5 Control flow
5.1 Choices (Exercises 5.2.4)
Q1. What type of vector does each of the following calls to ifelse() return?
Read the documentation and write down the rules in your own words.
A1. Here are the rules about what a call to ifelse() might return:
- It is type unstable, i.e. the type of return will depend on the type of which condition is true (
yesorno, i.e.):
ifelse(TRUE, 1, "no") # `numeric` returned
#> [1] 1
ifelse(FALSE, 1, "no") # `character` returned
#> [1] "no"- It works only for cases where
testargument evaluates to alogicaltype:
- If
testis argument is of logical type, butNA, it will returnNA:
ifelse(NA, 1, "no")
#> [1] NA- If the
testargument doesn’t resolve tologicaltype, it will try to coerce the output to alogicaltype:
# will work
ifelse("TRUE", 1, "no")
#> [1] 1
ifelse("false", 1, "no")
#> [1] "no"
# won't work
ifelse("tRuE", 1, "no")
#> [1] NA
ifelse(NaN, 1, "no")
#> [1] NAThis is also clarified in the docs for this function:
A vector of the same length and attributes (including dimensions and
"class") astestand data values from the values ofyesorno. The mode of the answer will be coerced from logical to accommodate first any values taken from yes and then any values taken fromno.
Q2. Why does the following code work?
x <- 1:10
if (length(x)) "not empty" else "empty"
#> [1] "not empty"
x <- numeric()
if (length(x)) "not empty" else "empty"
#> [1] "empty"A2. The code works because the conditional expressions in if() - even though of numeric type - can be successfully coerced to a logical type.
as.logical(length(1:10))
#> [1] TRUE
as.logical(length(numeric()))
#> [1] FALSE5.2 Loops (Exercises 5.3.3)
Q1. Why does this code succeed without errors or warnings?
A1. This works because 1:length(x) works in both positive and negative directions.
1:2
#> [1] 1 2
1:0
#> [1] 1 0
1:-3
#> [1] 1 0 -1 -2 -3In this case, since x is of length 0, i will go from 1 to 0.
Additionally, since out-of-bound (OOB) value for atomic vectors is NA, all related operations with OOB values will also produce NA.
x <- numeric()
out <- vector("list", length(x))
for (i in 1:length(x)) {
print(paste("i:", i, ", x[i]:", x[i], ", out[i]:", out[i]))
out[i] <- x[i]^2
}
#> [1] "i: 1 , x[i]: NA , out[i]: NULL"
#> [1] "i: 0 , x[i]: , out[i]: "
out
#> [[1]]
#> [1] NAA way to do avoid this unintended behavior is to use seq_along() instead:
x <- numeric()
out <- vector("list", length(x))
for (i in seq_along(x)) {
out[i] <- x[i]^2
}
out
#> list()Q2. When the following code is evaluated, what can you say about the vector being iterated?
A2. The iterator variable x initially takes all values of the vector xs. We can check this by printing x for each iteration:
xs <- c(1, 2, 3)
for (x in xs) {
cat("x:", x, "\n")
xs <- c(xs, x * 2)
cat("xs:", paste(xs), "\n")
}
#> x: 1
#> xs: 1 2 3 2
#> x: 2
#> xs: 1 2 3 2 4
#> x: 3
#> xs: 1 2 3 2 4 6It is worth noting that x is not updated after each iteration; otherwise, it will take increasingly bigger values of xs, and the loop will never end executing.
Q3. What does the following code tell you about when the index is updated?
for (i in 1:3) {
i <- i * 2
print(i)
}
#> [1] 2
#> [1] 4
#> [1] 6A3. In a for() loop the index is updated in the beginning of each iteration. Otherwise, we will encounter an infinite loop.
for (i in 1:3) {
cat("before: ", i, "\n")
i <- i * 2
cat("after: ", i, "\n")
}
#> before: 1
#> after: 2
#> before: 2
#> after: 4
#> before: 3
#> after: 6Also, worth contrasting the behavior of for() loop with that of while() loop:
5.3 Session information
sessioninfo::session_info(include_base = TRUE)
#> ─ Session info ───────────────────────────────────────────
#> setting value
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#> os Ubuntu 22.04.5 LTS
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#> ui X11
#> language (EN)
#> collate C.UTF-8
#> ctype C.UTF-8
#> tz UTC
#> date 2024-12-29
#> pandoc 3.6.1 @ /opt/hostedtoolcache/pandoc/3.6.1/x64/ (via rmarkdown)
#>
#> ─ Packages ───────────────────────────────────────────────
#> package * version date (UTC) lib source
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#> utils * 4.4.2 2024-10-31 [3] local
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#>
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#>
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