The loop decrements the loop counter if it finds an element that shouldn't be there and removes it.
var iMax = numListViews
for var i = 0; i < numListViews; i += 1 {
if (columnsSortTypesArray[i] == "") {
columnsSortTypesArray.removeAtIndex(i)
i--
iMax--
} else {
listViews[i].sortList(columnsSortTypesArray[i])
}
}
You shouldn't have used a for loop for that purpose in the first place.
Modifying the loop variable within the loop is error-prone and a good
example why C-style for loops have been deprecated.
To remove empty strings from an array use filter():
columnsSortTypesArray = columnsSortTypesArray.filter { $0 != "" }
To populate another array based on this array, use map(), e.g.
listViews = columnsSortTypesArray.map { sortList($0) }
Also: If you have two (or more) arrays which always have to be kept
in sync, consider to define a struct WhatEver with two (or more) properties
and use a single array of these WhatEver elements instead.
Related
I want to compare two Arrays with each other, that means each single item of them.
I need to run some code if two items in this arrays are the same.
I've done that so far with two For-Loops, but someone in the comments says that's not that good (because I get an Error too.
Has anybody an Idea which Code I can use to reach that?
Btw: That's the code I used before:
var index = 0
for Item1 in Array1 {
for Item2 in Array2 {
if (Item1 == Item2) {
// Here I want to put some code if the items are the same
}
}
// Index is in this case the counter which counts on which place in the Array1 I am.
index += 1
}
Okay I'll try again to describe what I mean:
I want to compare two Arrays. If there are some Items the same, I want to delete that Item from Array A / Array 1.
Okay if that is working, I want to add a few more statements that alow me to delete a Item only if an parameter of this item has a special worth, but I think I can do this step alone.
If you want to compare items from different array you need to add Equatable protocol for your Item
For example:
struct Item: Equatable {
let name: String
static func ==(l: Item, r: Item) -> Bool {
return l.name == r.name
}
}
You need to decide by which attributes you want to compare your Item. In my case I compare by name.
let array1: [Item] = [
.init(name: "John"),
.init(name: "Jack"),
.init(name: "Soer"),
.init(name: "Kate")
]
let array2: [Item] = [
.init(name: "John"),
]
for item1 in array1 {
if array2.contains(item1) {
// Array 2 contains item from the array1 and you can perform your code.
}
}
If you want to support this
Okay I'll try again to describe what I mean: I want to compare two
Arrays. If there are some Items the same, I want to delete that Item
from Array A / Array 1. Okay if that is working, I want to add a few
more statements that alow me to delete a Item only if an parameter of
this item has a special worth, but I think I can do this step alone.
I guess it can fit for you
You need to make your array1 mutable
var array1: [Item] = ...
Filter the array1 like this
let filteredArray1 = array1.filter { (item) -> Bool in
return !array2.contains(item)
}
And redefine your array1 with filtered array.
array1 = filteredArray1
Let me know it it works for you.
var itemsToRemove = array1.filter { array2.contains($0) }
for item in itemsToRemove {
if let i = array1.firstIndex(of: item) {
// run your conditional code here.
array1.remove(at: i)
}
}
Update
The question has been restated that elements are to be removed from one of the arrays.
You don't provide the actual code where you get the index out of bounds error, but it's almost certainly because you don't account for having removed elements when using the index, so you're probably indexing into a region that was valid at the start, but isn't anymore.
My first advice is don't do the actual removal inside the search loop. There is a solution to achieve the same result, which I'll give, but apart from having to be very careful about indexing into the shortened array, there is also a performance issue: Every deletion requires Array to shift all the later elements down one. That means that every deletion is an O(n) operation, which makes an the overall algorithim O(n^3).
So what do you do instead? The simplest method is to create a new array containing only the elements you wish to keep. For example, let's say you want to remove from array1 all elements that are also in array2:
array1 = array1.filter { !array2.contains($0) }
I should note that one of the reasons I kept my original answer below is because you can use those methods to replace array2.contains($0) to achieve better performance in some cases, and the original answer describes those cases and how to achieve it.
Also even though the closure is used to determine whether or not to keep the element, so it has to return a Bool, there is nothing that prevents you from putting additional code in it to do any other work you might want to:
array1 = array1.filter
{
if array2.contains($0)
{
doSomething(with: $0)
return false // don't keep this element
}
return true
}
The same applies to all of the closures below.
You could just use the removeAll(where:) method of Array.
array1.removeAll { array2.contains($0) }
In this case, if the closure returns true, it means "remove this element" which is the opposite sense of the closure used in filter, so you have to take that into account if you do additional work in it.
I haven't looked up how the Swift library implements removeAll(where:) but I'm assuming it does its job the efficient way rather than the naive way. If you find the performance isn't all that good you could roll your own version.
extension Array where Element: Equatable
{
mutating func efficientRemoveAll(where condition: (Element) -> Bool)
{
guard count > 0 else { return }
var i = startIndex
var iEnd = endIndex
while i < iEnd
{
if condition(self[i])
{
iEnd -= 1
swapAt(i, iEnd)
continue // note: skips incrementing i,iEnd has been decremented
}
i += 1
}
removeLast(endIndex - iEnd)
}
}
array1.efficientRemoveAll { array2.contains($0) }
Instead of actually removing the elements inside the loop, this works by swapping them with the end element, and handling when to increment appropriately. This collects the elements to be removed at the end, where they can be removed in one go after the loop finishes. So the deletion is just one O(n) pass at the end, which avoids increasing the algorithmic complexity that removing inside the loop would entail.
Because of the swapping with the current "end" element, this method doesn't preserve the relative order of the elements in the mutating array.
If you want to preserve the order you can do it like this:
extension Array where Element: Equatable
{
mutating func orderPreservingRemoveAll(where condition: (Element) -> Bool)
{
guard count > 0 else { return }
var i = startIndex
var j = i
repeat
{
swapAt(i, j)
if !condition(self[i]) { i += 1 }
j += 1
} while j != endIndex
removeLast(endIndex - i)
}
}
array1.orderPreservingRemoveAll { array2.contains($0) }
If I had to make a bet, this would be very close to how standard Swift's removeAll(where:) for Array is implemented. It keeps two indices, one for the current last element to be kept, i, and one for the next element to be examined, j. This has the effect of accumulating elements to be removed at the end (those past i).
Original answer
The previous solutions are fine for small (yet still surprisingly large) arrays, but they are O(n^2) solutions.
If you're not mutating the arrays, they can be expressed more succinctly
array1.filter { array2.contains($0) }.forEach { doSomething($0) }
where doSomething doesn't actually have to be a function call - just do whatever you want to do with $0 (the common element).
You'll normally get better performance by putting the smaller array inside the filter.
Special cases for sorted arrays
If one of your arrays is sorted, then you might get better performance in a binary search instead of contains, though that will require that your elements conform to Comparable. There isn't a binary search in the Swift Standard Library, and I also couldn't find one in the new swift-algorithms package, so you'd need to implement it yourself:
extension RandomAccessCollection where Element: Comparable, Index == Int
{
func sortedContains(_ element: Element) -> Bool
{
var range = self[...]
while range.count > 0
{
let midPoint = (range.startIndex + range.endIndex) / 2
if range[midPoint] == element { return true }
range = range[midPoint] > element
? self[..<midPoint]
: self[index(after: midPoint)...]
}
return false
}
}
unsortedArray.filter { sortedArray.sortedContains($0) }.forEach { doSomething($0) }
This will give O(n log n) performance. However, you should test it for your actual use case if you really need performance, because binary search is not especially friendly for the CPU's branch predictor, and it doesn't take that many mispredictions to result in slower performance than just doing a linear search.
If both arrays are sorted, you can get even better performance by exploiting that, though again you have to implement the algorithm because it's not supplied by the Swift Standard Library.
extension Array where Element: Comparable
{
// Assumes no duplicates
func sortedIntersection(_ sortedOther: Self) -> Self
{
guard self.count > 0, sortedOther.count > 0 else { return [] }
var common = Self()
common.reserveCapacity(Swift.min(count, sortedOther.count))
var i = self.startIndex
var j = sortedOther.startIndex
var selfValue = self[i]
var otherValue = sortedOther[j]
while true
{
if selfValue == otherValue
{
common.append(selfValue)
i += 1
j += 1
if i == self.endIndex || j == sortedOther.endIndex { break }
selfValue = self[i]
otherValue = sortedOther[j]
continue
}
if selfValue < otherValue
{
i += 1
if i == self.endIndex { break }
selfValue = self[i]
continue
}
j += 1
if j == sortedOther.endIndex { break }
otherValue = sortedOther[j]
}
return common
}
}
array1.sortedIntersection(array2).forEach { doSomething($0) }
This is O(n) and is the most efficient of any of the solutions I'm including, because it makes exactly one pass through both arrays. That's only possible because both of the arrays are sorted.
