Recently started learning Swift and is now faced with the problem (code below will insert):
I create an array of structure from 3 arrays. When creating an instance of the structure, I need to do it by random (randomElement as I understood) - all 3 parameters must be unique. How do I check for uniqueness in the function?
arrayOfHumans = createRandomHuman()
struct Human {
let name: String
let surname: String
let age: String
var email: String
}
var arrayOfHumans = [Human] ()
var humans: [Human] = []
let nameA = ["Tim", "Mike", "Stan"]
let surnameA = ["Burk", "Sims", "Stoch"]
let ageA = ["12", "30", "25"]
let emailA = ["one#live.com", "two#gmail.com", "three#outlook.com"]
func createRandomHuman() -> [Human] {
for _ in 1...3 {
if nameA.isEmpty == false {
let human = Human(name: nameA.randomElement()!,
surname: surnameA.randomElement()!,
age: ageA.randomElement()!,
email: emailA.randomElement()!)
humans.append(human)
}
}
return humans
}
Actual result:
first Struct {
name: Tim
surname: Sims
age: 12
email: three#outlook.com
}
second Struct {
name: Mike
surname: Stoch
age: 25
email: one#live.com
}
third Struct {
name: Stan
surname: Burk
age: 30
email: two#gmail.com
}
A solution is to shuffle the surname, age and email arrays separately to get a random but unique order.
let nameA = ["Tim", "Mike", "Stan"]
let surnameA = ["Burk", "Sims", "Stoch"]
let ageA = ["12", "30", "25"]
let emailA = ["one#live.com", "two#gmail.com", "three#outlook.com"]
func createRandomHuman() -> [Human] {
let shuffledSurnameA = surnameA.shuffled()
let shuffledAgeA = ageA.shuffled()
let shuffledEmailA = emailA.shuffled()
var humans: [Human] = []
for i in 0..<nameA.count {
let human = Human(name: nameA[i],
surname: shuffledSurnameA[i],
age: shuffledAgeA[i],
email: shuffledEmailA[i])
humans.append(human)
}
return humans
}
let arrayOfHumans = createRandomHuman()
Another way is to shuffle the indices
func createRandomHuman() -> [Human] {
let indices = nameA.indices
let shuffledIndices = (0..<3).map{ _ in indices.shuffled()}
var humans: [Human] = []
for i in 0..<nameA.count {
let human = Human(name: nameA[i],
surname: surnameA[shuffledIndices[0][i]],
age: ageA[shuffledIndices[1][i]],
email: emailA[shuffledIndices[2][i]])
humans.append(human)
}
return humans
}
If I understand you right, you want to ensure that all randomly created humans are different.
If so, the question is when are 2 humans different. This depends on your problem. Thus you as the programmer have to define it.
You can do it by adopting the Equatable protocol.
As soon as your structs conform to this protocol, you can loop through all previously initiated humans, and check if the new randomly created one is equal to one of them. If so, create a new one.
Related
I'm trying to perform a filter on an array of structs whereby if the element exists return the element, otherwise return the first element in the array. At the moment I have a func to do this, was wondering if there was a more efficient way. Predicate or Array extension?
For example, if this is my array of structures and method:
struct Person {
let name: String
let age: Int
}
let items = [Person(name: "Fred", age: 12),
Person(name: "Bill", age: 14),
Person(name: "Jane", age: 15),
Person(name: "Mary", age: 12)]
// Find the person based on name, if no match, return first item
func filterOrFirst(with name: String? = "") -> Person?
{
if (items.contains(where: {$0.name == name}))
{
return items.first(where: {$0.name == name})
}
return items.first
}
print(filterOrFirst(with: "Bill")) // prints Bill
print(filterOrFirst()) // prints Fred
You can do it like
func filterOrFirst(with name: String? = "") -> Person? {
if let item = items.first(where: {$0.name == name}) {
return item
}
return items.first
}
So you need not to traverse complete array two times. You can use ternary operator here. But it increases compilation time.
You can get idea from this:
var arrData:[String] = ["Cat","Dog","Horse","Sheep"]
func respondTheValue(strSearch:String) -> String{
let stringToSearch = strSearch
if let index = arrData.index(of: stringToSearch){ //Check if data is present or not
return arrData[index]
}else{ // Data not present then return first
return arrData.first!
}
}
And use it Like:
let responseStr = respondTheValue(strSearch: "Dog")
print(responseStr) //Case True : Output -- Dog
let responseStr1 = respondTheValue(strSearch: "Fish")
print(responseStr1) //Case False : Output -- Cat
Hope this helps.
I made up a method to add a variable rankto an array of structs.
