options(digits=3)
set.seed(3)
n <- 20
sex <- factor(sample(c("m","f"), n, rep=TRUE))
age <- rnorm(n, 50, 5)
treatment <- factor(sample(c("Drug","Placebo"), n, rep=TRUE))
# Generate a 3-choice variable; each of 3 variables has 5 possible levels
symp <- c('Headache','Stomach Ache','Hangnail',
'Muscle Ache','Depressed')
symptom1 <- sample(symp, n, TRUE)
symptom2 <- sample(symp, n, TRUE)
symptom3 <- sample(symp, n, TRUE)
cbind(symptom1, symptom2, symptom3)[1:5,]
Symptoms <- mChoice(symptom1, symptom2, symptom3, label='Primary Symptoms')
Symptoms
print(Symptoms, long=TRUE)
format(Symptoms[1:5])
inmChoice(Symptoms,'Headache')
inmChoicelike(Symptoms, 'head', ignore.case=TRUE)
levels(Symptoms)
inmChoice(Symptoms, 3)
# Find all subjects with either of two symptoms
inmChoice(Symptoms, c('Headache','Hangnail'))
# Note: In this example, some subjects have the same symptom checked
# multiple times; in practice these redundant selections would be NAs
# mChoice will ignore these redundant selections
# Find all subjects with both symptoms
inmChoice(Symptoms, c('Headache', 'Hangnail'), condition='all')
meanage <- N <- numeric(5)
for(j in 1:5) {
meanage[j] <- mean(age[inmChoice(Symptoms,j)])
N[j] <- sum(inmChoice(Symptoms,j))
}
names(meanage) <- names(N) <- levels(Symptoms)
meanage
N
# Manually compute mean age for 2 symptoms
mean(age[symptom1=='Headache' | symptom2=='Headache' | symptom3=='Headache'])
mean(age[symptom1=='Hangnail' | symptom2=='Hangnail' | symptom3=='Hangnail'])
summary(Symptoms)
#Frequency table sex*treatment, sex*Symptoms
summary(sex ~ treatment + Symptoms, fun=table)
# Check:
ma <- inmChoice(Symptoms, 'Muscle Ache')
table(sex[ma])
# could also do:
# summary(sex ~ treatment + mChoice(symptom1,symptom2,symptom3), fun=table)
#Compute mean age, separately by 3 variables
summary(age ~ sex + treatment + Symptoms)
summary(age ~ sex + treatment + Symptoms, method="cross")
f <- summary(treatment ~ age + sex + Symptoms, method="reverse", test=TRUE)
f
# trio of numbers represent 25th, 50th, 75th percentile
print(f, long=TRUE)
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