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I have a dictionary with string as key and an array of int as values , i need to sort the dictionary based on keys and the array value should also be sorted.
var dicNumArray : Dictionary<String , [Int]> = ["q":[4,3,2,1,5],"a":[2,3,4,5,5],"s":[123,123,132,43,4],"t":[0,88,66,542,321]]
The result i need is the dictionary itself where it is sorted by keys and respective values are also sorted.
You can apply sorted to each the value of the key-value pair of a dictionary using mapValues
And then, you can just use sorted with a predicate comparing the keys of the dictionary.
let result = dicNumArray.mapValues { $0.sorted() }
.sorted { $0.key < $1.key }
This will return an array of key-value pair tuples.
Since, dictionaries can't be trusted with order, working with an array of the key-value pairs is the next best approach.
We can use .key and .value to get the respective values.
result.first?.key // First key
result.first?.value // First value
Dictionaries don't have an order in Swift. Having said that, you can do something like this
var dicNumArray : Dictionary<String , [Int]> = ["q":[4,3,2,1,5],"a":[2,3,4,5,5],"s":[123,123,132,43,4],"t":[0,88,66,542,321]]
func sortData() {
for (key, value) in dicNumArray {
dicNumArray[key] = value.sorted(by: { $0 < $1 })
}
}
sortData()
This will sort array for each key. Once that's done, you can do something like
let keys = Array(dicNumArray.keys).sorted(by: { $0 < $1 })
This will give you a sorted array of dictionary keys. You can test it as follows
TEST
for key in keys {
print("\(key): \(dicNumArray[key]!)")
}
OUTPUT
a: [2, 3, 4, 5, 5]
q: [1, 2, 3, 4, 5]
s: [4, 43, 123, 123, 132]
t: [0, 66, 88, 321, 542]
you can apply sort key , then map to sorted by array
let sortedKeysAndValues = dicNumArray.sorted(by: {$0.0 < $1.0}).map { [$0.key:$0.value.sorted(by: <)]}.flatMap({$0})
I've a dictionary that changes dependent on the user's selected options. It could be for example:
["Item1": 7, "Item2": 4, "Item3": 4, "Item4": 7, "Item5": 6]
The number next to each Item is the count for each item. I want to know how I can convert this dictionary into a table? So in the left hand column is the item and in the right hand column is the item count?
What's the best way to do this?
The data is originally an array with format: ["Item1", "Item2", "Item3", "Item1", "Item1", "Item2"].
But I used
var counts: [String: Int] = [:] , myArray.forEach { counts[$0, default: 0] += 1 }
to count each item in the array
You can transform your dictionary into a sorted array like this:
let array = data.map { $0 }.sorted { $0.key < $1.key }
This will result in an array of (key: String, value: Int) sorted alphabetically by the keys.
Now, in your tableView delegate methods you can return array.count to get the number, and if you want to configure your cell you can do something like this:
let element = array[indexPath.row]
cell.textLabel.text = element.key
cell.detailLabel.text = "\(element.value)"
So, for a tableview data source, you want an array.
var dic = ["Item1": 7, "Item2": 4, "Item3": 4, "Item4": 7, "Item5": 6]
var tableInfo = [(String, Int)]()
for (k, v) in dic {
let temp = (k, v)
tableInfo.append(temp)
}
Revised per OP's comments.
Now you have an array of tuples suitable for sorting and use in your tableview delegate.
I'm using Swift 2.3 and I have the following type array of arrays of my custom object called Player
`var playing = [[obj-one, obj-two],[obj-three, obj-four]]`
How would I use a for-in loop or something else so I can get the array index and the object?
I have the following:
for (index, p) in playing { -- Expression type [[Player]] is ambigious
I've also tried
for in (index, p: Player) in playing { -- same result.
and
for in (index, p) in playing as! Player { -- doesn't conform to squence type
I want to just be able to print out which array the object belongs to and then work with that current object
Use enumerated() to pair up an index and an element, like this:
let a = [["hello", "world"], ["quick", "brown", "fox"]]
for outer in a.enumerated() {
for inner in outer.element.enumerated() {
print("array[\(outer.offset)][\(inner.offset)] = \(inner.element)")
}
}
This produces the following output:
array[0][0] = hello
array[0][1] = world
array[1][0] = quick
array[1][1] = brown
array[1][2] = fox
Functional approach:
let items = [["0, 0", "0, 1"], ["1, 0", "1, 1", "1, 2"]]
items.enumerated().forEach { (firstDimIndex, firstDimItem) in
firstDimItem.enumerated().forEach({ (secondDimIndex, secondDimItem) in
print("item: \(secondDimItem), is At Index: [\(firstDimIndex), \(secondDimIndex)]")
})
}
prints:
item: 0, 0, is At Index: [0, 0]
item: 0, 1, is At Index: [0, 1]
item: 1, 0, is At Index: [1, 0]
item: 1, 1, is At Index: [1, 1]
item: 1, 2, is At Index: [1, 2]
I wouldn't use a for loop, I would do something like this:
import Foundation
var playing = [["one", "two"], ["three", "four"]]
if let index = playing.index(where: { $0.contains("two") }) {
print(index)
} else {
print("Not found")
}
This prints:
0
Or to get the entire subarray containing what you want:
if let subarray = playing.first(where: { $0.contains("three") }) {
print(subarray)
} else {
print("Not found")
}
Prints:
["three", "four"]
I'm trying to implement a groupBy functionality where all the numbers of a nested list are grouped. My code so far:
struct MyClass {
var numbers: [Int]
...
}
var dict: [String : MyClass] = ...
let numbers = dict
.filter{ $0.0.containsString(searchString) }
.flatMap{ $0.1.numbers }
This yields me an Array of Ints. However I'd like to have a dictionary [Int : Int] with each unique number and the count of its occurence. So for example:
[1,2,3,4,1,2,2,1]
should be:
[1 : 2, 2 : 3, 3 : 1, 4 : 1]
I know there's a groupBy operator, but Swift doesn't seem to have one. I've tried with reduce:
func reducer(accumulator: [Int: Int], num: Int) -> [Int : Int] {
var acc = accumulator
acc[num]! += 1
return acc
}
filtered.reduce([:], combine: reducer)
But it crashes when I want to run it. Not sure why, I get a EXC_BAD_INSTRUCTION.
I'd appreciate any help.
let numbers = [1,2,3,4,1,2,2,1]
var results = [Int: Int]()
Set(numbers).forEach { number in results[number] = numbers.filter { $0 == number }.count }
print(results) // [2: 3, 3: 1, 1: 3, 4: 1]
Actually I'm not very sure if this is what you want. I just looked at your examples.
Using NSCountedSet:
var objects = [1,2,3,4,1,2,2,1]
let uniques = NSCountedSet(array: objects)
uniques.forEach { results[$0 as! Int] = uniques.countForObject($0) }
print(results) // [2: 3, 3: 1, 1: 3, 4: 1]
I would expect the crash to be ocurring on this line:
acc[num]! += 1
The first time this is called for a number, the entry doesn't exist in the dictionary yet so acc[num] is nil. Forcefully unwrapping it would cause a crash.
Not sure if this is the best solution but you can simple check for this case:
if (acc[num]) {
acc[num]! += 1
} else {
acc[num] = 1
}
Cleaner code from #vacawama in the comments:
acc[num] = (acc[num] ?? 0) + 1
Here's an extension to Array that does what you're asking:
extension Array where Element: Hashable {
var grouped: [Element:Int] {
var dict = [Element:Int]()
self.forEach { dict[$0] = (dict[$0] ?? 0) + 1 }
return dict
}
}
The key is the closure: { dict[$0] = (dict[$0] ?? 0) + 1 }.
It takes the current value in the array, tests to see if it's a key in the dictionary, returns the value for that key if it exists or 0 if it doesn't, then adds one and sets the key:value to be the pair of the current value and occurrences so far.
Example use:
[1,2,3,4,1,2,2,1].grouped // => [2: 3, 3: 1, 1: 3, 4: 1]
You need something like this:
if let _ = acc.indexForKey(num) {
acc[num]! += 1
}
else {
acc[num] = 1
}
It's sort of unclear what you're asking for, but here's a function that will take an array of ints and return a dictionary with the number as the key, and the count as the value:
func getDictionaryOfCounts(accumulator: [Int]) -> [Int : Int] {
var countingDictionary: [Int : Int] = [:]
accumulator.forEach { (value) in
if countingDictionary[value] != nil {
countingDictionary[value]! += 1
}
else{
countingDictionary[value] = 1
}
}
return countingDictionary
}
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
}
}
}