The array friutsArrayis created like in the function makeFriuts(). After that, the data gets sorted and according to this, every item gets a rank, respectively index.
In the end I need the FriutsWithRankstruct like it is.
But I´m wondering if there is a better, more effective way to to that. Maybe by even skipping the whole Friuts struct:
struct Friuts {
var name: String
var price: Double
}
struct FriutsWithRank {
var name: String
var price: Double
var rank: Int
}
var friutsArray = [Friuts]()
func makeFriuts() {
friutsArray.append(Friuts(name: "mango", price: 1.2))
friutsArray.append(Friuts(name: "banana", price: 0.79))
friutsArray.append(Friuts(name: "orange", price: 2.2))
}
func makeFriutsWithRank(data: [Friuts]) -> [FriutsWithRank] {
let dataSorted = data.sorted { $1.price < $0.price }
var datatoappend = [FriutsWithRank]()
var i = 0
dataSorted.forEach { fruit in
i += 1
let name = fruit.name
let price = fruit.price
let rank = i
let result = FriutsWithRank(name: name, price: price, rank: rank)
datatoappend.append(result)
}
return datatoappend
}
let friutsArrayWithRank = makeFriutsWithRank(data: friutsArray)
With more effective i mean not necessarily less code. I think the two arrays are now created with two iterations. Is it possible to skip the whole Fruits struct and work just with one struct and one iteration?.
I have applied some modification on your code, please read the inline comments. Not much more optimised, but more readable for sure.
// Typo fixed + it is a single Fruit, not Fruits
struct Fruit {
var name: String
var price: Double
}
struct FruitsWithRank {
// You already have a variable holding name and price, Friut
// Lets just reuse Fruit object
var fruit: Fruits
var rank: Int
}
var fruits = [Fruit]()
func makeFruits() {
fruits.append(Fruit(name: "mango", price: 1.2))
fruits.append(Fruit(name: "banana", price: 0.79))
fruits.append(Fruit(name: "orange", price: 2.2))
}
func makeFruitsWithRank(data: [Fruits]) -> [FruitsWithRank] {
let dataSorted = data.sorted { $1.price < $0.price }
var datatoappend = [FruitsWithRank]()
// Use `enumerated` to get index and the object straight away
for (index, fruit) in dataSorted.enumerated() {
// Just init your `FruitsWithRank` with the fruit and the index
let rankedFruit = FruitsWithRank(fruit: fruit, rank: index)
// Add to your array
datatoappend.append(rankedFruit)
}
return datatoappend
}
let fruitsWithRank = makeFruitsWithRank(data: fruitsArray)
EDIT:
Following the edits of your question, i have applied some changes. If you need FruitsWithRank having name and price, you can just create tuples with name and price, and create an array straight away without any loops or appends. You can omit the makeFruitsWithRank function, and sort, enumerate and create your types straight on the tuple array.
struct FruitsWithRank {
var rank: Int
var name: String
var price: Double
}
let rankedFruits: [FruitsWithRank] = [
(name: "mango", price: 1.2),
(name: "banana", price: 0.79),
(name: "orange", price: 2.2)]
.sorted { $0.price < $1.price }
.enumerated()
.map({ return FruitsWithRank(rank: $0 + 1, name: $1.name, price: $1.price) })
In the end this isn't much more efficient than your code, but it is more compact:
func makeFriutsWithRank(data: [Friuts]) -> [FriutsWithRank] {
let dataMapped = data.sorted { $1.price < $0.price }
.enumerated()
.map { FriutsWithRank(name: $1.name, price: $1.price, rank: $0 + 1) }
return dataMapped
}
Is it really necassary to have Struct? Because you already have sorted, the index can serve as rank.
I need to mutate the following array:
struct Person {
var name: String
var age = 0
}
func showPersonArray() -> [Person] {
var dataArray = [Person]()
dataArray.append(Person(name: "Sarah_Yayvo", age: 29))
dataArray.append(Person(name: "Shanda_Lear", age: 45))
dataArray.append(Person(name: "Heidi_Clare", age: 45))
return dataArray
}
How could I split the "name"-key into two keys: "givenName" and "familyName".
Some nice person gave me this code before:
let arraySeparated1 = dataArray.map { $0.substring(to: $0.range(of: "_")!.lowerBound) }
let arraySeparated2 = dataArray.map { $0.substring(from: $0.range(of: "_")!.upperBound) }
Is it possible to make the mutation inside the struct?
The function showPersonArray() is only for demo and test.
Maybe there is a way to work with a target struct, like this:
struct Persontarget {
var familyname: String
var givenName: String
var age = 0
}
struct Person: Array -> [Persontarget] {
var name: String
var age = 0
// Here the split/mutating code
return [PersonWithTwoNames]
}
Or with an String extension. Possibly my question sounds pretty newby-like, but i´m trying the whole day...
Thanks everyone!
I would write an initializer on the new Person type, which initializes it from the old person type (which I called LegacyPerson):
import Foundation
struct LegacyPerson {
let name: String
let age: Int
}
func getPersonArray() -> [LegacyPerson] {
return [
LegacyPerson(name: "Sarah_Yayvo", age: 29),
LegacyPerson(name: "Shanda_Lear", age: 45),
LegacyPerson(name: "Heidi_Clare", age: 45)
]
}
struct Person {
let familyName: String
let givenName: String
let age: Int
}
extension Person {
init(fromLegacyPerson person: LegacyPerson) {
let index = person.name.range(of: "_")!
self.init(
familyName: person.name.substring(from: index.upperBound),
givenName: person.name.substring(to: index.lowerBound),
age: person.age
)
}
}
let people: [Person] = getPersonArray().map(Person.init)
people.forEach{ print($0) }
Create a method or computer variable for your Person class that returns what you want.
func firstName() -> String {
return self.substring(to: $0.range(of: "_")!.lowerBound)
}
You should not force cast though
with help of computed property defined in an extension
struct Person {
let name: String
let age: Int
}
let person = Person(name: "Maria_Terezia", age: 300)
extension Person {
var names:[String] {
get {
return name.characters.split(separator: "_").map(String.init)
}
}
}
for name in person.names {
print(name)
}
prints
Maria
Terezia
With this Swift 3.0 lines:
struct Person {
var name: String
var surname: String
var phone: Int
var isCustomer: Bool
}
var contacts: [Person] = []
contacts.append(Person(name: "Jack", surname: "Johnson", phone: 2, isCustomer: false))
contacts.append(Person(name: "Mike", surname: "Morris", phone: 3, isCustomer: true))
I have created an array that includes two structures which include 4 variables each.
I can print a single object of the array like this: print(contacts[0].name)
but is there any way to print all the Strings of the name section at once?
Learn how to use map. I use it all the time.
print(contacts.map({ $0.name }))
Search for map in this Apple Documentation about Closures
You'll have to iterate over the array, either printing the values as you go, or capturing them into a string and printing them all at once.
Here's one way:
for contact in contacts {
print(contact.name)
}
Here's another:
contacts.forEach { print($0.name) }
Finally, you could join all the strings into one value with a separator and just print once. When you do it this way the joinWithSeparator function iterates the array for you:
let names = contacts.map { $0.name }
let joinedNames = names.joinWithSeparator(" ")
print(joinedNames)
You should implement the protocol CustomStringConvertible by defining the computed property description:
struct Person: CustomStringConvertible {
var name: String
var surname: String
var phone: Int
var isCustomer: Bool
var description: String {
return
"Name: \(name)\n" +
"Surname: \(surname)\n" +
"Phone: \(phone)\n" +
"Is Customer? \(isCustomer)\n"
}
}
And then:
var contacts: [Person] = []
contacts.append(Person(name: "Jack", surname: "Johnson", phone: 2, isCustomer: false))
contacts.append(Person(name: "Mike", surname: "Morris", phone: 3, isCustomer: true))
print(contacts)
Obviously you can define description as you want.
I'm trying to find the place of a value in a an array containing structures.
My array looks like this
struct User {
var firstName: String?
var lastName: String?
}
var allThePeople = [User(firstName: "John", lastName: "Doe"), User(firstName: "Jane", lastName: "Doe"), User(firstName: "John", lastName: "Travolta")];
Is there a way to get the places for all "Doe"'s in the array? (in this case 0 and 1)
You can filter allThePeople with a condition to get all the people with the last name "Doe".
let allTheDoes = allThePeople.filter { $0.lastName == "Doe" }
You can enumerate the array and flat map it to an array of indices.
let allTheDoeIndexes = allThePeople.enumerated().flatMap { $0.element.lastName == "Doe" ? $0.offset : nil }
= allThePeople.enumerated().flatMap { $1.lastName == "Doe" ? $0 : nil }
If you want the actual indices, use something like
struct User {
var firstName: String?
var lastName: String?
}
var allThePeople = [User(firstName: "John", lastName: "Doe"), User(firstName: "Jane", lastName: "Doe"), User(firstName: "John", lastName: "Travolta")]
var indices = [Int]()
for i in 0 ..< allThePeople.count {
if allThePeople[i].lastName == "Doe" {
indices.append(i)
}
}
indices // [0,1]
otherwise use filter as #Callam suggested.