My problem is that this code is case-sensitive. If I have "Sam" and "sam", they will be sorted into different keys. Any way that I can think of doing this is by converting the string into all lowercase, but I want it to stay as normal while being sorted without case-sensitivity:
var dict: [String: [String]] = [:]
for string in array {
if (dict[string] != nil) {
dict[string]?.append(string)
}
else {
dict[string] = [string]
}
}
As it is right now my code would result in:
["Sam": ["Sam"], "sam", ["sam"]]
Instead of what I want:
["Sam": ["Sam", "sam"]]
How can I accomplish this?
You can use reduce(into:) method and assign each element capitalized to the result:
let array = ["Sam", "sam", "SAM"]
let dict: [String: [String]] = array.reduce(into: [:]) {
$0[$1.capitalized, default: []].append($1)
}
print(dict) // ["Sam": ["Sam", "sam", "SAM"]]
If you just want to have case insensitive keys and case sensitive values, from given array, the shortest solution could be something like this:
var dict: [String: [String]] = [:]
array.forEach { dict[$0.lowercased(), default: []] += [$0] }
I have a dictionary with this structure:
a: [1,2]
b: [3,4]
c: [5,6]
and I need to return a string with this structure.
a,b,c\n1,3,5\n2,4,6
I solved the first part of the string. But to get the rest of the String. I try to iterate into my dictionary to get the first elements for each key in my dictionary and then get the rest for each value into the array.
Is there an easier way to get this?
Once you know what's the order of the keys (alpha ?), you can use this:
let dict: [String: [Int]] = ["a": [1,2], "b": [3, 4], "c": [5, 6]]
let keys = dict.keys.sorted() //Or do whatever you want here to get your target order
var matrix: [[String]] = []
keys.forEach {
guard let arrayAsInt = dict[$0] else { return }
let arrayAsString = arrayAsInt.map{ "\($0)" }
matrix.append( [$0] + arrayAsString)
}
print("Matrix: \(matrix)")
let transposed = matrix.transposed()
print("Transposed Matrix: \(transposed)")
let output = transposed.map { $0.joined(separator: ",")}.joined(separator: "\n")
print(output)
The outputs:
$>Matrix: [["a", "1", "2"], ["b", "3", "4"], ["c", "5", "6"]]
$>Transposed Matrix: [["a", "b", "c"], ["1", "3", "5"], ["2", "4", "6"]]
$>a,b,c
1,3,5
2,4,6
Obvisouly the "\n" might be invisible and be an actual new line
a,b,c
1,3,5
2,4,6
Being
a,b,c\n1,3,5\n2,4,6
What's the idea behind that? Create a matrix and use the transpose (it's used in maths with matrix, it's one of the basic modification of a matrix).
First transform the [String: [Int]] into a [[String]], where each element would be key followed by its values. I transformed it there as String for simpler code later.
Why doing that? Because the matrix value is easy to get from your initial dict. the transposed value is harder (not impossible) to get from dict but easier from matrix, and the transposed is quickly transformed into your format.
So my thinking was the reverse:
Get a structure from your output, then how to get it, it's a transpose, so I need to get the initial input as it, etc.
With the help of a code for Transpose Matrix (that accept String elements).
extension Collection where Self.Iterator.Element: RandomAccessCollection {
// PRECONDITION: `self` must be rectangular, i.e. every row has equal size.
func transposed() -> [[Self.Iterator.Element.Iterator.Element]] {
guard let firstRow = self.first else { return [] }
return firstRow.indices.map { index in
self.map{ $0[index] }
}
}
}
Any code (there a various) working ones, should the trick. I took it from here.
As pointed by #Leo Dabus, you can remove the Self.Iterator.Element
from the extension code (twice). I just wanted to it as such, not modifying the initial answer since it's not mind.
What you are looking for, besides composing the final string, is how to transpose a collection (this would work with collections of different sizes as well):
extension Sequence {
func max<T: Comparable>(_ predicate: (Element) -> T) -> Element? {
self.max(by: { predicate($0) < predicate($1) })
}
}
extension Collection where Element: RandomAccessCollection, Element.Indices == Range<Int> {
func transposed() -> [[Element.Element]] {
(0..<(max(\.count)?.count ?? .zero)).map {
index in compactMap { $0.indices ~= index ? $0[index] : nil }
}
}
}
let dict = ["a": [1,2,3],
"b": [4,5,6],
"c": [7,8,9]]
let sorted = dict.sorted(by: {$0.key < $1.key})
let result = sorted.map(\.key).joined(separator: ",") + "\n" +
sorted.map(\.value).transposed().map {
$0.map(String.init).joined(separator: ",")
}.joined(separator: "\n")
result // "a,b,c\n1,4,7\n2,5,8\n3,6,9"
A dictionary is an unordered collection so you need to sort it according to any specific key. Here I sort the dictionary according to the key if you don't care about an order you can just remove sort.
