如何在 Swift 的布尔数组中找到多个 True 语句

问题描述

我是一名新开发人员，似乎无法弄清楚如何在布尔数组中找到 True 语句的数量.我知道如何按索引而不是按值查找.任何帮助将不胜感激.

I am a new developer and cannot seem to figure out how to find the number of True statements in an Array of Booleans. I know how to find by index but not by value. Any assistance would be appreciated.

let arrayElement = [Bool](repeating: false, count: 10)
var before: [[Bool]] = [[Bool]](repeating: arrayElement, count:10)
for i in 0 ..< 10 {
    for j in 0 ..< 10 {
        if arc4random_uniform(3) == 1 {
            before[i][j] = true  
        }        
    }
}

推荐答案

计算一维数组中true条目数的方法

一种方法是过滤您的 Bool 元素数组(对于 true)并简单地计算过滤后数组中剩余元素的数量

Methods for counting number of true entries in a 1D array

One method would be to filter your array of Bool elements (for true) and simply count the number of remaining elements in the filtered array

let arr = [false, true, true, false, true]
let numberOfTrue = arr.filter{$0}.count
print(numberOfTrue) // 3

另一种方法是reduce(展开)数组并为每个等于true

Another approach is to reduce (unfold) the array and increment a counter for each element that equals true

let arr = [false, true, true, false, true]
let numberOfTrue = arr.reduce(0) { $0 + ($1 ? 1 : 0) }
print(numberOfTrue) // 3

或者，传统的 for 循环(带有条件循环签名)方法，与 reduce 方法相当:

Or, a traditional for loop (with conditional in loop signature) approach, comparable top the reduce method:

let arr = [false, true, true, false, true]
var trueCounter = 0
for bElem in arr where bElem { trueCounter += 1 }
print(trueCounter) // 3

应用于你的例子:使用joined()实现一维数组

通过简单地应用 .joined() 在 [[Bool]] 数组上依次构造一个 [Bool] 数组.

Applied to your example: use joined() to achieve a 1D array

The methods above can easily be applied to an array of arrays (of Bool elements: type [[Bool]]) by simply applying .joined() on the [[Bool]] array to sequentially construct a [Bool] array.

/* 'before' is of type [[Bool]], constructed as described
   in the question */
let numberOfTrueAlt1 = before.joined().filter{$0}.count

let numberOfTrueAlt2 = before.joined().reduce(0) { $0 + ($1 ? 1 : 0) }

var numberOfTrueAlt3 = 0
for bElem in before.joined() where bElem { numberOfTrueAlt3 += 1 }

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