Question

我有两个这样的数组：

var arrayA = ["Mike", "James", "Stacey", "Steve"]
var arrayB = ["Steve", "Gemma", "James", "Lucy"]

如您所见，James和Steve匹配，我希望能够将其从arrayA中删除。我怎么写这个？

Answer 1

@ francesco-vadicamo的回答 Swift 2/3/4 +

 arrayA = arrayA.filter { !arrayB.contains($0) }

Answer 2

最简单的方法是使用新的Set容器（在Swift 1.2 / Xcode 6.3中添加）：

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)

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：

arrayA = Array(intersection)

缺点是您必须创建2个新数据集。请注意，intersect不会改变它所调用的实例，它只返回一个新集。

有类似的方法可以添加，减去等，你可以看看它们

Answer 3

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"]

Answer 4

我同意Antonio的回答，但是对于小数组减法，你也可以使用这样的过滤器闭包：

let res = arrayA.filter { !contains(arrayB, $0) }

Answer 5

matt和freytag的解决方案是唯一能够解决重复问题的解决方案，并且应该比其他答案获得更多的+1。

以下是对于Swift 3.0的matt的答案的更新版本：

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)
    }
}

Answer 6

这也可以实现为减号：

u'2016\u5e741\u67081\u65e5'

现在你可以使用

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
}

Answer 7

使用安东尼奥提到的Array → Set → Array方法，并且操作员的方便，正如弗雷塔格所指出的那样，我对此非常满意：

// Swift 3.x/4.x
func - <Element: Hashable>(lhs: [Element], rhs: [Element]) -> [Element]
{
    return Array(Set<Element>(lhs).subtracting(Set<Element>(rhs)))
}

Answer 8

使用索引数组删除元素：

字符串和索引数组

let animals = ["cats", "dogs", "chimps", "moose", "squarrel", "cow"]
let indexAnimals = [0, 3, 4]
let arrayRemainingAnimals = animals
    .enumerated()
    .filter { !indexAnimals.contains($0.offset) }
    .map { $0.element }

print(arrayRemainingAnimals)

//result - ["dogs", "chimps", "cow"]

整数和索引数组

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($0.offset) }
    .map { $0.element }

print(numbers)

//result - [0, 1, 2, 4, 6, 7, 9, 10, 11]

使用其他数组的元素值删除元素

整数数组

let arrayResult = numbers.filter { element in
    return !indexesToRemove.contains(element)
}
print(arrayResult)

//result - [0, 1, 2, 4, 6, 7, 9, 10, 11]

字符串数组

let arrayLetters = ["a", "b", "c", "d", "e", "f", "g", "h", "i"]
let arrayRemoveLetters = ["a", "e", "g", "h"]
let arrayRemainingLetters = arrayLetters.filter {
    !arrayRemoveLetters.contains($0)
}

print(arrayRemainingLetters)

//result - ["b", "c", "d", "f", "i"]

Answer 9

对于较小的数组，我使用：

let date = new Date();

从基于另一个数组的数组中删除对象

9 个答案: