Ios 如何计算 Swift 数组中元素的出现次数

Swift 3 and Swift 2:个

您可以使用类型为[String: Int]的字典为[String]中的每个项目建立计数:

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:个

SWIFT 4 introduces (SE-0165)能够在字典查找中包含默认值，结果值可以通过+=和-=等操作进行变异，因此:

counts[item] = (counts[item] ?? 0) + 1

变成:

counts[item, default: 0] += 1

这使得使用forEach在一行简明的代码中执行计数操作变得很容易:

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: 100

SWIFT 4引入了新版本的reduce，它使用inout变量来累加结果.使用这一点，计数的创建就真正变成了一行:

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 }

最后，您可以将其作为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 }
    }
}

这个 idea 是从this question借来的，不过我改成了computed property.感谢@LeoDabus建议扩展Sequence而不是Array来获取额外的类型.

Examples:个

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]