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Removing duplicate elements from an array in Swift
(49 answers)
Closed 5 years ago.
I have an [Int] array and I need to check for all elements, that don't have identical value with some another element. Those non- identical elements I would like to insert to a new array.
var spojeni1 = [1, 1, 2, 3, 4, 4] // Here it is values 2,3
var NewArray = [Int]()
for i in spojeni1 {
if { // the value hasn't another identical value in the array
NewArray.append(i)
}
}
Hope it is clear, thank you
This can be done in a single line:
let numbers = [1, 1, 2, 3, 4, 4] // Here it is values 2,3
let uniques = Set(numbers).filter{ (n) in numbers.filter{$0==n}.count == 1 }
UPDATE
With Swift 4, you could also use a dictionary constructor to do it:
let uniques = Dictionary(grouping:numbers){$0}.filter{$1.count==1}.map{$0.0}
I would make use of 2 sets:
var seenOnce: Set<Int> = []
var seenMorethanOnce: Set<Int> = []
let input = [1, 1, 2, 3, 4, 4]
for number in input
{
if seenMorethanOnce.contains(number)
{
// Skip it since it's already been detected as a dupe.
}
else if seenOnce.contains(number)
{
// We saw it once, and now we have a dupe.
seenOnce.remove(number)
seenMorethanOnce.insert(number)
}
else
{
// First time seeing this number.
seenOnce.insert(number)
}
}
When that for loop finishes, the seenOnce set will have your values, and you can easily convert that to an array if you wish.
I want to find multiple elements of a value in an array in Matlab code.
I found the function mod and find, but these return the indices of elements and
not the elements. Moreover, I wrote the following code:
x=[1 2 3 4];
if (mod(x,2)==0)
a=x;
end
but this does not work. How can I solve this problem?
Looks like you what to find all multiples of 2 (or any number), you can achieve this using :
a = x( mod(x,2) == 0 ) ;
When you do a = x, x is still x=[1 2 3 4] regardless if (mod(x,2)==0) is true or false;
you can assign a value to (mod(x,2)==0), e.g. val = (mod(x,2)==0), then append/add this value to a new array.
Given a vector numberList = [ 1, 2, 3, 4, 5, 6]; and a number number = 2; you can find indices (position in a vector) of the numbers in the numberList that are a multiple of number using indices = find(mod(numberList, number) ==0);.
If necessary you may display a list of this multiples calling: multiples = numberList(indices).
multiples =
2 4 6
When I use the for loop in Playground, everything worked fine, until I changed the first parameter of for loop to be the highest value. (iterated in descending order)
Is this a bug? Did any one else have it?
for index in 510..509
{
var a = 10
}
The counter that displays the number of iterations that will be executions keeps ticking...
Xcode 6 beta 4 added two functions to iterate on ranges with a step other than one:
stride(from: to: by:), which is used with exclusive ranges and stride(from: through: by:), which is used with inclusive ranges.
To iterate on a range in reverse order, they can be used as below:
for index in stride(from: 5, to: 1, by: -1) {
print(index)
}
//prints 5, 4, 3, 2
for index in stride(from: 5, through: 1, by: -1) {
print(index)
}
//prints 5, 4, 3, 2, 1
Note that neither of those is a Range member function. They are global functions that return either a StrideTo or a StrideThrough struct, which are defined differently from the Range struct.
A previous version of this answer used the by() member function of the Range struct, which was removed in beta 4. If you want to see how that worked, check the edit history.
Apply the reverse function to the range to iterate backwards:
For Swift 1.2 and earlier:
// Print 10 through 1
for i in reverse(1...10) {
println(i)
}
It also works with half-open ranges:
// Print 9 through 1
for i in reverse(1..<10) {
println(i)
}
Note: reverse(1...10) creates an array of type [Int], so while this might be fine for small ranges, it would be wise to use lazy as shown below or consider the accepted stride answer if your range is large.
To avoid creating a large array, use lazy along with reverse(). The following test runs efficiently in a Playground showing it is not creating an array with one trillion Ints!
Test:
var count = 0
for i in lazy(1...1_000_000_000_000).reverse() {
if ++count > 5 {
break
}
println(i)
}
For Swift 2.0 in Xcode 7:
for i in (1...10).reverse() {
print(i)
}
Note that in Swift 2.0, (1...1_000_000_000_000).reverse() is of type ReverseRandomAccessCollection<(Range<Int>)>, so this works fine:
var count = 0
for i in (1...1_000_000_000_000).reverse() {
count += 1
if count > 5 {
break
}
print(i)
}
For Swift 3.0 reverse() has been renamed to reversed():
for i in (1...10).reversed() {
print(i) // prints 10 through 1
}
Updated for Swift 3
The answer below is a summary of the available options. Choose the one that best fits your needs.
reversed: numbers in a range
Forward
for index in 0..<5 {
print(index)
}
// 0
// 1
// 2
// 3
// 4
Backward
for index in (0..<5).reversed() {
print(index)
}
// 4
// 3
// 2
// 1
// 0
reversed: elements in SequenceType
let animals = ["horse", "cow", "camel", "sheep", "goat"]
Forward
for animal in animals {
print(animal)
}
// horse
// cow
// camel
// sheep
// goat
Backward
for animal in animals.reversed() {
print(animal)
}
// goat
// sheep
// camel
// cow
// horse
reversed: elements with an index
Sometimes an index is needed when iterating through a collection. For that you can use enumerate(), which returns a tuple. The first element of the tuple is the index and the second element is the object.
let animals = ["horse", "cow", "camel", "sheep", "goat"]
Forward
for (index, animal) in animals.enumerated() {
print("\(index), \(animal)")
}
// 0, horse
// 1, cow
// 2, camel
// 3, sheep
// 4, goat
Backward
for (index, animal) in animals.enumerated().reversed() {
print("\(index), \(animal)")
}
// 4, goat
// 3, sheep
// 2, camel
// 1, cow
// 0, horse
Note that as Ben Lachman noted in his answer, you probably want to do .enumerated().reversed() rather than .reversed().enumerated() (which would make the index numbers increase).
stride: numbers
Stride is way to iterate without using a range. There are two forms. The comments at the end of the code show what the range version would be (assuming the increment size is 1).
startIndex.stride(to: endIndex, by: incrementSize) // startIndex..<endIndex
startIndex.stride(through: endIndex, by: incrementSize) // startIndex...endIndex
Forward
for index in stride(from: 0, to: 5, by: 1) {
print(index)
}
// 0
// 1
// 2
// 3
// 4
Backward
Changing the increment size to -1 allows you to go backward.
for index in stride(from: 4, through: 0, by: -1) {
print(index)
}
// 4
// 3
// 2
// 1
// 0
Note the to and through difference.
stride: elements of SequenceType
Forward by increments of 2
let animals = ["horse", "cow", "camel", "sheep", "goat"]
I'm using 2 in this example just to show another possibility.
for index in stride(from: 0, to: 5, by: 2) {
print("\(index), \(animals[index])")
}
// 0, horse
// 2, camel
// 4, goat
Backward
for index in stride(from: 4, through: 0, by: -1) {
print("\(index), \(animals[index])")
}
// 4, goat
// 3, sheep
// 2, camel
// 1, cow
// 0, horse
Notes
#matt has an interesting solution where he defines his own reverse operator and calls it >>>. It doesn't take much code to define and is used like this:
for index in 5>>>0 {
print(index)
}
// 4
// 3
// 2
// 1
// 0
Check out On C-Style For Loops Removed from Swift 3
Swift 4 onwards
for i in stride(from: 5, to: 0, by: -1) {
print(i)
}
//prints 5, 4, 3, 2, 1
for i in stride(from: 5, through: 0, by: -1) {
print(i)
}
//prints 5, 4, 3, 2, 1, 0
With Swift 5, according to your needs, you may choose one of the four following Playground code examples in order to solve your problem.
