我正在使用Swift中的默认排序算法。但是对于10000(10k)记录,iPad上需要将近1分钟的时间。
private var fd: [[String:AnyObject]]? = [[String:AnyObject]]()
这是我的下降代码
self.fd = self.fd!.sort ({ (r1, r2) -> Bool in
let t = self.headers![b.tag - 10]
if r1[t.title] is String {
let dateFormatter = NSDateFormatter()
dateFormatter.dateFormat = "dd-MMM-yyyy"
if dateFormatter.dateFromString("(r1[t.title]!)") != nil {
let d1 = dateFormatter.dateFromString("(r1[t.title]!)")
let d2 = dateFormatter.dateFromString("(r2[t.title]!)")
return d1 > d2
} else {
// return "(r1[t.title])".localizedStandardCompare("(r2[t.title])") == .OrderedDescending
return "(r1[t.title])" > "(r2[t.title])"
}
} else if r1[t.title] is Int {
return Int(String(r1[t.title]!)) > Int(String(r2[t.title]!))
} else {
// return "(r1[t.title])".localizedStandardCompare("(r2[t.title])") == .OrderedDescending
return "(r1[t.title])" > "(r2[t.title])"
}})
上升
self.fd = self.fd!.sort ({
(r1, r2) -> Bool in
let t = self.headers![b.tag - 10]
if r1[t.title] is String {
let dateFormatter = NSDateFormatter()
dateFormatter.dateFormat = "dd-MMM-yyyy"
if dateFormatter.dateFromString("(r1[t.title]!)") != nil {
let d1 = dateFormatter.dateFromString("(r1[t.title]!)")
let d2 = dateFormatter.dateFromString("(r2[t.title]!)")
return d1 < d2
} else {
//return "(r1[t.title])".localizedStandardCompare("(r2[t.title])") == .OrderedAscending
return "(r1[t.title])" < "(r2[t.title])"
}
} else if r1[t.title] is Int {
return Int(String(r1[t.title]!)) < Int(String(r2[t.title]!))
} else {
// return "(r1[t.title])".localizedStandardCompare("(r2[t.title])") == .OrderedAscending
return "(r1[t.title])" < "(r2[t.title])"
}})
主要问题是它的数据大于10K的数据超过60秒。请给我一个解决方案。
您的排序主要性能问题是您在比较闭合中执行转换。这将将值格式化过程称为指数次数(基于数组的大小)。
这样做的一种更有效的方法是仅使用.map()执行转换并将排序函数应用于结果。
这是一个基于您的代码的示例。
// only initialize your date formatter once
let dateFormatter = DateFormatter()
dateFormatter.dateFormat = "dd-MMM-yyyy"
// Extract the appropriate value and returns it in one of 3 types
// sorting will compare the 3 typed values knowing that only one of the 3
// will actually differ for a given data type
//
// The element is returned in the tupple in order to rebuild the
// array content after sorting on the 3 values
func getSortValues(_ element:[String:Any], _ id:String) -> ([String:Any], String, Int, TimeInterval)
{
let value = element[id]
var stringOrder = ""
var numericOrder = 0
var dateOrder = TimeInterval(0)
if let stringValue = value as? String,
let dateValue = dateFormatter.date(from:stringValue)
{
dateOrder = dateValue.timeIntervalSinceReferenceDate
}
else if let intValue = value as? Int
{
numericOrder = intValue
}
else if value != nil
{
stringOrder = "(value!)"
}
return (element, stringOrder, numericOrder, dateOrder)
}
let columnId = headers![b.tag - 10]
fd = fd!.map{ getSortValues($0, columnId) }
.sorted{ $0.1 < $1.1 || $0.2 < $1.2 || $0.3 < $1.3 }
.map{ $0.0 }
它应该快10到30倍(取决于数组的大小)。