I have an Array which I have defined
var array: [Int] = Array(1...24)
I then add
array.insert(9999, atIndex: 0)
I would like to do something like
var array: [Int] = Array(9999...9999,1...24)
Is this possible ?
You could simply concatenate the arrays created from each range:
let array = Array(10 ... 14) + Array(1 ... 24)
Alternatively:
let array = [10 ... 14, 1 ... 4].flatMap { $0 }
which has the small advantage of not creating intermediate arrays
(as you can see in the open source implementation https://github.com/apple/swift/blob/master/stdlib/public/core/SequenceAlgorithms.swift.gyb).
As MartinR mentioned, you could simply concenate arrays using the + operator; and if this method is an answer for you, than this thread is a duplicate (see MartinR:s link), and should be closed.
If you explicitly wants to initialize an Int array using several ranges at once (see e.g. hola:s answer regarding array of ranges), you can make use of reduce as follows
let arr = [1...5, 11...15].reduce([]) { $0.0 + Array($0.1) }
Or, alternatively, flatten
var arr = Array([1...5, 11...15].flatten())
Both the above yields the following result
print(arr.dynamicType) // Array<Int>
print(arr) // [1, 2, 3, 4, 5, 11, 12, 13, 14, 15]
For an array of ranges you define the array as
let array: [Range<Int>] = [0...1, 5...100]
and so on and so forth.
Related
I have two arrays for which I am comparing [Int]
let filter = strongAgainstArray.filter{weakAgainstArray.contains($0)}
This returns an array of common values in the 2 arrays. I then want to go through and remove those values from each array, which I'm doing like so
for item in filter {
for var i = 0; i < strongAgainstArray.count; i += 1 {
if item == strongAgainstArray[i] {
strongAgainstArray.removeAtIndex(i)
print("Removing a Strong against index \(item)")
}
}
for var i = 0; i < weakAgainstArray.count; i += 1 {
if item == weakAgainstArray[i] {
weakAgainstArray.removeAtIndex(i)
print("Removing a Weak against index \(item)")
}
}
}
This works fine, but let's say one of my arrays contains two entries for 12 as an example. How do I only remove one of them? As it stands, all entries of 12 are being removed entirely.
EDIT
I'm now comparing my two arrays using
let commonValues = Array(Set(strongAgainstArray).intersect(weakAgainstArray))
and then those commonValues from each array with
cleanStrongAgainstArray = Array(Set(strongAgainstArray).subtract(Set(commonValues)).sort())
cleanWeakAgainstArray = Array(Set(weakAgainstArray).subtract(Set(commonValues)).sort())
This is a much better overall solution, but I'm still eventually running into the same issue, albeit slightly different than before.
In the playground, for example...
let array = [7,7,9]
let test = Array(Set(array))
test comes back containing [7, 9], and I need to keep that extra 7. How do I do that?
If the order of the arrays aren't important then you can easily achieve the whole solution using Sets:
let dirtyArray = [1,4,6,1,56,4,4,66,23,3,3,12]
let dirtyArray1 = [3,1,6,99,54]
let cleanArray = Array(Set(dirtyArray).union(Set(dirtyArray1)))
print (cleanArray)
[12, 54, 23, 4, 6, 66, 99, 56, 1, 3]
If order is important, use NSOrderedSet:
let strongAgainstArray = [1, 2, 3, 4]
let weakAgainstArray = [3, 4, 5, 6]
let result = NSOrderedSet(array: (strongAgainstArray + weakAgainstArray)).array
I have 2 Arrays. Say, array1 = [1,2,3,4,5] and array2 = [2,3]. How could I check in swift if array1 contains at least one item from array2?
You can do this by simply passing in your array2's contains function into your array1's contains function (or vice versa), as your elements are Equatable.
let array1 = [2, 3, 4, 5]
let array2 = [20, 15, 2, 7]
// this is just shorthand for array1.contains(where: { array2.contains($0) })
if array1.contains(where: array2.contains) {
print("Array 1 and array 2 share at least one common element")
} else {
print("Array 1 doesn't contains any elements from array 2")
}
This works by looping through array 1's elements. For each element, it will then loop through array 2 to check if it exists in that array. If it finds that element, it will break and return true – else false.
This works because there are actually two flavours of contains. One takes a closure in order to check each element against a custom predicate, and the other just compares an element directly. In this example, array1 is using the closure version, and array2 is using the element version. And that is the reason you can pass a contains function into another contains function.
Although, as correctly pointed out by #AMomchilov, the above algorithm is O(n2). A good set intersection algorithm is O(n), as element lookup is O(1). Therefore if your code is performance critical, you should definitely use sets to do this (if your elements are Hashable), as shown by #simpleBob.
