Filter and Find

Part of the Logic & Flow section of Coddy's R journey. Lesson 34 of 64.

Filter(f, x) keeps the elements for which the predicate f returns TRUE, in their original order:

nums <- c(12, 5, 30, 7, 18)
print(Filter(function(n) n > 10, nums))
print(Filter(function(n) n %% 2 == 1, nums))

Output:

[1] 12 30 18
[1] 5 7

For a vector, logical indexing does the same job: nums[nums > 10]. Filter() is useful when the test is not vectorized, such as a check that splits a string or calls if:

words <- c("level", "tea", "noon", "cat")
is_palindrome <- function(w) w == paste(rev(strsplit(w, "")[[1]]), collapse = "")
print(Filter(is_palindrome, words))

Output:

[1] "level" "noon"

Find(f, x) returns the first element that passes, and Position(f, x) its position. When nothing passes, both return NULL or NA:

nums <- c(3, 8, 15, 4)
print(Find(function(n) n > 5, nums))
print(Position(function(n) n > 5, nums))
print(is.null(Find(function(n) n > 100, nums)))

Output:

[1] 8
[1] 2
[1] TRUE

Filter() also works on lists, for example to keep the records that meet a condition:

people <- list(list(name = "Ada", age = 36), list(name = "Tim", age = 12))
adults <- Filter(function(p) p$age >= 18, people)
print(length(adults))
print(adults[[1]]$name)

Output:

[1] 1
[1] "Ada"

Challenge

Easy

Complete valid_emails(emails). An address is valid when it contains exactly one @, has no spaces, and has at least one character before and after the @. Return Valid: followed by the number of valid addresses, then one line per valid address in order.

Count the @ characters with nchar(e) - nchar(gsub("@", "", e, fixed = TRUE)).

The supplied code reads the addresses as a comma-separated line and prints each returned line.

Try it yourself

valid_emails <- function(emails) {
  # Write your code here
  "Valid: 0"
}

# Supplied input/output code: keep it as it is
input <- suppressWarnings(readLines(file("stdin")))
cat(valid_emails(strsplit(input[1], ",")[[1]]), sep = "\n")