I am looking for a way to see if a value (stored in a dictionary) is in an array. The array looks like this: var array = Array<Dictionary<String, Int>>()
I looked around here, on stackoverflow, and found this: How to check if an element is in an array. Only problem is I can't seem to use the contains method by writing arrray.contains.... What am I doing wrong?
You can use contains within contains to check if any of the dictionaries in an array contains the value you are looking for.
For example, to search array a for the value 1:
let a: [[String: Int]] = [["a": 1, "b": 2], ["c": 3], ["d": 4]]
Swift 1.2:
if contains(a, {contains($0.values, 1)}) {
println("it is in there")
}
Swift 2.0:
/* Thanks to #Qbyte for providing this solution */
if a.contains({ $0.values.contains(1) }) {
print("it is in there")
}
This works because each member of array a gets evaluated with the closure {contains($0.values), 1} to see if it is true. The closure takes the dictionary it is passed, gets the array of values from it and then uses contains to see if the value is in there.
You can use this extension:
extension Array {
func contains<T where T : Equatable>(obj: T) -> Bool {
return self.filter({$0 as? T == obj}).count > 0
}
}
Which you can test like:
var array1 = Array<Dictionary<String, Int>>()
array1 = [["zero": 0],["one": 1], ["two": 2]]
array1.contains(["zero": 0]) //true
array1.contains(["five": 5]) //false
Related
This must be a really basic question. In languages like R you can take an array (swift syntax here)
let x = [1,2,3,4,5]
and extract multiple elements using an array of indices. That is I would like to be able to do something like say (now in a pseudo-Swift syntax because it does not parse)
x[[0,2,3]]
to get a return value of
[1,3,4]
but this does not work directly in Swift. Is there a standard way of doing this? I am currently using Swift4.
I'm not aware of anything built into the Swift Array class that does this.
One possible solution is to define an extension to Array that filters the array to only include the elements at the provided indices.
extension Array {
func elements(at indices: [Int]) -> Array<Element> {
return self.enumerated().filter { indices.contains($0.0) }.map { $0.1 }
}
}
Example usage:
let x = [1,2,3,4,5]
let res = x.elements(at: [0,2,3])
print(res)
Output:
[1, 3, 4]
I have an array of arrays containg int, so let pairs:[[Int]].
I am looking for an elegant way to extract similar elements.
For example my pairs variable could contain something like: [ [1,2], [4,6], [1,2] ]
I would like to extract any array that occurs more than once like [1,2].
In the example [ [1,2], [4,6], [1,2], [3,7], [4,6] ] I would like to extract both [1,2] and [4,6].
This seemed trivial at first, but every go I had at it became very cumbersome with many "helper arrays" and nested "for loops". There surely is a simpler way in Swift, right?
Thanks
Here the way using just one helper Dictionary and one loop:
let pairs = [[1,2], [4,6], [1,2]] // The original array
var pairCount = [NSArray : Int]() // This is helper dictionary. The key is the array, the value is - how much time it appears in the source array. I use NSArray because swift array does not conform to Hashable protocol.
var newPairs = [[Int]]()
for pair in pairs { // Iterate over our pairs
var count: Int = pairCount[pair as NSArray] ?? 0 // If we already have a pair in our dictionary get count of it in array otherwise het 0
count += 1 // increase counter
pairCount[pair as NSArray] = count // save in dictionary
}
let pairsWithCountMoreThanOne = pairCount.flatMap({$1 > 1 ? $0 : nil}) // here we get the result
for pair in pairsWithCountMoreThanOne { // Just print it
print("\(pair)")
}
This code may not be memory efficient for large arrays or large objects but it is really trivial.
Please check the below :
let pairs = [ [1,2], [4,6], [1,2], [3,7], [4,6], [3,5], [4,6] ]
var repeats = [[Int]]()
pairs.forEach { (i) in
let count = (pairs.filter{ return $0 == i }).count
if count > 1 {
if !repeats.contains(where: { (pair) -> Bool in
return pair == i
}) {
repeats.append(i)
}
}
}
print(repeats) // Output is : [[1, 2], [4, 6]]
This question already has answers here:
In Swift, an efficient function that separates an array into 2 arrays based on a predicate
(7 answers)
Closed 6 months ago.
Problem
Given an array of values how can I split it into sub-arrays made of elements that are equal?