Although if you want to take advantage of the early exit that contains gives you, you'll want to do something like this:
extension Sequence where Iterator.Element : Hashable {
func intersects<S : Sequence>(with sequence: S) -> Bool
where S.Iterator.Element == Iterator.Element
{
let sequenceSet = Set(sequence)
return self.contains(where: sequenceSet.contains)
}
}
if array1.intersects(with: array2) {
print("Array 1 and array 2 share at least one common element")
} else {
print("Array 1 doesn't contains any elements from array 2")
}
This works much the same as the using the array's contains method – with the significant difference of the fact that the arraySet.contains method is now O(1). Therefore the entire method will now run at O(n) (where n is the greater length of the two sequences), with the possibility of exiting early.
With Swift 5, you can use one of the following paths in order to find if two arrays have common elements or not.
#1. Using Set isDisjoint(with:) method
Set has a method called isDisjoint(with:). isDisjoint(with:) has the following declaration:
func isDisjoint(with other: Set<Element>) -> Bool
Returns a Boolean value that indicates whether the set has no members in common with the given sequence.
In order to test if two arrays have no common elements, you can use the Playground sample code below that implements isDisjoint(with:):
let array1 = [1, 3, 6, 18, 24]
let array2 = [50, 100, 200]
let hasNoCommonElement = Set(array1).isDisjoint(with: array2)
print(hasNoCommonElement) // prints: true
#2. Using Set intersection(_:) method
Set has a method called intersection(_:). intersection(_:) has the following declaration:
func intersection<S>(_ other: S) -> Set<Element> where Element == S.Element, S : Sequence
Returns a new set with the elements that are common to both this set and the given sequence.
In order to test if two arrays have no common elements or one or more common elements, you can use the Playground sample code below that implements intersection(_:):
let array1 = [1, 3, 6, 18, 24]
let array2 = [2, 3, 18]
let intersection = Set(array1).intersection(array2)
print(intersection) // prints: [18, 3]
let hasCommonElement = !intersection.isEmpty
print(hasCommonElement) // prints: true
An alternative way would be using Sets:
let array1 = [1,2,3,4,5]
let array2 = [2,3]
let set1 = Set(array1)
let intersect = set1.intersect(array2)
if !intersect.isEmpty {
// do something with the intersecting elements
}
Swift 5
Just make an extension
public extension Sequence where Element: Equatable {
func contains(anyOf sequence: [Element]) -> Bool {
return self.filter { sequence.contains($0) }.count > 0
}
}
Use:
let someArray = ["one", "two", "three"]
let string = "onE, Cat, dog"
let intersects = string
.lowercased()
.replacingOccurrences(of: " ", with: "")
.components(separatedBy: ",")
.contains(anyOf: someArray)
print(intersects) // true
let a1 = [1, 2, 3]
let a2 = [2, 3, 4]
Option 1
a2.filter { a1.contains($0) }.count > 1
Option 2
a2.reduce(false, combine: { $0 || a1.contains($1) })
Hope this helps.
//
// Array+CommonElements.swift
//
import Foundation
public extension Array where Element: Hashable {
func set() -> Set<Array.Element> {
return Set(self)
}
func isSubset(of array: Array) -> Bool {
self.set().isSubset(of: array.set())
}
func isSuperset(of array: Array) -> Bool {
self.set().isSuperset(of: array.set())
}
func commonElements(between array: Array) -> Array {
let intersection = self.set().intersection(array.set())
return intersection.map({ $0 })
}
func hasCommonElements(with array: Array) -> Bool {
return self.commonElements(between: array).count >= 1 ? true : false
}
}
I have an basic Int array
Array = [8, 9, 8]
How do i sum all of its values progressively so that the end result would look like this
EndResult = [8, 17, 25]
Tried using for and while loops, but to no avail.
NB: Basic array[0] + array[1] advices will not work. I'm looking for something automatic like a loop solution.
Looking forward to your advices.
Thanks!
May be this:
var arr = [8, 9, 8]
for i in 1..<arr.count {
arr[i] += arr[i-1]
}
print(arr)
Probably there are better ways than this one, but it works
var array = [8, 9, 8]
var result = [Int]()
for i in 0..<array.count{
var temp = 0;
for j in 0...i{
temp+=array[j]
}
result.append(temp)
}
print(result) //[8, 17, 25]
You could use a reduce function in Swift to accomplish this. Note that you can't really do it with map, because you would need to know what the previous call to the map function returned, keep state in a variable outside the map function (which seems dirty), or loop over your array for every map function call.
let array = [8, 9, 8]
let results = array.reduce((0, []), combine: { (reduction: (lastValue: Int, values: Array<Int>), value: Int) in
let newValue = reduction.lastValue + value
return (newValue, reduction.values + [newValue])
}).1
I have an array of 977 of data, I've been trying to split it to 20 section in UITableView but i cannot get it correct, i am trying to do it dynamically. like if i got an array of 312 or 32 or 545 the equation should divide it, and add the last odd elements in array, I'm placing the new data in array of arrays.