let dict: [String: Any] = ["a": [1,2], "b": [3,4], "c": [5,6]]
let sortedKey = dict.keys.sorted(by: <)
let key = sortedKey.joined(separator: ",")
var firstValues: [String] = []
var secondValues: [String] = []
sortedKey.forEach { (key) in
if let dictValue = dict[key] as? [Int],
let firstValue = dictValue.first,
let secondValue = dictValue.last {
firstValues.append("\(firstValue)")
secondValues.append("\(secondValue)")
}
}
let finalString = key + "\n" + firstValues.joined(separator: ",") + "\n" + secondValues.joined(separator: ",")
print(finalString) // "a,b,c\n1,3,5\n2,4,6"
In Swift 4 I could the code below worked but in Swift 5 I get the following error: Type 'Dictionary<String, String>.Values.Iterator' does not conform to protocol 'Sequence'
guard let userIds = users.values.makeIterator() else { return }
for userId in userIds {
// User setup
}
What is the right way in Swift 5 now?
let dictionary: [String: Int] = ["a": 1, "b": 2]
for (key, value) in dictionary {
print(key, value)
}
You can do simply
for (_, userId) in users {
// User setup
}
You may try iterator like this:
let users = ["a":11, "b":12]
var userIds = users.values.makeIterator()
while let next = userIds.next() {
print(next) // 11 \n 12
}
Also:
let x = [
"kitty": 7,
"bob": 2,
"orange": 44
]
x.forEach { t in
print("key = \(t.key); value = \(t.value)")
}
This has apparently been available since Swift 3. Link to standard library docs on Dictionary type.
Swift 4, swift5 and above
let dict: [String: Any] = ["a": 1, "b": "hello", "c": 3]
for (key, value) in dict {
print(key, value)
}
Is there are analog of - (NSArray *)keysSortedByValueUsingSelector:(SEL)comparator in swift?
How to do this without casting to NSDictionary?
I tried this, but it seems to be not a good solution.
var values = Array(dict.values)
values.sort({
$0 > $1
})
for number in values {
for (key, value) in dict {
if value == number {
println(key + " : \(value)");
dict.removeValueForKey(key);
break
}
}
}
Example:
var dict = ["cola" : 10, "fanta" : 12, "sprite" : 8]
dict.sortedKeysByValues(>) // fanta (12), cola(10), sprite(8)
Just one line code to sort dictionary by Values in Swift 4, 4.2 and Swift 5:
let sortedByValueDictionary = myDictionary.sorted { $0.1 < $1.1 }
Try:
let dict = ["a":1, "c":3, "b":2]
extension Dictionary {
func sortedKeys(isOrderedBefore:(Key,Key) -> Bool) -> [Key] {
return Array(self.keys).sort(isOrderedBefore)
}
// Slower because of a lot of lookups, but probably takes less memory (this is equivalent to Pascals answer in an generic extension)
func sortedKeysByValue(isOrderedBefore:(Value, Value) -> Bool) -> [Key] {
return sortedKeys {
isOrderedBefore(self[$0]!, self[$1]!)
}
}
// Faster because of no lookups, may take more memory because of duplicating contents
func keysSortedByValue(isOrderedBefore:(Value, Value) -> Bool) -> [Key] {
return Array(self)
.sort() {
let (_, lv) = $0
let (_, rv) = $1
return isOrderedBefore(lv, rv)
}
.map {
let (k, _) = $0
return k
}
}
}
dict.keysSortedByValue(<)
dict.keysSortedByValue(>)
Updated:
Updated to the new array syntax and sort semantics from beta 3. Note that I'm using sort and not sorted to minimize array copying. The code could be made more compact, by looking at the earlier version and replacing sort with sorted and fixing the KeyType[] to be [KeyType]
Updated to Swift 2.2:
Changed types from KeyType to Key and ValueType to Value. Used new sort builtin to Array instead of sort(Array) Note performance of all of these could be slightly improved by using sortInPlace instead of sort
You could use something like this perhaps:
var dict = ["cola" : 10, "fanta" : 12, "sprite" : 8]
var myArr = Array(dict.keys)
var sortedKeys = sort(myArr) {
var obj1 = dict[$0] // get ob associated w/ key 1
var obj2 = dict[$1] // get ob associated w/ key 2
return obj1 > obj2
}
myArr // ["fanta", "cola", "sprite"]
This should give you the sorted keys based on value, and is a little more cleaner:
var sortedKeys = Array(dict.keys).sorted(by: { dict[$0]! < dict[$1]! })
I think this is the easiest way to sort Swift dictionary by value.
let dict = ["apple":1, "cake":3, "banana":2]
let byValue = {
(elem1:(key: String, val: Int), elem2:(key: String, val: Int))->Bool in
if elem1.val < elem2.val {
return true
} else {
return false
}
}
let sortedDict = dict.sort(byValue)
OneLiner :
let dict = ["b": 2, "a": 1, "c": 3]
(Array(dict).sorted { $0.1 < $1.1 }).forEach { (k,v) in print("\(k):\(v)") }
//Output: a:1, b:2, c:3
Swap out the .forEach with .map -> Functional programming
Syntactical sugar :
extension Dictionary where Value: Comparable {
var sortedByValue: [(Key, Value)] { return Array(self).sorted { $0.1 < $1.1} }
}
extension Dictionary where Key: Comparable {
var sortedByKey: [(Key, Value)] { return Array(self).sorted { $0.0 < $1.0 } }
}
["b": 2, "a": 1, "c": 3].sortedByKey // a:1, b:2, c:3
["b": 2, "a": 1, "c": 3].sortedByValue // a:1, b:2, c:3
Lots of answers, here's a one-liner. I like it because it makes full use of native Swift iterative functions and doesn't use variables. This should help the optimiser do its magic.
return dictionary.keys.sort({ $0 < $1 }).flatMap({ dictionary[$0] })
Note the use of flatMap, because subscripting a dictionary returns an optional value. In practice this should never return nil since we get the key from the dictionary itself. flatMap is there only to ensure that the result is not an array of optionals. If your array's associated value should BE an optional you can use map instead.
Sorting your keys by the dictionary's value is actually simpler than it appears at first:
let yourDict = ["One": "X", "Two": "B", "Three": "Z", "Four": "A"]
let sortedKeys = yourDict.keys.sort({ (firstKey, secondKey) -> Bool in
return yourDict[firstKey] < yourDict[secondKey]
})
And that's it! There's really nothing more to it. I have yet to find a quicker method, other than the same approach in form of a simple one-liner:
let yourDict = ["One": "X", "Two": "B", "Three": "Z", "Four": "A"]
let sortedKeys = yourDict.keys.sort { yourDict[$0] < yourDict[$1] }
Sorting a dictionary by key or value
Using Swift 5.2 internal handling of "sorted":
var unsortedDict = ["cola" : 10, "fanta" : 12, "sprite" : 8]
// sorting by value
let sortedDictByValue = unsortedDict.sorted{ $0.value > $1.value } // from lowest to highest using ">"
print("sorted dict: \(sortedDictByValue)")
// result: "sorted dict: [(key: "fanta", value: 12), (key: "cola", value: 10), (key: "sprite", value: 8)]\n"
// highest value
print(sortedDictByValue.first!.key) // result: fanta
print(sortedDictByValue.first!.value) // result: 12
// lowest value
print(sortedDictByValue.last!.key) // result: sprite
print(sortedDictByValue.last!.value) // result: 8
// by index
print(sortedDictByValue[1].key) // result: cola
print(sortedDictByValue[1].value) // result: 10
// sorting by key
let sortedDictByKey = unsortedDict.sorted{ $0.key < $1.key } // in alphabetical order use "<"
// alternative:
// let sortedDictByKey = unsortedDict.sorted{ $0 < $1 } // without ".key"
print("sorted dict: \(sortedDictByKey)")
// result: "sorted dict: [(key: "cola", value: 10), (key: "fanta", value: 12), (key: "sprite", value: 8)]\n"
// highest value
print(sortedDictByKey.first!.key) // result: cola
print(sortedDictByKey.first!.value) // result: 10
// lowest value
print(sortedDictByKey.last!.key) // result: sprite
print(sortedDictByKey.last!.value) // result: 8
// by index
print(sortedDictByKey[1].key) // result: fanta
print(sortedDictByKey[1].value) // result: 12
The following might be useful if you want the output to be an array of key value pairs in the form of a tuple, sorted by value.
var dict = ["cola" : 10, "fanta" : 12, "sprite" : 8]
let sortedArrByValue = dict.sorted{$0.1 > $1.1}
print(sortedArrByValue) // output [(key: "fanta", value: 12), (key: "cola", value: 10), (key: "sprite", value: 8)]
Since Swift 3.0 Dictionary has sorted(by:) function which returns an array of tuples ([(Key, Value)]).
let sorted = values.sorted(by: { (keyVal1, keyVal2) -> Bool in
keyVal1.value > keyVal2.value
})
Just cast it to NSDictionary and then call the method. Anywhere you use #selector in ObjC you can just use a String in Swift. So it would look like this:
var dict = ["cola" : 10, "fanta" : 12, "sprite" : 8]
let sortedKeys = (dict as NSDictionary).keysSortedByValueUsingSelector("compare:")
or
let sortedKeys2 = (dict as NSDictionary).keysSortedByValueUsingComparator
{
($0 as NSNumber).compare($1 as NSNumber)
}
As of Swift 3, to sort your keys based on values, the below looks promising:
var keys = Array(dict.keys)
keys.sortInPlace { (o1, o2) -> Bool in
return dict[o1]! as! Int > dict[o2]! as! Int
}
var dict = ["cola" : 10, "fanta" : 12, "sprite" : 8]
let arr = dic.sort{ (d1,d2)-> Bool in
if d1.value > d2.value {
retrn true
}
}.map { (key,value) -> Int in
return value
}
Take look a clean implementation way.
print("arr is :(arr)")
The following way in Swift 3 sorted my dictionary by value in the ascending order:
for (k,v) in (Array(dict).sorted {$0.1 < $1.1}) {
print("\(k):\(v)")
}
SWIFT 3:
Using a few resources I put this beautifully short code together.
dictionary.keys.sorted{dictionary[$0]! < dictionary[$1]!}
This returns an array of the dictionary keys sorted by their values. It works perfectly & doesn't throw errors when the dictionary is empty. Try this code in a playground:
//: Playground - noun: a place where people can play
import UIKit
let dictionary = ["four": 4, "one": 1, "seven": 7, "two": 2, "three": 3]
let sortedDictionary = dictionary.keys.sorted{dictionary[$0]! < dictionary[$1]!}
print(sortedDictionary)
// ["one", "two", "three", "four", "seven"]
let emptyDictionary = [String: Int]()
let emptyDictionarySorted = emptyDictionary.keys.sorted{emptyDictionary[$0]! < emptyDictionary[$1]!}
print(emptyDictionarySorted)
// []
If you'd like some help on why the heck the code uses $0, $1 and doesn't even have parentheses after the "sorted" method, check out this post - https://stackoverflow.com/a/34785745/7107094
This is how I did it - sorting in this case by a key called position. Try this in a playground:
var result: [[String: AnyObject]] = []
result.append(["name" : "Ted", "position": 1])
result.append(["name" : "Bill", "position": 0])
result
result = sorted(result, positionSort)
func positionSort(dict1: [String: AnyObject], dict2: [String: AnyObject]) -> Bool {
let position1 = dict1["position"] as? Int ?? 0
let position2 = dict2["position"] as? Int ?? 0
return position1 < position2
}
Sorting the dictionary with a dictionary as the value (Nested dictionary)
var students: [String: [String: Any?]] = ["16CSB40" : ["Name": "Sunitha", "StudentId": "16CSB40", "Total": 90], "16CSB41" : ["Name": "Vijay", "StudentId": "16CSB40", "Total": 80], "16CSB42" : ["Name": "Tony", "StudentId": "16CSB42", "Total": 95]] // Sort this dictionary with total value
let sorted = students.sorted { (($0.1["Total"] as? Int) ?? 0) < (($1.1["Total"] as? Int) ?? 0) }
print(sorted) //Sorted result
Use this, and then just loop through the dictionary again using the output keys.
extension Dictionary where Value: Comparable {
func sortedKeysByValue() -> [Key] {
keys.sorted { return self[$0]! < self[$1]! }
}
}
...or this if you hate force unwrapping :)
extension Dictionary where Value: Comparable {
func sortedKeysByValue() -> [Key] {
keys.sorted { (key1, key2) -> Bool in
guard let val1 = self[key1] else { return true }
guard let val2 = self[key2] else { return true }
return val1 < val2
}
}
}