问题
Is there no easy way to remove a specific element from an array, if it is equal to a given string? The workarounds are to find the index of the element of the array you wish to remove, and then removeAtIndex
, or to create a new array where you append all elements that are not equal to the given string. But is there no quicker way?
回答1:
You can use filter() to filter your array as follow
var strings = ["Hello","Playground","World"]
strings = strings.filter{$0 != "Hello"}
print(strings) // "[Playground, World]"
edit/update:
Xcode 10 • Swift 4.2 or later
You can use the new RangeReplaceableCollection
mutating method called removeAll(where:)
var strings = ["Hello","Playground","World"]
strings.removeAll { $0 == "Hello" }
print(strings) // "[Playground, World]"
If you need to remove only the first occurrence of an element we ca implement a custom remove method on RangeReplaceableCollection
constraining the elements to Equatable:
extension RangeReplaceableCollection where Element: Equatable {
@discardableResult
mutating func removeFirstElementEqual(to element: Element) -> Element? {
guard let index = firstIndex(of: element) else { return nil }
return remove(at: index)
}
}
Or using a predicate for non Equatable elements:
extension RangeReplaceableCollection {
@discardableResult
mutating func removeFirstElement(where predicate: @escaping ((Element) -> Bool)) -> Element? {
guard let index = firstIndex(where: predicate) else { return nil }
return remove(at: index)
}
}
var strings = ["Hello","Playground","World"]
strings.removeFirstElementEqual(to: "Hello")
print(strings) // "[Playground, World]"
strings.removeFirstElement { $0 == "Playground" }
print(strings) // "[World]"
回答2:
It's not clear if by quicker you mean in terms of execution time or amount of code.
In the latter case you can easily create a copy using the filter
method. For example, given the following array:
let array = ["1", "2", "3", "4", "5"]
you can create a copy with all elements but "2" as:
let filteredArray = array.filter { $0 != "2" }
回答3:
Using filter like suggested above is nice. But if you want to remove only one occurrence of a value or you assume there are no duplicates in the array and you want a faster algorithm, use this (in Swift3):
if let index = array.index(of: "stringToRemove") {
array.remove(at: index)
} else {
// not found
}
回答4:
You'll want to use filter()
. If you have a single element (called say obj
) to remove, then the filter()
predicate will be { $0 != obj }
. If you do this repeatedly for a large array this might be a performance issue. If you can defer removing individual objects and want to remove an entire sub-array then use something like:
var stringsToRemove : [String] = ...
var strings : [String] = ...
strings.filter { !contains(stringsToRemove, $0) }
for example:
1> ["a", "b", "c", "d"].filter { !contains(["b", "c"], $0) }
$R5: [String] = 2 values {
[0] = "a"
[1] = "d"
}
回答5:
You could use filter() in combination with operator overloading to produce an easily repeatable solution:
func -= (inout left: [String], right: String){
left = left.filter{$0 != right}
}
var myArrayOfStrings:[String] = ["Hello","Playground","World"]
myArrayOfStrings -= "Hello"
print(myArrayOfStrings) // "[Playground, World]"
回答6:
if you need to delete subArray from array then this is a perfect solution using Swift3:
var array = ["a", "b", "c", "d", "e", "f", "g", "h", "i", "j"]
let subArrayToDelete = ["c", "d", "e", "ee"]
array = array.filter{ !subArrayToDelete.contains($0) }
print(array) // ["a", "b", "f", "g", "h", "i", "j"]
this is better for your performance rather than deleting one by one.
btw even faster solution is (but it will rearrange items in the final array):
array = Array(Set(array).subtracting(subArrayToDelete))
来源:https://stackoverflow.com/questions/27878798/remove-specific-array-element-equal-to-string-swift