So here is what I'm doing :
var dataof977 = Mydata()
var mutA = NSMutableArray()
for (var i = 0; i < 19; i++)
{
var halfArray : NSArray!
var theRange = NSRange()
theRange.location = i*19;
theRange.length = dataof977.afa.count / 19
halfArray = dataof977.afa.subarrayWithRange(theRange)
mutA.addObject(halfArray)
}
Note : dataof977 is reference of class and afa is a String array.
What am i missing here ?s
Three things:
You need to start each location where the previous left off. To do this, introduce a location variable to keep track of where you are in the original array.
Some of your sections will need more items since your count might not be a multiple of 20. I think your best best is to give the first n sections an extra item to make up for the leftovers.
Your loop needs to iterate 20 times, not 19. I have changed it to use for in which is better Swift style.
var mutA = NSMutableArray()
let sublength = dataof977.afa.count / 20
let leftovers = dataof977.afa.count % 20
// location of each new subarray
var location = 0
for i in 0..<20
{
var length = sublength
// The first sections will have 1 more each
if i < leftovers {
length++
}
let theRange = NSMakeRange(location, length)
let halfArray = dataof977.afa.subarrayWithRange(theRange)
mutA.addObject(halfArray)
// move location up past the items we just added
location += length
}
If you have the possibility to work with Swift arrays instead of NSArray, you can use stride to iterate by steps and divide an array in equal parts with the remainder elements in the last array:
let arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
extension Array {
func splitBy(subSize: Int) -> [[Element]] {
return 0.stride(to: self.count, by: subSize).map { startIndex in
let endIndex = startIndex.advancedBy(subSize, limit: self.count)
return Array(self[startIndex ..< endIndex])
}
}
}
let chunks = arr.splitBy(5)
print(chunks) // [[1, 2, 3, 4, 5], [6, 7, 8, 9, 10], [11, 12]]
Note: this is for Swift 2.
You are only doing 19 iterations, not 20.
for var i = 0; i < 20; i++
You are taking one 19th of the data count, not one 20th. You are repeating the length calculation 19 times in the loop, which is redundant.
It would be easier to dispense with NSRange. Just iterate though all array elements, keep a counter up to the pre-calculate subarray size, and once it reaches the critical number, reset it to zero and start a new subarray to add to your result array.
This will presumably perform better if you use Swift arrays, but this is not strictly necessary.
Swift has a great function .reduce that provides a running total for an entire array:
let array = [1, 2, 3, 4, 5]
let addResult = array.reduce(0) { $0 + $1 }
// addResult is 15
let multiplyResult = array.reduce(1) { $0 * $1 }
// multiplyResult is 120
But is there a simple way to use this for a running total only up to a specific element? For example for an array of [1, 2, 3, 4, 5] is there a way to use the .reduce function to return a running total for only up to the 3rd item (1+2+3=6)?
(I know I can always do a loop, just trying to learn all the possibilities with swift)
Besides using the reduce method of arrays, as in #MartinR's answer, it's also possible to use the global reduce function along with enumerate
The enumerate function returns a sequence of tuples, whose first element (named index) is the element index and the second (named element) is the element value.
So you can achieve the same result with the following code:
let sum = reduce(enumerate(array), 0) {
$0 + ($1.index < 3 ? $1.element : 0)
}
The advantage of this method is that you can apply more complex rules to exclude elements - for example, to sum all elements having even index:
let sum = reduce(enumerate(array), 0) {
$0 + ($1.index % 2 == 0 ? $1.element : 0)
}
The closure called by reduce() has no information about the
current index.
Theoretically it is possible to create a closure that
uses a captured variable to "remember" how often it has been called:
func addFirst(var count : Int) -> (Int, Int) -> Int {
return {
count-- > 0 ? $0 + $1 : $0
}
}
let array = [1, 2, 3, 4, 5]
let sumOfFirstThreeItems = array.reduce(0, addFirst(3) )
But this would still run over the whole array and not just the
first 3 elements. There is no way to "stop" the process.
A much simpler way is to operate on an array slice:
let array = [1, 2, 3, 4, 5]
let sumOfFirstThreeItems = array[0 ..< 3].reduce(0) { $0 + $1 }
A slice is an Array-like type that represents a sub-sequence of any
Array, ContiguousArray, or other Slice.