我有一個列表，我的目標是確定該列表中的值超過特定值的次數。例如，如果我的列表是： List = {0, 0, 3, 3, 4, 0, 4, 4, 4} 我想知道有兩個實例，我在列表中的值大于 2 并保持在上面2. 所以在這種情況下有 2 個實例，因為它一度低于 2 并再次高于它。 private void Report_GeneratorButton_Click(object sender, EventArgs e) { //Lists var current = _CanDataGraph._DataPoints[CanDataGraph.CurveTag.Current].ToList(); var SOC = _CanDataGraph._DataPoints[CanDataGraph.CurveTag.Soc].ToList(); var highcell = _CanDataGraph._DataPoints[CanDataGraph.CurveTag.HighestCell].ToList(); var lowcell = _CanDataGraph._DataPoints[CanDataGraph.CurveTag.LowestCell].ToList(); //Seperates current list into charging, discharging, and idle List<double> charging = current.FindAll(i => i > 2); List<double> discharging = current.FindAll(i => i < -2); List<double> idle = current.FindAll(i => i < 2 && i > -2); //High cell List<double> overcharged = highcell.FindAll(i => i > 3.65); int ov = overcharged.Count; if (ov > 1) { Console.WriteLine("This Battery has gone over Voltage!"); } else { Console.WriteLine("This battery has never been over Voltage."); } //Low cell List<double> overdischarged = lowcell.FindAll(i => i > 3.65); int lv = overdischarged.Count; if (lv > 1) { Console.WriteLine("This Battery has been overdischarged!"); } else { Console.WriteLine("This battery has never been overdischarged."); } //Each value is 1 second int chargetime = charging.Count; int dischargetime = discharging.Count; int idletime = idle.Count; Console.WriteLine("Charge time: " + chargetime + "s" + "\n" + "Discharge time: " + dischargetime + "s" + "\n" + "Idle time: " + idletime); }我當前的代碼是這樣并輸出： This battery has never been over Voltage. This battery has never been overdischarged. Charge time: 271s Discharge time: 0s Idle time: 68
4 回答
蠱毒傳說
TA貢獻1895條經驗 獲得超3個贊
有很多方法可以解決這個問題；我的建議是將它分解成許多較小的問題，然后編寫一個簡單的方法來解決每個問題。
這是一個更簡單的問題：給定一個序列T，給我返回一個刪除了“雙倍”項目的序列T：
public static IEnumerable<T> RemoveDoubles<T>(
this IEnumerable<T> items)
{
T previous = default(T);
bool first = true;
foreach(T item in items)
{
if (first || !item.Equals(previous)) yield return item;
previous = item;
first = false;
}
}
偉大的。這有什么幫助？因為現在解決您的問題是：
int count = myList.Select(x => x > 2).RemoveDoubles().Count(x => x);
跟著。
如果您有myListas{0, 0, 3, 3, 4, 0, 4, 4, 4}那么結果Select是{false, false, true, true, true, false, true, true, true}.
的結果RemoveDoubles是{false, true, false, true}。
的結果Count是 2，這是期望的結果。
盡可能使用現成的部件。如果你不能，試著解決一個簡單的、一般的問題，讓你得到你需要的東西；現在您有了一個工具，您可以將其用于需要您刪除序列中重復項的其他任務。
藍山帝景
TA貢獻1843條經驗 獲得超7個贊
該解決方案應該可以達到預期的效果。
List<int> lsNums = new List<int>() {0, 0, 3, 3, 4, 0, 4, 4, 4} ;
public void MainFoo(){
int iChange = GetCritcalChangeNum(lsNums, 2);
Console.WriteLine("Critical change = %d", iChange);
}
public int GetCritcalChangeNum(List<int> lisNum, int iCriticalThreshold) {
int iCriticalChange = 0;
int iPrev = 0;
lisNum.ForEach( (int ele) => {
if(iPrev <= iCriticalThreshold && ele > iCriticalThreshold){
iCriticalChange++;
}
iPrev = ele;
});
return iCriticalChange;
}
30秒到達戰場
TA貢獻1828條經驗 獲得超6個贊
您可以創建一個擴展方法，如下所示。
public static class ListExtensions
{
public static int InstanceCount(this List<double> list, Predicate<double> predicate)
{
int instanceCount = 0;
bool instanceOccurring = false;
foreach (var item in list)
{
if (predicate(item))
{
if (!instanceOccurring)
{
instanceCount++;
instanceOccurring = true;
}
}
else
{
instanceOccurring = false;
}
}
return instanceCount;
}
}
并像這樣使用您新創建的方法
current.InstanceCount(p => p > 2)
手掌心
TA貢獻1942條經驗 獲得超3個贊
執行此操作的另一種方法System.Linq是遍歷列表，同時選擇itemitself 和 it's index，然后返回true每一項大于value且前一項小于或等于 的項value，然后選擇結果數true。當然索引有一個特例0，我們不檢查前一項：
public static int GetSpikeCount(List<int> items, int threshold)
{
return items?
.Select((item, index) =>
index == 0
? item > threshold
: item > threshold && items[index - 1] <= threshold)
.Count(x => x == true) // '== true' is here for readability, but it's not necessary
?? 0; // return '0' if 'items' is null
}
示例用法：
private static void Main()
{
var myList = new List<int> {0, 0, 3, 3, 4, 0, 4, 4, 4};
var count = GetSpikeCount(myList, 2);
// count == 2
}
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