R - iteratively apply a function of a list of variables - r

Question

My goal is to create a function that, when looped over multiple variables of a data frame, will return a new data frame containing the percents and 95% confidence intervals for each level of each variable.
As an example, if I applied this function to "cyl" and "am" from the mtcars data frame, I would want this as the final result:
variable level ci.95
1 cyl 4 34.38 (19.50, 53.11)
2 cyl 6 21.88 (10.35, 40.45)
3 cyl 8 43.75 (27.10, 61.94)
4 am 0 59.38 (40.94, 75.5)
5 am 1 40.62 (24.50, 59.06)
So, far I have function that seems to work for a single variable; however, I have two issues that I'm hoping the community can help me with:
General R-ifying my code. I'm still an R novice. I've read enough posts to know that R enthusiasts generally discourage using for loops, but I still really struggle with using the apply functions (which seems to be the alternative to for loops in most cases).
Applying this function to a list of variables - resulting in a single data frame containing the returned values from the function for each level of each variable.
Here's where I'm at with my code so far:
t1.props <- function(x, data = NULL) {
# Grab dataframe and/or variable name
if(!missing(data)){
var <- data[,deparse(substitute(x))]
} else {
var <- x
}
# Grab variable name for use in ouput
var.name <- substitute(x)
# Omit observations with missing data
var.clean <- na.omit(var)
# Number of nonmissing observations
n <- length(var.clean)
# Grab levels of variable
levels <- sort(unique(var.clean))
# Create an empty data frame to store values
out <- data.frame(variable = NA,
level = NA,
ci.95 = NA)
# Estimate prop, se, and ci for each level of the variable
for(i in seq_along(levels)) {
prop <- paste0("prop", i)
se <- paste0("se", i)
log.prop <- paste0("log.trans", i)
log.se <- paste0("log.se", i)
log.l <- paste0("log.l", i)
log.u <- paste0("log.u", i)
lcl <- paste0("lcl", i)
ucl <- paste0("ucl", i)
# Find the proportion for each level of the variable
assign(prop, sum(var.clean == levels[i]) / n)
# Find the standard error for each level of the variable
assign(se, sd(var.clean == levels[i]) /
sqrt(length(var.clean == levels[i])))
# Perform a logit transformation of the original percentage estimate
assign(log.prop, log(get(prop)) - log(1 - get(prop)))
# Transform the standard error of the percentage to a standard error of its
# logit transformation
assign(log.se, get(se) / (get(prop) * (1 - get(prop))))
# Calculate the lower and upper confidence bounds of the logit
# transformation
assign(log.l,
get(log.prop) -
qt(.975, (length(var.clean == levels[i]) - 1)) * get(log.se))
assign(log.u,
get(log.prop) +
qt(.975, (length(var.clean == levels[i]) - 1)) * get(log.se))
# Finally, perform inverse logit transformations to get the confidence bounds
assign(lcl, exp(get(log.l)) / (1 + exp(get(log.l))))
assign(ucl, exp(get(log.u)) / (1 + exp(get(log.u))))
# Create a combined 95% CI variable for easy copy/paste into Word tables
ci.95 <- paste0(round(get(prop) * 100, 2), " ",
"(", sprintf("%.2f", round(get(lcl) * 100, 2)), ",", " ",
round(get(ucl) * 100, 2), ")")
# Populate the "out" data frame with values
out <- rbind(out, c(as.character(var.name), levels[i], ci.95))
}
# Remove first (empty) row from out
# But only in the first iteration
if (is.na(out[1,1])) {
out <- out[-1, ]
rownames(out) <- 1:nrow(out)
}
out
}
data(mtcars)
t1.props(cyl, mtcars)
I appreciate any help or advice you have to offer.

