Question

我正在制作一个应用程序，它将为我提供用户插入的数字的均值，中值和范围。
但我似乎无法添加数字，然后将它们除以2 这就是我的尝试：

    public static String Find_Mean()
    {
        int Number = 0;
        for (int size = 0; size < list.Count; size++)
        {
            Number = Convert.ToInt16(list[size].ToString());
            Number += Number;
        }
        int Final_Number = Number / 2;
        return Convert.ToString(Final_Number);
    }

我想要做的是将所有数字从arraylist中加起来然后除以2。

Answer 1

尝试使用Linq：

 int[] x;
 x.Average();
 x.Max();
 x.Min();

Answer 2

Number = Convert.ToInt32(list[size].ToString());

您在此处每次迭代都会覆盖数字的值。

Answer 3

每次在循环中将Number设置为数组列表元素并覆盖总数时，这就是为什么你没有得到总数。您需要使用单独的变量来维护总计。类似的东西：

int Number = 0;
int Total = 0;
for (int size = 0; size < list.Count; size++)
{
    Number = Convert.ToInt16(list[size].ToString());
    Total += Number;
}
int Final_Number = Total / 2;

如果您使用 .Net 2.0或更高版本，那么如果您可以使用generic列表List<int>则会更好。

您还可以将循环中的转换更改为：

Number = Convert.ToInt32(list[0]);

由于Convert.ToInt32也有对象类型的重载，如果您的数字类型为int，则其为Int32而不是Int16。

Answer 4

您将在此处为Number重新分配值：

for (int size = 0; size < list.Count; size++)
{
    Number = Convert.ToInt16(list[size].ToString());
    Number += Number;
}

试试这个：

for (int size = 0; size < list.Count; size++)
{
    Number += Convert.ToInt16(list[size].ToString());
}

Answer 5

以下是我目前用于计算一些简单统计数据的内容：

private void CalculateStatistics(IEnumerable<Double> valuesToAggregate)
{
    // We need to iterate multiple times over the values, so it makes
    // sense to create a list to improve performance.
    var aggregateMe = valuesToAggregate.ToList();

    if (aggregateMe.Count > 0)
    {
        // To calculate the median, the simplest approach
        // is to sort the list.
        aggregateMe.Sort();

        // Cause we already sorted the list,
        // the min value must be available within the first element.
        Min = aggregateMe[0];

        // the max value must be available within the last element.
        Max = aggregateMe[aggregateMe.Count - 1];

        // The average has really to be calculated, by another iteration run.
        Mean = aggregateMe.Average();

        // Taking the median from a sorted list is easy.
        var midpoint = (aggregateMe.Count - 1) / 2;
        Median = aggregateMe[midpoint];

        // If the list contains a even number of element,
        // the median is the average of the two elements around the midpoint.
        if (aggregateMe.Count % 2 == 0)
            Median = (Median + aggregateMe[midpoint + 1]) / 2;
    }
    else
    {
        // There is no value available to calculate some statistic.
        Min = Double.NaN;
        Max = Double.NaN;
        Mean = Double.NaN;
        Median = Double.NaN;
    }
}

请注意，您可以改进此特定实现，具体取决于您获取的数据类型（字符串，双精度等）以及它们是如何存储的（它们是否已作为列表进入，您可以操作等等？）。

Answer 6

public static String Find_Mean()
{
    List<int> integerList = list.Select(q => int.Parse(q)).ToList();
    var result = integerList.Sum() / 2;
    return result.ToString();
}

从arraylist中添加数字

6 个答案: