Is it possible to iterate through an NSDictionary in a specific order so I can save an index for an key-value pair in CoreData, based on the order that the data is originally typed? i.e in the code below Set 1 would have an index of 1, set 2 - 2 and set 3 - 3 rather than at random, as is normal NSDictionary behaviour? Thanks in advance if anyone can either give a solution, or tell me its not possible!
let string1 = "rain, wait, train".wordsInArray
let string2 = "oil, join, coin".wordsInArray
let string3 = "made, came, same".wordsInArray
let lists: [String: [String]] =
["Set 1: List 1": string1,
"Set 1: List 2": string2,
"Set 1: List 3": string3]
var index = 0
For list in lists {
list.listIndex = index
index = index + 1
coreDataStack.saveMainContext()
}
extension String {
var wordsInArray:[String] {
return componentsSeparatedByCharactersInSet(NSCharacterSet.punctuationCharacterSet()).joinWithSeparator("").componentsSeparatedByString(" ")
}
In your example, your keys happen to be added in alphanumeric order. This could be fortuitous but if you meant to get the data in key sorted order, that is not the same request as creation order and it would be easy to do:
for (key,wordlist) in lists.sort({$0.0 < $1.0})
{
// you will be getting the dictionary entries in key order
}
// trickier to access by index though
let aKey = lists.keys.sort()[2]
let aWordList = lists[aKey]
// but it lets you get the wordlist from the key
let S1L3Words = lists["Set 1: List 3"]
On the other hand, if you're meaning to ONLY use a creation order, and do not need to access the elements by their keys, you could declare your structure as an array of tuples:
let lists: [(String, [String])] =
[
("Set 1: List 1", string1),
("Set 1: List 2", string2),
("Set 1: List 3", string3)
]
// your for loop will get them in the same order
for (key,wordlist) in lists
{
// array order, no matter what the key values are
}
// also accessible directly by index
let (aKey, aWordList) = lists[2] // ("Set 1: List 3", ["made", "came", "same"])
And finally, if you don't need keys at all, you can just make it an array of arrays:
let lists: [[String]] = [ string1 , string2 , string3 ]
for (key,wordlist) in lists
{
// array order
}
// also accessible directly by index
let aWordList = lists[2] // ["made", "came", "same"]
Related
I have two dictionaries in Swift with few similar values which are in dynamic mode:
dict1 = ["a1":"value 1", "b1":"value2", "c1":"value 3"]
dict2 = ["b1": "value2", "d1": "value4"]
If I want to compare these two dictionaries and want to extract only the matching keys even nested, how do I about to do that?
If you want the common keys with the value in one of them :
let intersectionDict = dict1.filter { dict2.keys.contains($0.key) }
//Or
let intersectionDict2 = dict2.filter { dict1.keys.contains($0.key) }
If you want the values to match too:
let intersectionDict3 = dict1.filter { dict2[$0.key] == $0.value }
And the result is:
print(intersectionDict3) //["b1": "value2"]
As others have shown, you can do this using a filter statement. You can make it even quicker by always filtering the smaller of the two dicts, improving the time complexity from O(dict1.size) to O(min(dict1.size, dict2.size).
extension Dictionary {
func intersectingByKeys(with other: Dictionary) -> Dictionary {
let (smallerDict, largerDict) = (self.count < other.count) ? (self, other) : (other, self)
return smallerDict.filter { key, _ in largerDict.keys.contains(key) }
}
}
let dict1 = ["a1":"value 1", "b1":"value2", "c1":"value 3"]
let dict2 = ["b1": "value2", "d1": "value4"]
print(dict1.intersectingByKeys(with: dict2))
You can create a Set from the keys of one of the dictionaries and call intersection on the Set with the keys of the other dictionary.
let matchingKeys = Set(dict1.keys).intersection(dict2.keys) // {"b1"}
I have a dictionary which contains languages as values and the initial character of every language name (A, B, C, ...) as for the key.
var dictionary = [Character: [Language]]()
I would like to get all the languages out of the dictionary in the form of an array. To do so, I do
let languages = dictionary.values // Dictionary<Character, [Language]>.Values
That's not an array. I try to get the array like this
let languages = Array(tableViewSource.values) // [[Language]]
This returns an array of an array. How do I just get an array of languages? I saw the merge(_:uniquingKeysWith:) but I don't need to merge dictionaries.
You can try
let allLangs = nestedArr.reduce([], +)
Martin R's and Sh_Khan's answers outline the standard ways of doing this (you basically just want to flatten an array). However I'd like to show a more "manual" approach:
var langs = [String]()
for lang_list in dictionary.values {
langs.append(contentsOf: lang_list)
}
Alternatively, you can use the += operator instead of .append(contentsOf:). An easier way of doing this would be using flatMap(_:):
dictionary.values.flatMap { $0 }
If you want a single array (concatenating all dictionary values) then
Array(dictionary.values.joined())
does the trick. (This creates the final array without creating any additional intermediate arrays.)
Example:
let dictionary: [Character: [String]] = ["E": ["espanol", "english"], "G": ["german", "greek"]]
let langs = Array(dictionary.values.joined())
print(langs) // ["german", "greek", "espanol", "english"]
Note that the order of key/value pairs in a dictionary is unspecified.
An alternative is
let dictionary: [Character: [String]] = ["E": ["espanol", "english"], "G": ["german", "greek"]]
let langs = dictionary.keys.sorted().flatMap { dictionary[$0]! }
print(langs) // ["espanol", "english", "german", "greek"]
which gives the languages sorted by the corresponding keys.
I'm trying to figure out the best way in Swift to add values to an Array that is a Value in a Dictionary. I want to build a dictionary of contacts sorted by the first letter of their first name. For example [A : [Aaron, Adam, etc...], B : [Brian, Brittany, ect...], ...]
I found this function:
updateValue(_:forKey:)
And tried using it in a loop:
for contact in self.contacts.sorted() {
self.contactDictionary.updateValue([contact], forKey: String(describing: contact.characters.first))
}
But when I tried to use that it replaced the existing array with a new one. I know I can manually check to see if the key in the dictionary exists, if it does, retrieve the array and then append a new value, otherwise add the new key/value pair but I'm not sure if Swift provides an easier/better way to do this.
Any insight would be much appreciated!
You can use reduce(into:) method (Swift4) and as follow:
let contacts = ["Aaron", "Adam", "Brian", "Brittany", ""]
let dictionary = contacts.reduce(into: [String:[String]]()) { result, element in
// make sure there is at least one letter in your string else return
guard let first = element.first else { return }
// create a string with that initial
let initial = String(first)
// initialize an array with one element or add another element to the existing value
result[initial] = (result[initial] ?? []) + [element]
}
print(dictionary) // ["B": ["Brian", "Brittany"], "A": ["Aaron", "Adam"]]
If you are using Swift3 or earlier you would need to create a mutable result dictionary inside the closure:
let contacts = ["Aaron", "Adam", "Brian", "Brittany", ""]
let dictionary = contacts.reduce([String:[String]]()) { result, element in
var result = result
guard let first = element.first else { return result }
let initial = String(first)
result[initial] = (result[initial] ?? []) + [element]
return result
}
print(dictionary) // ["B": ["Brian", "Brittany"], "A": ["Aaron", "Adam"]]
Note that the result is not sorted. A dictionary is an unordered collection. If you need to sort your dictionary and return an array of (key, Value) tuples you can use sorted by key as follow:
let sorted = dictionary.sorted {$0.key < $1.key}
print(sorted)
"[(key: "A", value: ["Aaron", "Adam"]), (key: "B", value: ["Brian", "Brittany"])]\n"
Swift 4's new dictionary initializers can do it all for you:
let contactInitials = contacts.filter{!$0.isEmpty}.map{ ($0.first!,[$0]) }
let dict = [Character:[String]](contactInitials, uniquingKeysWith:+)
I'm trying to figure out the best way in Swift to add values to an Array that is a Value in a Dictionary. I want to build a dictionary of contacts sorted by the first letter of their first name. For example [A : [Aaron, Adam, etc...], B : [Brian, Brittany, ect...], ...]
