简体   繁体   English

列表中的最大值和最小值

[英]maximum and minimum value in list

I'm currently doing my current project and I had a problem. 我目前正在执行当前的项目,但遇到了问题。 Here's what the project needs to do: 这是项目需要做的事情:

Find the maximum and the minimum temperature from a certain range of date. 找到某个日期范围内的最高和最低温度。 The range of the date will be inputted by the user. 日期范围将由用户输入。

So, I make a form as the main menu for inputting the items and finding the maximum and minimum value (both in the new form). 因此,我将一个表单作为输入项目并查找最大值和最小值(均为新表单)的主菜单。 I also make a class to store the items: 我还开设了一个课程来存储项目:

public class TempDate
{
    public double Temp { get; set; }

    public DateTime Date { get; set; }
}

In the first form, just call it FormAddData , from here items will be stored into the list using a textbox and here's the code: 在第一种形式中,只需将其FormAddData ,此处将使用文本框将项目存储到列表中,这是代码:

private void buttonSubmit_Click(object sender, EventArgs e)
    {
        FormMenu formMenu = (FormMenu)this.Owner;

        DateTime date = dateTimePickerDate.Value.Date;
        double temp = double.Parse(textBoxTemp.Text);

        TempDate tempDate = new TempDate();

        tempDate.Date = date;
        tempDate.Temp = temp;

        formMenu.listOfTempDate.Add(tempDate);
        listBoxInfo.Items.Add(date + "\t" + temp + "°C");
    }

In the second form that called FormMaxMinRange . 在第二种形式中,该形式称为FormMaxMinRange In this form, I use two DateTimePicker the first one for the starting date and the second for the ending date. 在这种形式中,我使用两个DateTimePicker ,第一个用于开始日期,第二个用于结束日期。 From here I need to make a button that will select all the items from the range that I used from starting and ending date. 从这里,我需要创建一个按钮,该按钮将从开始和结束日期使用的范围中选择所有项目。 Here's my code: 这是我的代码:

private void buttonMaxMin_Click(object sender, EventArgs e)
    {
        FormMenu formMenu = (FormMenu)this.Owner;

        DateTime start = dateTimePickerStart.Value.Date;
        DateTime end = dateTimePickerEnd.Value.Date;
        int highest = 0;
        double max = formMenu.listOfTempDate[0].Temp;
        int lowest = 0;
        double min = formMenu.listOfTempDate[0].Temp;

        for (int i = 1; i < formMenu.listOfTempDate.Count; i++)
        {
            if (formMenu.listOfTempDate[i].Date >= start
                && formMenu.listOfTempDate[i].Date <= end)
            {
                if (formMenu.listOfTempDate[i].Temp > max)
                {
                    highest = i;
                    max = formMenu.listOfTempDate[i].Temp;
                }

                if (formMenu.listOfTempDate[i].Temp < min)
                {
                    lowest = i;
                    min = formMenu.listOfTempDate[i].Temp;
                }
            }
        }
        listBoxMaxMin.Items.Add("");
        listBoxMaxMin.Items.Add("Lowest temp: " + min + ", on " + formMenu.listOfTempDate[lowest].Date);
        listBoxMaxMin.Items.Add("Highest temp: " + max + ", on " + formMenu.listOfTempDate[highest].Date);
    }

Here's the main form that i declared the class (which include the list): 这是我声明该类(包括列表)的主要形式:

public partial class FormMenu : Form
{
    public List<TempDate> listOfTempDate = new List<TempDate>();

    public FormMenu()
    {
        InitializeComponent();
    }

    private void fromCertainRangeToolStripMenuItem_Click(object sender, EventArgs e)
    {
        FormMaxMinRange formMaxMinRange = new FormMaxMinRange();
        formMaxMinRange.Owner = this;
        formMaxMinRange.ShowDialog();
    }
}

But, the problem is, the minimum value was not selected inside the range of selection. 但是,问题是,未在选择范围内选择最小值。 Also I want the max and min value was printed in the listbox. 我也希望最大值和最小值被打印在列表框中。 Sorry for the long and weird question. 很抱歉这个长期而奇怪的问题。 I hope someone can understand what I means with this question to complete my project. 我希望有人能够理解我对这个问题的理解以完成我的项目。 Thank you. 谢谢。

See this code snippet. 请参见此代码段。 You can use Linq to select the reduced list (with Start/Enddate) and order it by Temp. 您可以使用Linq选择简化列表(带有“开始/结束日期”),并按“临时”顺序对其进行排序。 Now you can easy select the first (min) and the last (max) object. 现在,您可以轻松选择第一个(最小)和最后一个(最大)对象。

List<TempDate> loTempDateList = new List<TempDate>()
{
    new TempDate() {Date = DateTime.Now.AddDays(-10), Temp = 10.01 },
    new TempDate() {Date = DateTime.Now.AddDays(-5), Temp = 20.01 },
    new TempDate() {Date = DateTime.Now.AddDays(-3), Temp = 30.01 },
    new TempDate() {Date = DateTime.Now, Temp = 40.01 }
};
DateTime ldStart = DateTime.Now.AddDays(-6);
DateTime ldEnd = DateTime.Now.AddDays(-1);

var loDateList = loTempDateList.Where(item => item.Date <= ldEnd && item.Date >= ldStart)
    .OrderBy(item => item.Temp);

TempDate loMin = loDateList.First();
TempDate loMax = loDateList.Last();

Console.WriteLine("{0}: {1} with max temp", loMax.Date, loMax.Temp);
Console.WriteLine("{0}: {1} with min temp", loMin.Date, loMin.Temp);

Output (for today): 输出(今天):

9/26/2017 3:17:09 PM: 30.01 with max temp
9/24/2017 3:17:09 PM: 20.01 with min temp

Update (with your variable names): 更新(使用您的变量名):

Copy this under DateTime end = dateTimePickerEnd.Value.Date; 将其复制到DateTime end = dateTimePickerEnd.Value.Date; in your Form 在您的表格中

var loDateList = listOfTempDate.Where(item => item.Date <= end && item.Date >= start)
    .OrderBy(item => item.Temp);

TempDate loMin = loDateList.FirstOrDefault();
TempDate loMax = loDateList.LastOrDefault();

if (loMin != null && loMax !=  null)
{
    listBoxMaxMin.Items.Add("");
    listBoxMaxMin.Items.Add("Lowest temp: " + loMin.Temp + ", on " + loMin.Date);
    listBoxMaxMin.Items.Add("Highest temp: " + loMax.Temp + ", on " + loMax.Date);
}

I would suggest you use Linq Max and Min methods. 我建议您使用Linq MaxMin方法。

// filter out only the dates in the range you need
var items = formMenu.listOfTempDateWhere(
    item => ((TempDate)item).Date >= start && ((TempDate)item).Date <= end
);

// get the maximum value
var max = items.Max(item => item.Temp);
// get the minimum value
var min = items.Min(item => item.Temp);

Just remember to add using System.Linq on the top of your .cs file 只需记住在.cs文件顶部using System.Linq添加即可


try this online 在线尝试

If you don't like a LINQ approach (I never use LINQ, for some, possibly invalid reason, I think it's evil), you can override the List class and extend it with methods of your own. 如果您不喜欢LINQ方法(出于某些可能无效的原因,我从不使用LINQ,我认为这很邪恶),则可以覆盖List类并使用您自己的方法进行扩展。

public class TempDataList<T> : List<TempData>
{
    public TempDataList() : base()
    {
    }

    public TempDataList(IEnumerable<TempData> collection) : base(collection)
    {
    }

    public TempData GetMaxTemp(DateTime startDate, DateTime endDate)
    {
        TempData highestTempData = null;
        for (int i = 0; i < this.Count; i++)
        {
            if (this[i].Date >= startDate && this[i].Date <= endDate)
            {
                if (highestTempData == null || this[i].Temp > highestTempData.Temp)
                {
                    highestTempData = this[i];
                } 
            }
        }
        return highestTempData;
    }

    public TempData GetMinTemp(DateTime startDate, DateTime endDate)
    {
        TempData lowestTempData = null;
        for (int i = 0; i < this.Count; i++)
        {
            if (this[i].Date >= startDate && this[i].Date <= endDate)
            {
                if (lowestTempData == null || this[i].Temp < lowestTempData.Temp)
                {
                    lowestTempData = this[i];
                }
            }
        }
        return lowestTempData;
    }
}

And fill the extended list and call the methods: 并填写扩展列表并调用方法:

TempDataList<TempData> tempDataList = new TempDataList<TempData>();
tempDataList.Add(new TempData(10, DateTime.UtcNow));
tempDataList.Add(new TempData(20, DateTime.UtcNow));
tempDataList.Add(new TempData(15, DateTime.MinValue));
tempDataList.Add(new TempData(25, DateTime.MaxValue));

Console.WriteLine(tempDataList.GetMaxTemp(DateTime.UtcNow.AddDays(-1), DateTime.UtcNow.AddDays(1)).Temp);
Console.WriteLine(tempDataList.GetMinTemp(DateTime.UtcNow.AddDays(-1), DateTime.UtcNow.AddDays(1)).Temp);

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM