xcode 基于另一个数组从数组中删除对象
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Removing objects from an array based on another array
提问by Henry Brown
I have two arrays like this:
我有两个这样的数组：
var arrayA = ["Mike", "James", "Stacey", "Steve"]
var arrayB = ["Steve", "Gemma", "James", "Lucy"]
As you can see, Jamesand Stevematch and I want to be able to remove them from arrayA. How would I write this?
正如你所看到的，James和Steve比赛，我希望能够从删除它们arrayA。我会怎么写这个？
采纳答案by matt
Like this:
像这样：
var arrayA = ["Mike", "James", "Stacey", "Steve"]
var arrayB = ["Steve", "Gemma", "James", "Lucy"]
for word in arrayB {
if let ix = find(arrayA, word) {
arrayA.removeAtIndex(ix)
}
}
// now arrayA is ["Mike", "Stacey"]
回答by Federico Zanetello
@francesco-vadicamo's answer in Swift 2/3/4+
@francesco-vadicamo 在Swift 中的回答2/3/4+
arrayA = arrayA.filter { !arrayB.contains(var setA = Set(arrayA)
var setB = Set(arrayB)
// Return a set with all values contained in both A and B
let intersection = setA.intersect(setB)
// Return a set with all values in A which are not contained in B
let diff = setA.subtract(setB)
) }
回答by Antonio
The easiest way is by using the new Setcontainer (added in Swift 1.2 / Xcode 6.3):
最简单的方法是使用新Set容器（在 Swift 1.2 / Xcode 6.3 中添加）：
arrayA = Array(intersection)
If you want to reassign the resulting set to arrayA, simply create a new instance using the copy constructor and assign it to arrayA:
如果要将结果集重新分配给arrayA，只需使用复制构造函数创建一个新实例并将其分配给arrayA：
let res = arrayA.filter { !contains(arrayB, var arrayA = ["Mike", "James", "Stacey", "Steve"]
var arrayB = ["Steve", "Gemma", "James", "Lucy"]
for word in arrayB {
if let ix = arrayA.index(of: word) {
arrayA.remove(at: ix)
}
}
) }
The downside is that you have to create 2 new data sets.
Note that intersectdoesn't mutate the instance it is invoked in, it just returns a new set.
缺点是您必须创建 2 个新数据集。请注意，intersect它不会改变调用它的实例，它只会返回一个新集合。
There are similar methods to add, subtract, etc., you can take a look at them
还有类似的加减法等方法，大家可以看看
回答by Francesco Vadicamo
I agree with Antonio's answer, however for small array subtractions you can also use a filter closure like this:
我同意安东尼奥的回答，但是对于小数组减法，您也可以使用这样的过滤器闭包：
func -<T:RangeReplaceableCollectionType where T.Generator.Element:Equatable>( lhs:T, rhs:T ) -> T {
var lhs = lhs
for element in rhs {
if let index = lhs.indexOf(element) { lhs.removeAtIndex(index) }
}
return lhs
}
回答by Ruiz
matt and freytag's solutions are the ONLY ones that account for duplicates and should be receiving more +1s than the other answers.
matt 和 freytag 的解决方案是唯一考虑重复的解决方案，并且应该比其他答案获得更多的 +1。
Here is an updated version of matt's answer for Swift 3.0:
这是马特对 Swift 3.0 的回答的更新版本：
arrayA - arrayB
回答by freytag
Original answer
原答案
This can also be implemented as a minus func:
这也可以实现为减函数：
func -<T: RangeReplaceableCollection>(lhs: T, rhs: T) -> T where T.Iterator.Element: Equatable {
var lhs = lhs
for element in rhs {
if let index = lhs.firstIndex(of: element) { lhs.remove(at: index) }
}
return lhs
}
Now you can use
现在你可以使用
// Swift 3.x/4.x
func - <Element: Hashable>(lhs: [Element], rhs: [Element]) -> [Element]
{
return Array(Set<Element>(lhs).subtracting(Set<Element>(rhs)))
}
Updated implementation for Swift 5
Swift 5 的更新实现
let animals = ["cats", "dogs", "chimps", "moose", "squarrel", "cow"]
let indexAnimals = [0, 3, 4]
let arrayRemainingAnimals = animals
.enumerated()
.filter { !indexAnimals.contains(var numbers = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
let indexesToRemove = [3, 5, 8, 12]
numbers = numbers
.enumerated()
.filter { !indexesToRemove.contains(let animals = ["cats", "dogs", "chimps", "moose", "squarrel", "cow"]
let indexAnimals = [0, 3, 4]
let arrayRemainingAnimals = animals
.enumerated()
.filter { !indexAnimals.contains(var numbers = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
let indexesToRemove = [3, 5, 8, 12]
numbers = numbers
.enumerated()
.filter { !indexesToRemove.contains(let arrayResult = numbers.filter { element in
return !indexesToRemove.contains(element)
}
print(arrayResult)
//result - [0, 1, 2, 4, 6, 7, 9, 10, 11]
.offset) }
.map { let arrayLetters = ["a", "b", "c", "d", "e", "f", "g", "h", "i"]
let arrayRemoveLetters = ["a", "e", "g", "h"]
let arrayRemainingLetters = arrayLetters.filter {
!arrayRemoveLetters.contains(let arrayResult = numbers.filter { element in
return !indexesToRemove.contains(element)
}
print(arrayResult)
//result - [0, 1, 2, 4, 6, 7, 9, 10, 11]
)
}
print(arrayRemainingLetters)
//result - ["b", "c", "d", "f", "i"]
.element }
print(numbers)
//result - [0, 1, 2, 4, 6, 7, 9, 10, 11]
.offset) }
.map { let arrayLetters = ["a", "b", "c", "d", "e", "f", "g", "h", "i"]
let arrayRemoveLetters = ["a", "e", "g", "h"]
let arrayRemainingLetters = arrayLetters.filter {
!arrayRemoveLetters.contains(/* poormans sub for Arrays */
extension Array where Element: Equatable {
static func -=(lhs: inout Array, rhs: Array) {
rhs.forEach {
if let indexOfhit = lhs.firstIndex(of: ##代码##) {
lhs.remove(at: indexOfhit)
}
}
}
static func -(lhs: Array, rhs: Array) -> Array {
return lhs.filter { return !rhs.contains(##代码##) }
}
}
)
}
print(arrayRemainingLetters)
//result - ["b", "c", "d", "f", "i"]
.element }
print(arrayRemainingAnimals)
//result - ["dogs", "chimps", "cow"]
.offset) }
.map { ##代码##.element }
print(numbers)
//result - [0, 1, 2, 4, 6, 7, 9, 10, 11]
.offset) }
.map { ##代码##.element }
print(arrayRemainingAnimals)
//result - ["dogs", "chimps", "cow"]
回答by AmitaiB
Using the Array → Set → Arraymethod mentioned by Antonio, and with the convenience of an operator, as freytag pointed out, I've been very satisfied using this:
使用Array → Set → ArrayAntonio 提到的方法，并借助操作员的便利，正如 freytag 指出的那样，我对使用此方法感到非常满意：
回答by Krunal
Remove elements using indexes array:
使用索引数组删除元素：
Array of Strings and indexes
##代码##Array of Integers and indexes
##代码##
字符串和索引数组
##代码##整数和索引数组
##代码##
Remove elements using element value of another array
使用另一个数组的元素值删除元素
Arrays of integers
##代码##Arrays of strings
##代码##
整数数组
##代码##字符串数组
##代码##
回答by JGS
For smaller arrays I use:
对于较小的数组，我使用：
##代码##