我看过一些类似的例子,但所有这些例子似乎都依赖于知道要统计出现次数的元素。我的数组是动态生成的,所以我无法知道要统计哪些元素的出现次数(我想统计它们全部的出现次数)。有人能给予建议吗?
编辑:
也许我应该更清楚一点,数组将包含多个不同的字符串(例如
["FOO", "FOO", "BAR", "FOOBAR"]
)
如何在不事先知道它们是什么的情况下统计foo、bar和foobar的出现次数呢?
我看过一些类似的例子,但所有这些例子似乎都依赖于知道要统计出现次数的元素。我的数组是动态生成的,所以我无法知道要统计哪些元素的出现次数(我想统计它们全部的出现次数)。有人能给予建议吗?
编辑:
也许我应该更清楚一点,数组将包含多个不同的字符串(例如
["FOO", "FOO", "BAR", "FOOBAR"]
)
如何在不事先知道它们是什么的情况下统计foo、bar和foobar的出现次数呢?
Swift 3 和 Swift 2:
您可以使用类型为 [String: Int]
的字典来累计每个 [String]
中的项目计数:
let arr = ["FOO", "FOO", "BAR", "FOOBAR"]
var counts: [String: Int] = [:]
for item in arr {
counts[item] = (counts[item] ?? 0) + 1
}
print(counts) // "[BAR: 1, FOOBAR: 1, FOO: 2]"
for (key, value) in counts {
print("\(key) occurs \(value) time(s)")
}
输出:
BAR occurs 1 time(s)
FOOBAR occurs 1 time(s)
FOO occurs 2 time(s)
Swift 4:
Swift 4 引入 (SE-0165)了向字典查找中添加默认值的能力,而且结果值可通过 +=
和 -=
这样的操作进行改变,因此:
counts[item] = (counts[item] ?? 0) + 1
变成:
counts[item, default: 0] += 1
这使得使用 forEach
在一行中完成计数操作变得容易:
let arr = ["FOO", "FOO", "BAR", "FOOBAR"]
var counts: [String: Int] = [:]
arr.forEach { counts[$0, default: 0] += 1 }
print(counts) // "["FOOBAR": 1, "FOO": 2, "BAR": 1]"
Swift 4: reduce(into:_:)
Swift 4引入了一个新版本的reduce
,它使用一个inout
变量来累积结果。使用它,计数器的创建真正成为了一行代码:
let arr = ["FOO", "FOO", "BAR", "FOOBAR"]
let counts = arr.reduce(into: [:]) { counts, word in counts[word, default: 0] += 1 }
print(counts) // ["BAR": 1, "FOOBAR": 1, "FOO": 2]
或使用默认参数:
let counts = arr.reduce(into: [:]) { $0[$1, default: 0] += 1 }
Sequence
的扩展,以便可以在包含Hashable
项(包括Array
、ArraySlice
、String
和String.SubSequence
)的任何Sequence
上调用它:extension Sequence where Element: Hashable {
var histogram: [Element: Int] {
return self.reduce(into: [:]) { counts, elem in counts[elem, default: 0] += 1 }
}
}
Sequence
而不是Array
以获取其他类型。
示例:
print("abacab".histogram)
["a": 3, "b": 2, "c": 1]
print("Hello World!".suffix(6).histogram)
["l": 1, "!": 1, "d": 1, "o": 1, "W": 1, "r": 1]
print([1,2,3,2,1].histogram)
[2: 2, 3: 1, 1: 2]
print([1,2,3,2,1,2,1,3,4,5].prefix(8).histogram)
[1: 3, 2: 3, 3: 2]
print(stride(from: 1, through: 10, by: 2).histogram)
[1: 1, 3: 1, 5: 1, 7: 1, 9: 1]
0
。第一次遇到字符串时,它不会出现在 counts
字典中,因此 count["FOO"]
将返回 nil
,而 ?? 将其转换为 0
。下次我们遇到 "FOO"
时,count["FOO"]
将返回 Optional(1)
,而 ?? 将其解开为 1
。 - vacawamaarray.filter{$0 == element}.count
var count = 0 array.forEach { x in if x == element { count += 1 }}
。 - Cristina De Rito使用 Swift 5,根据您的需求,您可以选择以下7个Playground示例代码之一来计算数组中可哈希项的出现次数。
Array
的reduce(into:_:)
和Dictionary
的subscript(_:default:)
下标let array = [4, 23, 97, 97, 97, 23]
let dictionary = array.reduce(into: [:]) { counts, number in
counts[number, default: 0] += 1
}
print(dictionary) // [4: 1, 23: 2, 97: 3]
repeatElement(_:count:)
函数、zip(_:_:)
函数和Dictionary
的init(_:uniquingKeysWith:)
初始化器let array = [4, 23, 97, 97, 97, 23]
let repeated = repeatElement(1, count: array.count)