Example
Given this array
let numbers = [1, 1, 1, 3, 3, 4]
I want this output
[[1,1,1], [3, 3], [4]]
What I am NOT looking for
A possible way of solving this would be creating some sort of index to indicate the occurrences of each element like this.
let indexes = [1:3, 3:2, 4:1]
And finally use the index to rebuild the output array.
let subsequences = indexes.sort { $0.0.0 < $0.1.0 }.reduce([Int]()) { (res, elm) -> [Int] in
return res + [Int](count: elm.1, repeatedValue: elm.0)
}
However with this solution I am losing the original values. Of course in this case it's not a big problem (an Int value is still and Inteven if recreated) but I would like to apply this solution to more complex data structures like this
struct Starship: Equatable {
let name: String
let warpSpeed: Int
}
func ==(left:Starship, right:Starship) -> Bool {
return left.warpSpeed == right.warpSpeed
}
Final considerations
The function I am looking for would be some kind of reverse of flatten(), infact
let subsequences: [[Int]] = [[1,1,1], [3, 3], [4]]
print(Array(subsequences.flatten())) // [1, 1, 1, 3, 3, 4]
I hope I made myself clear, let me know should you need further details.
// extract unique numbers using a set, then
// map sub-arrays of the original arrays with a filter on each distinct number
let numbers = [1, 1, 1, 3, 3, 4]
let numberGroups = Set(numbers).map{ value in return numbers.filter{$0==value} }
print(numberGroups)
[EDIT] changed to use Set Initializer as suggested by Hamish
[EDIT2] Swift 4 added an initializer to Dictionary that will do this more efficiently:
let numberGroups = Array(Dictionary(grouping:numbers){$0}.values)
For a list of objects to be grouped by one of their properties:
let objectGroups = Array(Dictionary(grouping:objects){$0.property}.values)
If you could use CocoaPods/Carthage/Swift Package Manager/etc. you could use packages like oisdk/SwiftSequence which provides the group() method:
numbers.lazy.group()
// should return a sequence that generates [1, 1, 1], [3, 3], [4].
or UsrNameu1/TraverSwift which provides groupBy:
groupBy(SequenceOf(numbers), ==)
If you don't want to add external dependencies, you could always write an algorithm like:
func group<S: SequenceType where S.Generator.Element: Equatable>(seq: S) -> [[S.Generator.Element]] {
var result: [[S.Generator.Element]] = []
var current: [S.Generator.Element] = []
for element in seq {
if current.isEmpty || element == current[0] {
current.append(element)
} else {
result.append(current)
current = [element]
}
}
result.append(current)
return result
}
group(numbers)
// returns [[1, 1, 1], [3, 3], [4]].
Let's assume that you have an unsorted array of items. You will need to sort the initial array then you will have something like this:
[1, 1, 1, 3, 3, 4]
After that you will initialize two arrays: one for storing arrays and another one to use it as a current array.
Loop through the initial array and:
if the current value isn't different from the last one, push it to the current array
otherwise push the current array to the first one then empty the current array.
Hope it helps!
Worth mentioning, using Swift Algorithms this is now a one-liner:
import Algorithms
let numbers = [1, 1, 1, 3, 3, 4]
let chunks: [[Int]] = numbers.chunked(by: ==).map { .init($0) }
print(chunks) // [[1, 1, 1], [3, 3], [4]]
I am kinda stumped on figuring this out. I want to populate an array with the string values that comes from a for-in loop.
Here's an example.
let names = ["Anna", "Alex", "Brian", "Jack"]
for x in names {
println(x)
}
The current x value would generate 4 string values (Anna, Alex, Brian, Jack).
However I need some advice in going about getting these four values back into an array. Thank you in advance.
Whatever is on the right side of a for - in expression must be a SequenceType. Array, as it happens, can be initialised with any SequenceType. So if you're just doing something like this:
var newArray: [String] = []
for value in exoticSequence {
newArray.append(value)
}
The same thing can be accomplished (faster), by doing this:
let newArray = Array(exoticSequence)
And it doesn't matter what type exoticSequence is: if the for-in loop worked, Array() will work.
However, if you're applying some kind of transformation to your exoticSequence, or you need some kind of side effect, .map() might be the way to go. .map() over any SequenceType can return an array. Again, this is faster, and more clear:
let exoticSequence = [1, 2, 3]
let newArray = exoticSequence.map {
value -> Int in
// You can put whatever would have been in your for-in loop here
print(value)
// In a way, the return statement will replace the append function
let whatYouWouldHaveAppended = value * 2
return whatYouWouldHaveAppended
}
newArray // [2, 4, 6]
And it's equivalent to:
let exoticSequence = [1, 2, 3]
var newArray: [Int] = []
for value in exoticSequence {
print(value)
let whatYouWouldHaveAppended = value * 2
newArray.append(whatYouWouldHaveAppended)
}
newArray // [2, 4, 6]
How can I easily add elements to an array inside a dictionary?
It's always complaining with could not find member 'append' or could not find an overload for '+='
var dict = Dictionary<String, Array<Int>>()
dict["key"] = [1, 2, 3]
// all of these fail
dict["key"] += 4
dict["key"].append(4) // xcode suggests dict["key"].?.append(4) which also fails
dict["key"]!.append(4)
dict["key"]?.append(4)
// however, I can do this:
var arr = dict["key"]!
arr.append(4) // this alone doesn't affect dict because it's a value type (and was copied)
dict["key"] = arr
if I just assign the array to a var, modify it and then reassign it to the dict, won't I be copying everything? that wouldn't be efficient nor elegant.
Swift beta 5 has added this functionality, and you've nailed the new method in a couple of your attempts. The unwrapping operators ! and ? now pass through the value to either operators or method calls. That is to say, you can add to that array in any of these ways:
dict["key"]! += [4]
dict["key"]!.append(4)
dict["key"]?.append(4)
As always, be careful about which operator you use -- force unwrapping a value that isn't in your dictionary will give you a runtime error:
dict["no-key"]! += [5] // CRASH!
Whereas using optional chaining will fail silently:
dict["no-key"]?.append(5) // Did it work? Swift won't tell you...
Ideally you'd be able to use the new null coalescing operator ?? to address this second case, but right now that's not working.
Answer from pre-Swift beta 5:
It's a quirk of Swift that it's not possible to do what you're trying to do. The issue is that the value of any Optional variable is in fact a constant -- even when forcibly unwrapping. If we just define an Optional array, here's what we can and can't do:
var arr: Array<Int>? = [1, 2, 3]
arr[0] = 5
// doesn't work: you can't subscript an optional variable
arr![0] = 5
// doesn't work: constant arrays don't allow changing contents
arr += 4
// doesn't work: you can't append to an optional variable
arr! += 4
arr!.append(4)
// these don't work: constant arrays can't have their length changed
The reason you're having trouble with the dictionary is that subscripting a dictionary returns an Optional value, since there's no guarantee that the dictionary will have that key. Therefore, an array in a dictionary has the same behavior as the Optional array, above:
var dict = Dictionary<String, Array<Int>>()
dict["key"] = [1, 2, 3]
dict["key"][0] = 5 // doesn't work
dict["key"]![0] = 5 // doesn't work
dict["key"] += 4 // uh uh
dict["key"]! += 4 // still no
dict["key"]!.append(4) // nope
If you need to change something in an array in the dictionary you'll need to get a copy of the array, change it, and reassign, like this:
if var arr = dict["key"] {
arr.append(4)
dict["key"] = arr
}
ETA: Same technique works in Swift beta 3, though constant arrays no longer allow changes to contents.
The accepted answer bypasses the following much simpler possibility, which also works for older Swift versions:
var dict = Dictionary<String, Array<Int>>()
dict["key"] = [1, 2, 3]
print(dict)
dict["key", default: [Int]()].append(4)
print(dict)
This will print:
["key": [1, 2, 3]]
["key": [1, 2, 3, 4]]
And this:
var dict = Dictionary<String, Array<Int>>()
dict["key", default: [Int]()].append(4)
print(dict)
will print:
["key": [4]]
As a simple workaround you can use a NSMutableArray:
import Foundation
var dict = Dictionary<String, NSMutableArray>()
dict["key"] = [1, 2, 3] as NSMutableArray
dict["key"]!.addObject(4)
I am using effectively such simple solution in my project:
https://github.com/gui-dos/Guigna/blob/5c02f7e70c8ee3b2265f6916c6cbbe5cd3963fb5/Guigna-Swift/Guigna/GuignaAppDelegate.swift#L1150-L1157
Here is what I was telling Nate Cook, in the comments for his quality answer. This is what I consider "easily [adding] elements to an array inside a dictionary":
dict["key"] = dict["key"]! + 4
dict["key"] = dict["key"] ? dict["key"]! + 4 : [4]
For now, we need to define the + operator ourselves.
#infix func +<T>(array: T[], element: T) -> T[] {
var copy = array
copy += element
return copy
}
I think this version removes too much safety; maybe define it with a compound operator?
#infix func +<T>(array: T[]?, element: T) -> T[] {
return array ? array! + element : [element]
}
dict["key"] = dict["key"] + 4
Finally, this is the cleanest I can get it, but I'm confused about how array values/references work in this example.
#assignment func +=<T>(inout array: T[]?, element: T) {
array = array + element
}
dict["key"] += 5
Use 'for in', for getting values from dictionary's inside array. Here's an example to help you
var a = ["x":["a","b","c"], "y":["d"]]
for b in a {
print(b)
}
Output:
("x", ["d"])
("y", ["a", "b", "c"])