Question

我有一个长度为3的数组说Array(3,4,5)我的目标长度是Int说的7.如何用Array填充Array元素直到Int的长度达到val A = Array(3,4,5) val T = 7 //Desired output Array(3,3,3,3,3,4,5)？

val difflength = T - A.size
val firstElement = A.head
val PadArray = (for(i <- 0 to difflength) yield firstElement).toArray
PadArray ++ A

我目前的方法：

namespace WinFormsApp
{
    public partial class Form1 : Form
    {
        private List<Category> categories;

        public Form1()
        {
            InitializeComponent();

            categories = new List<Category>();

            var categoryOne = new Category { Name = "Category 1"} ;
            categoryOne.Items.Add( new CategoryItem { Name = "Item 1"} );

            var categoryTwo = new Category { Name = "Category 2" };
            categoryTwo.Items.Add( new CategoryItem { Name = "Item 2" } );

            categories.Add( categoryOne );
            categories.Add( categoryTwo );
        }

        private void Form1_Load(object sender, System.EventArgs e)
        {
            categoryBindingSource.DataSource = categories;
        }
    }

    public class Category
    {
        public string Name { get; set; }

        public List<CategoryItem> Items { get; private set; }

        public Category()
        {
            Items = new List<CategoryItem>();
        }
    }

    public class CategoryItem
    {
        public string Name { get; set; }
    }
}

有更简单的方法吗？

Answer 1

Array's fill method可以派上用场：

val a = Array(3,4,5)
val b = {
  val t = 7
  val diffLength = t - a.size
  val firstElement = a.head
  Array.fill(diffLength)(firstElement) ++ a
}

结果：

scala> b
res0: Array[Int] = Array(3, 3, 3, 3, 3, 4, 5)

Scala如何重复Array的第一个元素，直到Array.size达到一定数量

1 个答案: