如何從 swift 中的浮點數數組中獲取前 2 個最大值？

Question

我需要從浮點數數組中獲取 2 個最大值，即最高和第二高。是否有任何簡單的方法來獲取它們的索引，或者我是否需要為此更改數組的結構？

Answer 1

只需對數組進行排序並獲取所需的值

var array1 = [2.1, 2.2, 2.5, 3.0, 4.2, 2]

var array2 = array1.sort(){ $0 > $1}

//One way 
let firstMax = array2[0]
let secondMax = array2[1]

//Second way
let firstMax = array2.removeFirst()
let secondMax = array2.removeFirst()

編輯

如果你想要索引只是讓它們像

let maxPos = array1.indexOf(firstMax)
let secMaxPos = array1.indexOf(secondMax)

如果您對這些東西感到困惑，請遵循以下正常基礎知識。

var max = -1.0, maxPos = -1, secMax = -1.0, secMaxPos = -1
for (index, value) in array1.enumerate() {
    if value > max {
        max = value
        maxPos = index
    } else if value > secMax {
        secMax = value
        secMaxPos = index
    }
}
print("max:\(max)->pos:\(maxPos)::secMax:\(secMax)->secMaxPos:\(secMaxPos)") 

Answer 2

您可以使用enumerate()創建一個包含 (index, value) 的元組數組，然后按值對該數組進行排序以找到兩個最大值：

let arr: [Float] = [1.2, 3.14, 1.609, 2.718, 0.3]

// Create an array of (index, value) tuples sorted by value
// in decreasing order
let result = arr.enumerate().sort { $0.1 > $1.1 }
print(result)

 [(1, 3.1400001), (3, 2.71799994), (2, 1.60899997), (0, 1.20000005), (4, 0.300000012)]

let (topIndex, top) = result[0]
print("top = \(top), index = \(topIndex)")

 top = 3.14, index = 1

let (secondIndex, second) = result[1]
print("second = \(second), index = \(secondIndex)")

 second = 2.718, index = 3

Answer 3

import Algorithms

[1, 3, 5, 6, 4, 2].max(count: 2) // [5, 6]

https://github.com/apple/swift-algorithms/blob/main/Guides/MinMax.md

Answer 4

試試你的游戲

var item = [2.01 ,3.95,1.85,2.65,1.6]

var sorted = item.sort({ $0 > $1 })

打印（排序）

Answer 5

基本思想是遍歷數組並挑選出最大和第二大的項目。

一些可能難以閱讀的短代碼：

let myFloatArray : [Float] = ...
let top2 = myFloatArray.enumerated().reduce(((-1, Float.nan), (-1, Float.nan)), combine: { t, v in
    // Check if value is larger than first item in tuple
    if (!(v.1 <= t.0.1)) {
        // Check if value is larger than second item in tuple
        if (!(v.1 <= t.1.1)) {
            // Return with new value as largest
            return (t.1, v)
        } else {
            // Return with new value as next largest
            return (v, t.1)
        }
    }
    // Return old result
    return t
})

或者使用更明確的變量名稱：

var largestIndex = -1;
var largestValue = Float.nan;
var secondLargestIndex = -1;
var secondLargestValue = Float.nan;
for index in 0..<myFloatArray.count {
    let value = myFloatArray[index];
    if (!(value <= secondLargestValue)) {
        if (!(value <= largestValue)) {
            secondLargestValue = largestValue;
            secondLargestIndex = largestIndex;
            largestValue = value;
            largestIndex = index;
        } else {
            secondLargestValue = value;
            secondLargestIndex = index;
        }
    }
}

Answer 6

var array1 = [1,2,3,4,5]

var array2 = array1.sorted(){ $0 > $1}

打印（數組2）

//一種方式

讓 firstMax = array2[0]

讓 secondMax = array2[1]

//回答

打印（firstMax）

打印（第二個最大）

Answer 7

func secondLargest(arr: [Int]){
var max1 = -1;
var max2 = -1;
for item in arr {
    if item > max1 {
        max2 = max1;
        max1 = item
    }else if item > max2 {
        max2 = item
    }
}
}

Answer 8

 import UIKit var array = [6,4,3,5,9,7,8,2,10] var maxValue = 0, temp = 0, maxIndex = 0 for (index, item) in array.enumerated() { if (item > temp){ maxIndex = index } } array.remove(at: maxIndex) for (_, item) in array.enumerated() { if (item > maxValue){ maxValue = item } } print("---------") print(maxValue) print("---------")

Answer 9

var numberArray = [4.4,5.3,3.2,2.1,6.0,1.2,9.6,8.0,9.4]
var secontLargestValue = 0.0
var firstLargestValue = 0.0

for value in numberArray {
    if value > firstLargestValue {
        secontLargestValue = firstLargestValue
        firstLargestValue = value
    } else if (value > secontLargestValue && value != firstLargestValue) 
    {
        secontLargestValue = value
    }
}

print("First largest value - \(firstLargestValue)")
print("Second largest value - \(secontLargestValue)")