#1. Using ClosedRange reversed() method
ClosedRange has a method called reversed(). reversed() method has the following declaration:
func reversed() -> ReversedCollection<ClosedRange<Bound>>
Returns a view presenting the elements of the collection in reverse order.
Usage:
let reversedCollection = (0 ... 5).reversed()
for index in reversedCollection {
print(index)
}
/*
Prints:
5
4
3
2
1
0
*/
As an alternative, you can use Range reversed() method:
let reversedCollection = (0 ..< 6).reversed()
for index in reversedCollection {
print(index)
}
/*
Prints:
5
4
3
2
1
0
*/
#2. Using sequence(first:next:) function
Swift Standard Library provides a function called sequence(first:next:). sequence(first:next:) has the following declaration:
func sequence<T>(first: T, next: #escaping (T) -> T?) -> UnfoldFirstSequence<T>
Returns a sequence formed from first and repeated lazy applications of next.
Usage:
let unfoldSequence = sequence(first: 5, next: {
$0 > 0 ? $0 - 1 : nil
})
for index in unfoldSequence {
print(index)
}
/*
Prints:
5
4
3
2
1
0
*/
#3. Using stride(from:through:by:) function
Swift Standard Library provides a function called stride(from:through:by:). stride(from:through:by:) has the following declaration:
func stride<T>(from start: T, through end: T, by stride: T.Stride) -> StrideThrough<T> where T : Strideable
Returns a sequence from a starting value toward, and possibly including, an end value, stepping by the specified amount.
Usage:
let sequence = stride(from: 5, through: 0, by: -1)
for index in sequence {
print(index)
}
/*
Prints:
5
4
3
2
1
0
*/
As an alternative, you can use stride(from:to:by:):
let sequence = stride(from: 5, to: -1, by: -1)
for index in sequence {
print(index)
}
/*
Prints:
5
4
3
2
1
0
*/
#4. Using AnyIterator init(_:) initializer
AnyIterator has an initializer called init(_:). init(_:) has the following declaration:
init(_ body: #escaping () -> AnyIterator<Element>.Element?)
Creates an iterator that wraps the given closure in its next() method.
Usage:
var index = 5
guard index >= 0 else { fatalError("index must be positive or equal to zero") }
let iterator = AnyIterator({ () -> Int? in
defer { index = index - 1 }
return index >= 0 ? index : nil
})
for index in iterator {
print(index)
}
/*
Prints:
5
4
3
2
1
0
*/
If needed, you can refactor the previous code by creating an extension method for Int and wrapping your iterator in it:
extension Int {
func iterateDownTo(_ endIndex: Int) -> AnyIterator<Int> {
var index = self
guard index >= endIndex else { fatalError("self must be greater than or equal to endIndex") }
let iterator = AnyIterator { () -> Int? in
defer { index = index - 1 }
return index >= endIndex ? index : nil
}
return iterator
}
}
let iterator = 5.iterateDownTo(0)
for index in iterator {
print(index)
}
/*
Prints:
5
4
3
2
1
0
*/
For Swift 2.0 and above you should apply reverse on a range collection
for i in (0 ..< 10).reverse() {
// process
}
It has been renamed to .reversed() in Swift 3.0
In Swift 4 and latter
let count = 50//For example
for i in (1...count).reversed() {
print(i)
}
Swift 4.0
for i in stride(from: 5, to: 0, by: -1) {
print(i) // 5,4,3,2,1
}
If you want to include the to value:
for i in stride(from: 5, through: 0, by: -1) {
print(i) // 5,4,3,2,1,0
}
If one is wanting to iterate through an array (Array or more generally any SequenceType) in reverse. You have a few additional options.
First you can reverse() the array and loop through it as normal. However I prefer to use enumerate() much of the time since it outputs a tuple containing the object and it's index.
The one thing to note here is that it is important to call these in the right order:
for (index, element) in array.enumerate().reverse()
yields indexes in descending order (which is what I generally expect). whereas:
for (index, element) in array.reverse().enumerate() (which is a closer match to NSArray's reverseEnumerator)
walks the array backward but outputs ascending indexes.
as for Swift 2.2 , Xcode 7.3 (10,June,2016) :
for (index,number) in (0...10).enumerate() {
print("index \(index) , number \(number)")
}
for (index,number) in (0...10).reverse().enumerate() {
print("index \(index) , number \(number)")
}
Output :
index 0 , number 0
index 1 , number 1
index 2 , number 2
index 3 , number 3
index 4 , number 4
index 5 , number 5
index 6 , number 6
index 7 , number 7
index 8 , number 8
index 9 , number 9
index 10 , number 10
index 0 , number 10
index 1 , number 9
index 2 , number 8
index 3 , number 7
index 4 , number 6
index 5 , number 5
index 6 , number 4
index 7 , number 3
index 8 , number 2
index 9 , number 1
index 10 , number 0
You can consider using the C-Style while loop instead. This works just fine in Swift 3:
var i = 5
while i > 0 {
print(i)
i -= 1
}
var sum1 = 0
for i in 0...100{
sum1 += i
}
print (sum1)
for i in (10...100).reverse(){
sum1 /= i
}
print(sum1)
You can use reversed() method for easily reverse values.
var i:Int
for i in 1..10.reversed() {
print(i)
}
The reversed() method reverse the values.
Reversing an array can be done with just single step .reverse()
var arrOfnum = [1,2,3,4,5,6]
arrOfnum.reverse()
This will works decrement by one in reverse order.
let num1 = [1,2,3,4,5]
for item in nums1.enumerated().reversed() {
print(item.offset) // Print the index number: 4,3,2,1,0
print(item.element) // Print the value :5,4,3,2,1
}
Or you can use this index, value property
let num1 = [1,2,3,4,5]
for (index,item) in nums1.enumerated().reversed() {
print(index) // Print the index number: 4,3,2,1,0
print(item) // Print the value :5,4,3,2,1
}
For me, this is the best way.
var arrayOfNums = [1,4,5,68,9,10]
for i in 0..<arrayOfNums.count {
print(arrayOfNums[arrayOfNums.count - i - 1])
}
When using Groovy's eachWithIndex method the index value starts at 0, I need it to start at 1. How can I do that?
The index will always start from 0
Your options are:
1) Add an offset to the index:
int offs = 1
list.eachWithIndex { it, idx ->
println "$it # pos ${idx + offs}"
}
2) Use something other than eachWithIndex (ie: transpose a list of integers starting at 1 with your original list, and then loop through this)
3) You can also use default parameters to hack this sort of thing in... If we pass eachWithIndex a closure with 3 parameters (the two that eachWithIndex is expecting and a third with a default value of index + 1):
[ 'a', 'b', 'c' ].eachWithIndex { item, index, indexPlusOne = index + 1 ->
println "Element $item has position $indexPlusOne"
}
We will give the output:
Element a has position 1
Element b has position 2
Element c has position 3
I wouldn't use eachWithIndex at all, a for loop with .indexed(1) will be more elegant/readable:
for (item in ['a','b','c','d','e'].indexed(1)) {
println("element $item.key = $item.value")
}
outputs:
element 1 = a
element 2 = b
element 3 = c
element 4 = d
element 5 = e
caveat -- indexed(n) is only available in Groovy 2.4.0 onwards
What about just using a Range?
def offset = 1
def arr = [1,2,3,4,5]
def len = arr.size() - 1
arr[offset..len].eachWithIndex{n,i -> println "${i+offset}:$n"}
You could use some metaprogramming hackery as well:
def eachWithMyIndex = { init, clos ->
delegate.eachWithIndex { it, idx -> clos(it, idx + init) }
}
List.metaClass.eachWithMyIndex = eachWithMyIndex
[1,2,3].eachWithMyIndex(10) {n, idx -> println("element $idx = $n") }
gives output:
element 10 = 1
element 11 = 2
element 12 = 3