n <- 20 # No. of top outliers to be displayed
KNN_Outlier <- kNNdist(x=dat, k = k)
rank_KNN_Outlier <- order(x=KNN_Outlier, decreasing = TRUE) # Sorting (descending)
KNN_Result <- data.frame(ID = rank_KNN_Outlier, score = KNN_Outlier[rank_KNN_Outlier])
head(KNN_Result, top_n)
graph <- g0a +
geom_point(data=dat[rank_KNN_Outlier[1:top_n],], mapping=aes(x=x1,y=x2), shape=19,
color="red", size=2) +
geom_text(data=dat[rank_KNN_Outlier[1:top_n],],
mapping=aes(x=(x1-0.5), y=x2, label=rank_KNN_Outlier[1:top_n]), size=2.5)
graph
## Use KNNdist() to calculate the weighted KNN outlier score
k <- 4 # KNN parameter
top_n <- 20 # No. of top outliers to be displayed
The WKNN_Outler function below is what is causing the error. From what I can gather, the apply function shouldn't be having any issues, as the data (dat) is converted into a data.frame, which should prevent the error, but doesn't.
WKNN_Outlier <- apply(kNNdist(x=dat, k = k), 1, mean) # Weighted KNN outlier score (mean)
rank_WKNN_Outlier <- order(x=WKNN_Outlier, decreasing = TRUE)
WKNN_Result <- data.frame(ID = rank_WKNN_Outlier, score = WKNN_Outlier[rank_WKNN_Outlier])
head(WKNN_Result, top_n)
ge1 <- g0a +
geom_point(data=dat[rank_WKNN_Outlier[1:top_n],], mapping=aes(x=x1,y=x2), shape=19,
color="red", size=2) +
geom_text(data=dat[rank_WKNN_Outlier[1:top_n],],
mapping=aes(x=(x1-0.5), y=x2, label=rank_WKNN_Outlier[1:top_n]), size=2.5)
ge1
The function kNNdist(x=dat, k = k) produces a vector not a matrix, which is why when you try to do the apply function it tells you dim(X) must have a positive length (vectors have a NULL dim).
Try:
WKNN_Outlier <- apply(kNNdist(x=dat, k = k, all=T), 1, mean)

Adapting the meansd moderator option in sjPlot interaction

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

How to change the position of axis values in nomogram, rms package?

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]

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