Comprehension

0.0.3

List Comprehension For Swift
dduan/Comprehension

What's New

0.0.3

2018-10-03T06:06:59Z

Swift 4.2 support

Comprehension

This library provides List Comprehension for Swift. Like so:

let numbers = [1, 2, 3]
let letters = "ab"

// produces ["a1", "a2", "a3", "b1", "b2", "b3"]
Array(letters, numbers) { "\($0)\($1)" }

// produces ["a1", "a3", "b1", "b3"]
Array(letters, numbers, where: { $1 % 2 != 0 }) { "\($0)\($1)" }

Installation

Swift Package Manager

.package(url: "git@github.com:dduan/Comprehension.git", .branch("master")),

(and "Comprehension" to the target's dependencies.)

CocoaPods

use_frameworks!

pod "Comprehension"

Carthage

github "dduan/Comprehension"

Direct Embed

Include Sources/Comprehension/Comprehension.swift in your project.

What's List Comprehension?

List comprehension is a powerful syntax for processing lists in languages such as Haskell, Python, etc. It's particularly handy when you need to deal with combinations of each elements in multiple lists (their Cartesian product).

The origin of list comprehension is often attribute to set comprehension in mathematics, whereas

{2·x|x ∈ ℕ, x ≤ 10}

… means "take all integers (element of set ℕ) that are less that 10, multiply each by 2 and form a new set with the results."

Here's an example of how list comprehension works in Haskell:

ghci> [x+y | x <- [1,2,3], y <- [10,100,1000]]
[11,101,1001,12,102,1002,13,103,1003]

Here, integers from one list is sequentially paired with that from the other list, resulting in 3 * 3 = 9 pairs. The sum of each pair forms the resulting list of integers.

Comprehension on a single list is equivalent of list.filter(f).map(g). When input is 2 lists:

// Comprehension syntax
let result = Array(list0, list1, where: f) { g($0, $1) }

… is equivalent of …

var result = [TypeOfElementInList0, TypeOfElementInList1]()
for e0 in list0 {
    for e1 in list1 {
        if f(e0, e1) {
            result.append(g(e0, e1))
        }
    }
}
// The fact that `result` is a var is not equivalent, but 🤷‍♀️.

… or in functional style …

let result = list0
    .map { e0 in
        list1.map { e1 in
            (e0, e1)
        }
    }
    .joined()
    .filter(f)
    .map(g)

Note how the list comprehension version is much more concise above. And imagine expanding the example to 3, 4, 5 lists!

License

MIT, see LICENSE.

Description

  • Swift Tools 4.2.0
View More Packages from this Author

Dependencies

  • None
Last updated: Mon Aug 29 2022 05:34:38 GMT-0500 (GMT-05:00)