//let repeated = Array(repeating: 1, count: array.count) // also works
let zipSequence = zip(array, repeated)
let dictionary = Dictionary(zipSequence, uniquingKeysWith: { (current, new) in
return current + new
})
//let dictionary = Dictionary(zipSequence, uniquingKeysWith: +) // also works
print(dictionary) // prints [4: 1, 23: 2, 97: 3]
#3.使用字典的初始化器init(grouping: by:)
和方法mapValues(_:)
let array = [4, 23, 97, 97, 97, 23]
let dictionary = Dictionary(grouping: array, by: { $0 })
let newDictionary = dictionary.mapValues { (value: [Int]) in
return value.count
}
print(newDictionary) // prints: [97: 3, 23: 2, 4: 1]
Dictionary
的init(grouping:by:)
初始化器和map(_:)
方法let array = [4, 23, 97, 97, 97, 23]
let dictionary = Dictionary(grouping: array, by: { $0 })
let newArray = dictionary.map { (key: Int, value: [Int]) in
return (key, value.count)
}
print(newArray) // prints: [(4, 1), (23, 2), (97, 3)]
#5. 使用 for 循环和 Dictionary 的 subscript(_:) 下标extension Array where Element: Hashable {
func countForElements() -> [Element: Int] {
var counts = [Element: Int]()
for element in self {
counts[element] = (counts[element] ?? 0) + 1
}
return counts
}
}
let array = [4, 23, 97, 97, 97, 23]
print(array.countForElements()) // prints [4: 1, 23: 2, 97: 3]
#6. 使用 NSCountedSet
和 NSEnumerator
的 map(_:)
方法(需要 Foundation)import Foundation
extension Array where Element: Hashable {
func countForElements() -> [(Element, Int)] {
let countedSet = NSCountedSet(array: self)
let res = countedSet.objectEnumerator().map { (object: Any) -> (Element, Int) in
return (object as! Element, countedSet.count(for: object))
}
return res
}
}
let array = [4, 23, 97, 97, 97, 23]
print(array.countForElements()) // prints [(97, 3), (4, 1), (23, 2)]
#7. 使用NSCountedSet
和AnyIterator
(需要Foundation库)import Foundation
extension Array where Element: Hashable {
func counForElements() -> Array<(Element, Int)> {
let countedSet = NSCountedSet(array: self)
var countedSetIterator = countedSet.objectEnumerator().makeIterator()
let anyIterator = AnyIterator<(Element, Int)> {
guard let element = countedSetIterator.next() as? Element else { return nil }
return (element, countedSet.count(for: element))
}
return Array<(Element, Int)>(anyIterator)
}
}
let array = [4, 23, 97, 97, 97, 23]
print(array.counForElements()) // [(97, 3), (4, 1), (23, 2)]
致谢:
我更新了oisdk的回答以适用于Swift2。
16/04/14 我将此代码更新为Swift2.2。
16/10/11 更新为Swift3。
可哈希:
extension Sequence where Self.Iterator.Element: Hashable {
private typealias Element = Self.Iterator.Element
func freq() -> [Element: Int] {
return reduce([:]) { (accu: [Element: Int], element) in
var accu = accu
accu[element] = accu[element]?.advanced(by: 1) ?? 1
return accu
}
}
}
可比较的:
extension Sequence where Self.Iterator.Element: Equatable {
private typealias Element = Self.Iterator.Element
