To sort the map, we use a series of operations executed in a single line:val result = capitals.toList().sortedBy { (_, value) -> value}.toMap()First, capitals is converted to a list using toList(). Then, sortedBy() is used to sort the list by value { (_, value) -> value...
How to sort a map by value in Kotlin: Kotlinmapis used to holdkey-valuepairs and using akey, we can access thevaluethat is associated with it. Thekeysof a Kotlin map should be unique. We can’t have duplicate keys, but different keys can have the same value. In this post, we will...
The simplest way to sort aLinkedHashMapby values is to convert it into a list of key-value pairs. After that, we sort the list based on the values. Then, we create a newLinkedHashMapusing the sorted list. Let’s have a look at an example: ...
funmain(args:Array<String>){val mapList=mutableListOf(1to"A",2to"B",5to"C",3to"D")mapList.sortBy{it.first}println(mapList)// [(1, A), (2, B), (3, D), (5, C)]mapList.sortBy{it.second}println(mapList)// [(1, A), (2, B), (5, C), (3, D)]} 以上代码的结...
sortBy和sortByDescendingMap的使用: java中,map提供了key和value, kotline写法:var map:Map<String,String> // 第一种 map= mapOf("a" to "65","b" to "66","c" to "67" ); // map= mapOf(Pair("A","96"),Pair("B","97"),Pair("C","99")) 如果是可变MutableMap,依旧可以通过add...
sortBy{it.length} (4)、Map/MutableMap A: 特点 以键值对的形式存储元素,键唯一 B: 初始化 调用mapOf / mutableMapOf 执行初始化,在组织键值对元素时都有如下两种方式: 键to 值 Pair(键,值) 代码语言:javascript 代码运行次数:0 运行 AI代码解释 val userMap:Map<String,String>=mapOf("姓名" to “...
var map: MutableMap<Int, String> = mutableMapOf( 1001 to "Tom", 1002 to "Mary" ) 3.2 增删改查3.2.1 增加元素 1)putfun main() { var map: MutableMap<Int, String> = mutableMapOf() map.put(1001, "Tom") } 2)map[key] = value...
lastChar() = get(length - 1) //为 Int 类声明一个扩展函数 doubleValue() ,用于返回其两倍值 //this 关键字代表了 Int 值本身 fun Int.doubleValue() = this * 2 之后,我们就可以像调用类本身内部声明的方法一样,直接调用扩展函数 fun main() { val name = "leavesC" println("$name lastChar...
map使用remove删除元素,remove传入想删除的键值 Java val map = mutableMapOf<Number,String>(1to"a",2to"b")map.remove(2)map.forEach{println("${it.key} ${it.value}")// 1 a} list集合相关操作 按索引取元素 list本质就可以理解为一个变长数组,最简单的就是像数组一样通过[]拿到值 ...
sortedByDescending,根据指定的值进行降序排序 AI检测代码解析 fun <T> Array<T>.toString1(): String { val sb = StringBuilder("[") for ((index, value) in this.withIndex()) { sb.append(value) sb.append(",") } sb.delete(sb.length - 1, sb.length) ...