Answer 1

You can also keep the function mainly intact and use lapply over it:
vars <- c("cyl", "am")
lapply(vars, t1.props, data=mtcars)
[[1]]
variable level ci.95
1 cyl 4 34.38 (19.50, 53.11)
2 cyl 6 21.88 (10.35, 40.45)
3 cyl 8 43.75 (27.10, 61.94)
[[2]]
variable level ci.95
1 am 0 59.38 (40.94, 75.5)
2 am 1 40.62 (24.50, 59.06)
And combine them all into one data frame with:
lst <- lapply(vars, t1.props, data=mtcars)
do.call(rbind,lst)
Data
You must simplify the var and var.name assignments to:
t1.props <- function(x, data = NULL) {
# Grab dataframe and/or variable name
if(!missing(data)){
var <- data[,x]
} else {
var <- x
}
# Grab variable name for use in ouput
var.name <- x
# Omit observations with missing data
var.clean <- na.omit(var)
# Number of nonmissing observations
n <- length(var.clean)
# Grab levels of variable
levels <- sort(unique(var.clean))
# Create an empty data frame to store values
out <- data.frame(variable = NA,
level = NA,
ci.95 = NA)
# Estimate prop, se, and ci for each level of the variable
for(i in seq_along(levels)) {
prop <- paste0("prop", i)
se <- paste0("se", i)
log.prop <- paste0("log.trans", i)
log.se <- paste0("log.se", i)
log.l <- paste0("log.l", i)
log.u <- paste0("log.u", i)
lcl <- paste0("lcl", i)
ucl <- paste0("ucl", i)
# Find the proportion for each level of the variable
assign(prop, sum(var.clean == levels[i]) / n)
# Find the standard error for each level of the variable
assign(se, sd(var.clean == levels[i]) /
sqrt(length(var.clean == levels[i])))
# Perform a logit transformation of the original percentage estimate
assign(log.prop, log(get(prop)) - log(1 - get(prop)))
# Transform the standard error of the percentage to a standard error of its
# logit transformation
assign(log.se, get(se) / (get(prop) * (1 - get(prop))))
# Calculate the lower and upper confidence bounds of the logit
# transformation
assign(log.l,
get(log.prop) -
qt(.975, (length(var.clean == levels[i]) - 1)) * get(log.se))
assign(log.u,
get(log.prop) +
qt(.975, (length(var.clean == levels[i]) - 1)) * get(log.se))
# Finally, perform inverse logit transformations to get the confidence bounds
assign(lcl, exp(get(log.l)) / (1 + exp(get(log.l))))
assign(ucl, exp(get(log.u)) / (1 + exp(get(log.u))))
# Create a combined 95% CI variable for easy copy/paste into Word tables
ci.95 <- paste0(round(get(prop) * 100, 2), " ",
"(", sprintf("%.2f", round(get(lcl) * 100, 2)), ",", " ",
round(get(ucl) * 100, 2), ")")
# Populate the "out" data frame with values
out <- rbind(out, c(as.character(var.name), levels[i], ci.95))
}
# Remove first (empty) row from out
# But only in the first iteration
if (is.na(out[1,1])) {
out <- out[-1, ]
rownames(out) <- 1:nrow(out)
}
out
}

Answer 2

The nice thing about all the functions you're using is that they are already vectorized (except sd and qt, but you can easily vectorize them for specific arguments with Vectorize). This means you can pass vectors to them without needing to write a single loop. I left out the parts of your function that deal with preparing the input and prettying up the output.
t1.props <- function(var, data=mtcars) {
N <- nrow(data)
levels <- names(table(data[,var]))
count <- unclass(table(data[,var])) # counts
prop <- count / N # proportions
se <- sqrt(prop * (1-prop)/(N-1)) # standard errors of props.
lprop <- log(prop) - log(1-prop) # logged prop
lse <- se / (prop*(1-prop)) # logged se
stat <- Vectorize(qt, "df")(0.975, N-1) # tstats
llower <- lprop - stat*lse # log lower
lupper <- lprop + stat*lse # log upper
lower <- exp(llower) / (1 + exp(llower)) # lower ci
upper <- exp(lupper) / (1 + exp(lupper)) # upper ci
data.frame(variable=var,
level=levels,
perc=100*prop,
lower=100*lower,
upper=100*upper)
}
So, the only explicit applying/looping comes when you apply the function to multiple variables as follows
## Apply your function to two variables
do.call(rbind, lapply(c("cyl", "am"), t1.props))
# variable level perc lower upper
# 4 cyl 4 34.375 19.49961 53.11130
# 6 cyl 6 21.875 10.34883 40.44691
# 8 cyl 8 43.750 27.09672 61.94211
# 0 am 0 59.375 40.94225 75.49765
# 1 am 1 40.625 24.50235 59.05775
As far as the loop in your code, it's not like that is particularly important in terms of efficiency, but you can see how much easier code can be to read when its concise - and apply functions offer a lot of simple one-line solutions.
I think the most important thing to change in your code is the use of assign and get. Instead, you can store variables in lists or another data structure, and use setNames, names<-, or names(...) <- to name the components when needed.