I found this function:
updateValue(_:forKey:)
And tried using it in a loop:
for contact in self.contacts.sorted() {
self.contactDictionary.updateValue([contact], forKey: String(describing: contact.characters.first))
}
But when I tried to use that it replaced the existing array with a new one. I know I can manually check to see if the key in the dictionary exists, if it does, retrieve the array and then append a new value, otherwise add the new key/value pair but I'm not sure if Swift provides an easier/better way to do this.
Any insight would be much appreciated!
You can use reduce(into:) method (Swift4) and as follow:
let contacts = ["Aaron", "Adam", "Brian", "Brittany", ""]
let dictionary = contacts.reduce(into: [String:[String]]()) { result, element in
// make sure there is at least one letter in your string else return
guard let first = element.first else { return }
// create a string with that initial
let initial = String(first)
// initialize an array with one element or add another element to the existing value
result[initial] = (result[initial] ?? []) + [element]
}
print(dictionary) // ["B": ["Brian", "Brittany"], "A": ["Aaron", "Adam"]]
If you are using Swift3 or earlier you would need to create a mutable result dictionary inside the closure:
let contacts = ["Aaron", "Adam", "Brian", "Brittany", ""]
let dictionary = contacts.reduce([String:[String]]()) { result, element in
var result = result
guard let first = element.first else { return result }
let initial = String(first)
result[initial] = (result[initial] ?? []) + [element]
return result
}
print(dictionary) // ["B": ["Brian", "Brittany"], "A": ["Aaron", "Adam"]]
Note that the result is not sorted. A dictionary is an unordered collection. If you need to sort your dictionary and return an array of (key, Value) tuples you can use sorted by key as follow:
let sorted = dictionary.sorted {$0.key < $1.key}
print(sorted)
"[(key: "A", value: ["Aaron", "Adam"]), (key: "B", value: ["Brian", "Brittany"])]\n"
Swift 4's new dictionary initializers can do it all for you:
let contactInitials = contacts.filter{!$0.isEmpty}.map{ ($0.first!,[$0]) }
let dict = [Character:[String]](contactInitials, uniquingKeysWith:+)
I am doing this to loop through my dictionary until until I match the key. My dictionary is defined as [Int:String]
var index = 0
for (key, value) in mylist! {
if key == property.propertyValue as! Int {
// use index here
}
index += 1
}
Is there a better way to do this? I see examples of filtering (something like the example below) but I am not sure how to make it work with a dictionary. Could I use something like this to find the index of the item? Or is there another way?
mylist.filter{$0.key == 1}
UPDATE:
This works:
let index = Array(mylist!.keys).index(of: 1)
But this doesn't:
let index = mylist!.index(forKey: 1)
It seems they both should work. I wonder why the 2nd one doesn't.
A dictionary is an unordered collection type and doesn't have an index.
You can get the value directly by the key
let value = mylist[property.propertyValue as! Int]
If I understand you correctly, you could do it like so:
let myList = [
2: "Hello",
4: "Goodbye",
8: "Whats up",
16: "Hey"
]
let index = Array(myList.keys).index(of: property.propertyValue)
And then to find the key you're looking for again...
let key = Array(myList.keys)[index!]
As said in other answers, a dictionary is probably not the data structure you're looking for. But this should answer the question you've asked.
Given your dictionary
let dict = [1:"a", 2:"b", 3: ""]
you can extract the index of a given key (e.g. `1) simply writing
let indexForKey1 = dict.index(forKey: 1)
Fetching all the indexes
You can also build a dictionary where they key is the index and the value is the key of dict
let indexes = dict.keys.map { dict.index(forKey: $0) }
BTW: what do you really need to do?