Swift 快速减少数组中的元组
是否有一种方法可以调用元组数组上的reduce函数来查找最大值 比如说Swift 快速减少数组中的元组,swift,Swift,是否有一种方法可以调用元组数组上的reduce函数来查找最大值 比如说 struct Gate{ var maxedOpenGates = 1 var GatesOpen : [(openOrdered : Int, gateNum : Int )] = [] init(defaultSelected : Int = 0){ GatesOpen.append(openOrdered : 1, gateNum : defaultSelected)
struct Gate{
var maxedOpenGates = 1
var GatesOpen : [(openOrdered : Int, gateNum : Int )] = []
init(defaultSelected : Int = 0){
GatesOpen.append(openOrdered : 1, gateNum : defaultSelected)
}
private func manipulateGates(gates : [SegmentButton]){
for i in GatesOpen{
gates[i.gateNum].toggleSelected()
}
}
mutating func openGate(index : Int, buttons : [SegmentButton]){
if GatesOpen.count < maxedOpenGates{
GatesOpen.append( openOrdered: GatesOpen.count , gateNum: index)
}else{
//////Finding the gate that was Opened the longest////
let lastGate = GatesOpen.reduce(Int.min,{ max($0,$1) })
}
manipulateGates(buttons)
}
}
是的,您可以像任何其他数组一样减少元组数组。 combine函数的参数是数组的元素,在 案例“门元组”。如果减少数组的结果也应该是门元组,那么初始元素和combine函数的结果 也必须是该类型:
let lastGate = gatesOpen.reduce(gatesOpen[0]) {
$0.openOrdered < $1.openOrdered ? $1 : $0
}
使用带有reduce函数的闭包
//////Finding the gate that was Opened the longest////
let intitalValue = (Int.min,Int.min)
let lastGate: (openOrdered: Int, gateNum: Int) = GatesOpen.reduce(intitalValue){
(longestOpenGate: (openOrdered: Int, gateNum: Int), gate) in
if(longestOpenGate.openOrdered < gate.openOrdered) {
return gate;
} else {
return longestOpenGate;
}
}
let lastGateNumber = lastGate.gateNum;
//查找打开时间最长的门////
设intitalValue=(Int.min,Int.min)
让lastGate:(openOrdered:Int,gateNum:Int)=gatespen.reduce(intitalValue){
(longestOpenGate:(openOrdered:Int,gateNum:Int),gate)在
if(longestOpenGate.openOrdered
或者
let lastGate:(openOrdered:Int,gateNum:Int)=gatespen.reduce(intitalValue){
$0.openOrdered<$1.openOrdered?$1:$0
}
让lastGateNumber=lastGate.gateNum;
将元组数组缩减为数组:
我发现
dropFirst(gatespen)
比手动切片操作好一点。如果您想用一个可选参数来解释空数组的可能性,那么可以使用first.map
在数组为空时返回nil
:let lastGate=gatesOpen.first.map{fst in dropFirst(gatesOpen).reduce(fst){$0.openOrdered1.openOrdered?$1:$0}}
@AirspeedVelocity:dropFirst的好主意我还考虑过处理一个空数组(你的建议非常优雅!),但是从给定的代码来看,似乎至少有一个“门”,所以我不想让事情复杂化。谢谢你的反馈。非常好!上面的解释和示例的补充极大地帮助我理解reduce函数和其他模拟这种设计的函数。您能否演示如何使用别名代替$0
/$1
?@javadba{acculator,element in…}
Thx。.0
在$1.0
中是什么意思?0$,1$表示第一个和第二个闭包元素,.0和.1表示闭包元素中的第一个和第二个项目。这是标准的swift语法tho。✌️
let lastGate = dropFirst(gatesOpen).reduce(gatesOpen[0]) {
$0.openOrdered < $1.openOrdered ? $1 : $0
}
//////Finding the gate that was Opened the longest////
let intitalValue = (Int.min,Int.min)
let lastGate: (openOrdered: Int, gateNum: Int) = GatesOpen.reduce(intitalValue){
(longestOpenGate: (openOrdered: Int, gateNum: Int), gate) in
if(longestOpenGate.openOrdered < gate.openOrdered) {
return gate;
} else {
return longestOpenGate;
}
}
let lastGateNumber = lastGate.gateNum;
let lastGate: (openOrdered: Int, gateNum: Int) = GatesOpen.reduce(intitalValue){
$0.openOrdered < $1.openOrdered ? $1 : $0
}
let lastGateNumber = lastGate.gateNum;
let arr:[Int] = [(0,""),(1,"")].reduce([]) { $0 + [$1.0] }//Ouput:[0, 1]