Special case for large array of Hashable elements
However, if your arrays meet some criteria, you can get O(n) performance by going through Set. Specifically if
The arrays are large
The array elements conform to Hashable
You're not mutating the arrays
Make the larger of the two arrays a Set:
let largeSet = Set(largeArray)
smallArray.filter { largeSet.contains($0) }.forEach { doSomething($0) }
For the sake of analysis if we assume both arrays have n elements, the initializer for the Set will be O(n). The intersection will involve testing each element of the smallArray's elements for membership in largeSet. Each one of those tests is O(1) because Swift implements Set as a hash table. There will be n of those tests, so that's O(n). Then the worst case for forEach will be O(n). So O(n) overall.
There is, of course, overhead in creating the Set, the worst of which is the memory allocation involved, so it's not worth it for small arrays.
Swift 4.2 introduced a new removeAll {where:} function. From what I have read, it is supposed to be more efficient than using filter {where:}. I have several scenarios in my code like this:
private func getListOfNullDates(list: [MyObject]) -> [MyObject] {
return list.filter{ $0.date == nil }
.sorted { $0.account?.name < $1.account?.name }
}
However, I cannot use removeAll{where:} with a param because it is a constant. So I would need to redefine it like this:
private func getListOfNullDates(list: [MyObject]) -> [MyObject] {
var list = list
list.removeAll { $0.date == nil }
return list.sorted { $0.account?.name < $1.account?.name }
}
Is using the removeAll function still more efficient in this scenario? Or is it better to stick with using the filter function?
Thank you for this question 🙏🏻
I've benchmarked both functions using this code on TIO:
let array = Array(0..<10_000_000)
do {
let start = Date()
let filtering = array.filter { $0 % 2 == 0 }
let end = Date()
print(filtering.count, filtering.last!, end.timeIntervalSince(start))
}
do {
let start = Date()
var removing = array
removing.removeAll { $0 % 2 == 0 }
let end = Date()
print(removing.count, removing.last!, end.timeIntervalSince(start))
}
(To have the removing and filtering identical, the closure passed to removeAll should have been { $0 % 2 != 0 }, but I didn't want to give an advantage to either snippet by using a faster or slower comparison operator.)
And indeed, removeAll(where:) is faster when the probability of removing elements (let's call it Pr)is 50%! Here are the benchmark results :
filter : 94ms
removeAll : 74ms
This is the same case when Pr is less than 50%.
Otherwise, filtering is faster for a higher Pr.
One thing to bear in mind is that in your code list is mutable, and that opens the possibility for accidental modifications.
Personally, I would choose performance over old habits, and in a sense, this use case is more readable since the intention is clearer.
Bonus : Removing in-place
What's meant by removing in-place is the swap the elements in the array in such a way that the elements to be removed are placed after a certain pivot index. The elements to keep are the ones before the pivot element :
var inPlace = array
let p = inPlace.partition { $0 % 2 == 0 }
Bear in mind that partition(by:) doesn't keep the original order.
This approach clocks better than removeAll(where:)
Beware of premature optimization. The efficiency of a method often depends on your specific data and configuration, and unless you're working with a large data set or performing many operations at once, it's not likely to have a significant impact either way. Unless it does, you should favor the more readable and maintainable solution.
As a general rule, just use removeAll when you want to mutate the original array and filter when you don't. If you've identified it as a potential bottleneck in your program, then test it to see if there's a performance difference.
var index: Int=0
for index in 1...3{
print(index)
}
print(index)//prints 0
If I run this code, the last print gives 0, which means the index inside the for-in is not the same as outside. Seems like this force declares a new constant.
I am looking for similar way to retain the last value in the sequence
I know I can do
var index_out: Int=0
for index in 1...3{
print(index)
index_out = index
}
print(index_out)
If you're gonna loop through something and want to know the end index just use the amount of times you looped through it:
let n = 3
for index in 1...n{
print(index)
}
print(n)
Or with an array:
let array = [Int](count: 10, repeatedValue: 2)
for index in 0..<array.count {
print(index)
}
print(array.count)
The way you know you can do it is how to do it, for loops create their own scope. Anything declared inside of a set of { } means that it's for use within that scope only.
You can also use the way for in is implemented with an outer scope generator and while:
var generator = sequence.generate()
var element = generator.next()
while let x = generator.next() {
element = x
}
// element is nil if the sequence is empty
print(element)
this is only another way to do this but personally I think you should avoid that.
A much nicer solution would be with reduce (in Swift 1.x: global function, in Swift 2: as method)
// with default value
let lastElement = sequence.reduce(default) { $1 }
// if default should be nil you have to provide the explicit type because it cannot be inferred; probably use an extension:
// as extension
extension SequenceType {
var last: Self.Generator.Element? {
return self.reduce(nil) { $1 }
}
}
There is no way to do it, since it is in different scope than your method. But you also SHOULD NOT do what you do, to assign to your index on each iteration - that is wasteful. Instead, just increase index after you are done:
var data = []
var lastIndex = 0
for (index, object) in enumerate(data) {
...
}
lastIndex += data.count
Or, if you need to break for cycle for some reason, just assign it from inside of the cycle just before you break.
I have a small debate at work: Is it a good practice to calculate the size of the Array in swift before running over it's items? What would be a better code practice:
Option A:
func setAllToFalse() {
for (var i = 0; i < mKeyboardTypesArray.count; i++ ) {
self.mKeyboardTypesArray[i] = false
}
}
or Option B:
func setAllToFalse() {
let typesCount = mKeyboardTypesArray.count
for (var i = 0; i < typesCount; i++ ) {
self.mKeyboardTypesArray[i] = false
}
}
All, of course, if I don't alter the Array during the loops.
I did go over the documentation, which states this:
The loop is executed as follows:
When the loop is first entered, the initialization expression is
evaluated once, to set up any constants or variables that are needed
for the loop. The condition expression is evaluated. If it evaluates
to false, the loop ends, and code execution continues after the for
loop’s closing brace (}). If the expression evaluates to true, code
execution continues by executing the statements inside the braces.
After all statements are executed, the increment expression is
evaluated. It might increase or decrease the value of a counter, or
set one of the initialized variables to a new value based on the
outcome of the statements. After the increment expression has been
evaluated, execution returns to step 2, and the condition expression
is evaluated again.
The idiomatic way to say this in Swift is:
func setAllToFalse() {
mKeyboardTypesArray = mKeyboardTypesArray.map {_ in false}
}
That way, there is nothing to evaluate and nothing to count.
In fact, this would make a nice Array method:
extension Array {
mutating func setAllTo(newValue:T) {
self = self.map {_ in newValue}
}
}
Now you can just say:
mKeyboardTypesArray.setAllTo(false)
Alternatively, you could do it this way (this involves taking the count, but only once):
mKeyboardTypesArray = Array(count:mKeyboardTypesArray.count, repeatedValue:false)
The loop condition is evaluated each time through the loop. Consider this experiment:
extension Array {
var myCount: Int {
get {
println("here")
return self.count
}
}
}
let a = [1, 2, 3, 4, 5]
for var i = 0; i < a.myCount; i++ {
println(a[i])
}
Output:
here 1 here 2 here 3 here 4 here 5 here
You could see a small speed improvement from Option B, but I expect that the count property on Array is not expensive if the array is unchanged. It is potentially a good code practice anyway because it communicates to the reader that you expect the array size to remain constant for duration of the loop.
It is possible that the compiler would optimize array.count by detecting that nothing in the loop modifies the array, but it wouldn't be able to do that for array.myCount because of the println side effect.
I have found that it is not, which can cause a crash if you're iterating through an array and removing items (for example). In current Swift syntax, this for loop
for i in 0..<m_pendingCommands.count
{
if m_pendingCommands[i].packetID < command.packetID
{
m_pendingCommands.remove(at: i)
}
}
crashed halfway through the array, with a bad index.
I switched this to a while loop:
var i: Int = 0
while i < m_pendingCommands.count
{
if m_pendingCommands[i].packetID < ID
{
m_pendingCommands.remove(at: i)
}
else
{
i += 1
}
}
One strange behaviour: I have several objects and arrays:
for image in images {
for nextID in image.parts {
if nextID.number != 0 {
if let n = primaryLookUp[nextID.number] {
image.parts[0].newID = 0
nextID.newID = 0 // cannot assign!!!
}
}
}
nextID is simply going through the array of .parts. Does ist become a "let" assignment so I couldn't change anything later on? The
image.parts[0].newID = 0
is valid!
I believe this is the explanation for what you are seeing:
The value of a loop variable is immutable, just as if it had been assigned with let.
In your first loop, images is an array of objects (defined by a class). Since a class object is a reference type, you can alter the fields of the object. Only the loop variable image in that case cannot be assigned.
In your second loop, image.parts is an array of structs. Since a struct is a value type, the entire nextID and its fields will be immutable within the loop.
If you add var to the second loop, you will be able to assign to nextID.newID:
for image in images {
for var nextID in image.parts {
if nextID.number != 0 {
if let n = primaryLookUp[nextID.number] {
image.parts[0].newID = 0
nextID.newID = 0 // this now works!!!
}
}
}
}
but, you are changing a copy of the nextID (since structures copies are by value) and you are not changing the nextID contained in the original image object.