swift:计算字典中的重复值

最直接的方法是使用为此目的而设计的构造:计数集。没有本机Swift计数集,但您可以使用 NSCountedSet .

let dict = [
    "key1": true,
    "key2": true,
    "key3": false
]

let countedSet = NSCountedSet()
for (_, value) in dict {
    countedSet.add(value)
}
print("Count for true: \(countedSet.count(for: true))")

使用 filter 方法删除不需要的值,然后调用 count

// Get the count of everything which is true
let trueCount = dict.filter { $0.value }.count

// Get the count of everything which is false
let falseCount = dict.filter { !$0.value }.count

// A more efficient way to get the count of everything which is false
let falseCount = dict.count - trueCount

您可以使用高阶函数,如 reduce :

let trueFalstDict: [String: Bool] = ["id1": false, "id2": true, "id3": false, "id4": false]

var trueFalseCount: (trues: Int, falses: Int)

trueFalseCount.trues = trueFalstDict.reduce(0) { $0 + ($1.value ? 1 : 0) }
trueFalseCount.falses = trueFalstDict.reduce(0) { $0 + ($1.value ? 0 : 1) }

print(trueFalseCount) // (trues: 1, falses: 3)

使用@DeanWomberne的建议:

let trueFalstDict: [String: Bool] = ["id1": false, "id2": true, "id3": false, "id4": false]

var trueFalseCount: (trues: Int, falses: Int)

trueFalseCount = trueFalstDict.reduce((trues: 0, falses: 0)) {
    return (
        true: $0.trues + ($1.value ? 1 : 0),
        false: $0.falses + ($1.value ? 0 : 1)
    )
}

print(trueFalseCount) // (trues: 1, falses: 3)

let dict = [
    1: true,
    2: false,
    3: false,
    4: false
]

使用 NSCountedSet :

let numTrue = NSCountedSet(dict.values).count(for: true)
let numFalse = dict.count - numTrue

使用 reduce

let numTrue = dict.values.reduce(0) { $0 + ($1 ? 1 : 0) }
let numFalse = dict.count - numTrue

这两种方法都利用了以下事实: Bool 仅允许两个不同的值,允许简单计算,例如 false true 计数(反之亦然)。

例子:

var currencyDB:[(name: String, value: Double, isShows: Bool, isMain: Bool)] = 
[("USD",1.1588, true, true), 
("JPY",129.30, true, false), 
("BGN",1.9558, true, false), 
("CZK",25.657, true, false), 
("DKK",7.4526, false, false), 
("GBP",0.8905, false, false)]

print(currencyDB.map{ $0.2 }.filter { $0 }.count))  // print: 4 -> trues
print(currencyDB.map{ $0.2 }.filter { !$0 }.count)) // print: 2 -> falses