func freqTuple() -> [(element: Element, count: Int)] {
let empty: [(Element, Int)] = []
return reduce(empty) { (accu: [(Element, Int)], element) in
var accu = accu
for (index, value) in accu.enumerated() {
if value.0 == element {
accu[index].1 += 1
return accu
}
}
return accu + [(element, 1)]
}
}
}
使用方法
let arr = ["a", "a", "a", "a", "b", "b", "c"]
print(arr.freq()) // ["b": 2, "a": 4, "c": 1]
print(arr.freqTuple()) // [("a", 4), ("b", 2), ("c", 1)]
for (k, v) in arr.freq() {
print("\(k) -> \(v) time(s)")
}
// b -> 2 time(s)
// a -> 4 time(s)
// c -> 1 time(s)
for (element, count) in arr.freqTuple() {
print("\(element) -> \(count) time(s)")
}
// a -> 4 time(s)
// b -> 2 time(s)
// c -> 1 time(s)
var
。我尝试着修复它,但到目前为止还没有成功... - MassivePenguinNSCountedSet* countedSet = [[NSCountedSet alloc] initWithArray:array];
for (NSString* string in countedSet)
NSLog (@"String %@ occurs %zd times", string, [countedSet countForObject:string]);
func freq<S: SequenceType where S.Generator.Element: Hashable>(seq: S) -> [S.Generator.Element:Int] {
return reduce(seq, [:]) {
(var accu: [S.Generator.Element:Int], element) in
accu[element] = accu[element]?.successor() ?? 1
return accu
}
}
freq(["FOO", "FOO", "BAR", "FOOBAR"]) // ["BAR": 1, "FOOBAR": 1, "FOO": 2]
这个操作是通用的,只要您的元素可以散列,它就可以使用:
freq([1, 1, 1, 2, 3, 3]) // [2: 1, 3: 2, 1: 3]
freq([true, true, true, false, true]) // [false: 1, true: 4]
如果您不能使元素可哈希,则可以使用元组来实现:
func freq<S: SequenceType where S.Generator.Element: Equatable>(seq: S) -> [(S.Generator.Element, Int)] {
let empty: [(S.Generator.Element, Int)] = []
return reduce(seq, empty) {
(var accu: [(S.Generator.Element,Int)], element) in
for (index, value) in enumerate(accu) {
if value.0 == element {
accu[index].1++
return accu
}
}
return accu + [(element, 1)]
}
}
freq(["a", "a", "a", "b", "b"]) // [("a", 3), ("b", 2)]
Swift 4
let array = ["FOO", "FOO", "BAR", "FOOBAR"]
// Merging keys with closure for conflicts
let mergedKeysAndValues = Dictionary(zip(array, repeatElement(1, count: array.count)), uniquingKeysWith: +)
// mergedKeysAndValues is ["FOO": 2, "BAR": 1, "FOOBAR": 1]
我喜欢避免使用内部循环,尽可能多地使用.map。因此,如果我们有一个字符串数组,我们可以按照以下方式计算出现次数:
var occurances = ["tuples", "are", "awesome", "tuples", "are", "cool", "tuples", "tuples", "tuples", "shades"]
var dict:[String:Int] = [:]
occurances.map{
if let val: Int = dict[$0] {
dict[$0] = val+1
} else {
dict[$0] = 1
}
}
["tuples": 5, "awesome": 1, "are": 2, "cool": 1, "shades": 1]
var numberOfOccurenses = countedItems.filter(
{
if $0 == "FOO" || $0 == "BAR" || $0 == "FOOBAR" {
return true
}else{
return false
}
}).count
func checkItemCount(arr: [String]) {
var dict = [String: Any]()
for x in arr {
var count = 0
for y in arr {
if y == x {
count += 1
}
}
dict[x] = count
}
print(dict)
}
let arr = ["FOO", "FOO", "BAR", "FOOBAR"]
checkItemCount(arr: arr)