Question

var nums = new[]{ 1, 2, 3, 4, 5, 6, 7};
var pairs  = /* some linq magic here*/ ;

=＆GT; 对= {{1,2}，{3,4}，{5,6}，{7,0}}

pairs的元素应该是两元素列表，或者是一些具有两个字段的匿名类的实例，类似于new {First = 1, Second = 2}。

Answer 1

默认的linq方法都不能通过单次扫描来懒散地执行此操作。用它自己压缩序列会进行2次扫描，并且分组并不完全是懒惰的。您最好的选择是直接实施它：

public static IEnumerable<T[]> Partition<T>(this IEnumerable<T> sequence, int partitionSize) {
    Contract.Requires(sequence != null)
    Contract.Requires(partitionSize > 0)

    var buffer = new T[partitionSize];
    var n = 0;
    foreach (var item in sequence) {
        buffer[n] = item;
        n += 1;
        if (n == partitionSize) {
            yield return buffer;
            buffer = new T[partitionSize];
            n = 0;
        }
    }
    //partial leftovers
    if (n > 0) yield return buffer;
}

Answer 2

试试这个：

int i = 0;
var pairs = 
  nums
    .Select(n=>{Index = i++, Number=n})
    .GroupBy(n=>n.Index/2)
    .Select(g=>{First:g.First().Number, Second:g.Last().Number});

Answer 3

这可能比您要求的更为通用 - 您可以设置自定义itemsInGroup：

int itemsInGroup = 2;
var pairs = nums.
            Select((n, i) => new { GroupNumber = i / itemsInGroup, Number = n }).
            GroupBy(n => n.GroupNumber).
            Select(g => g.Select(n => n.Number).ToList()).
            ToList();

修改

如果你想附加零（或其他一些数字）以防最后一组的大小不同：

int itemsInGroup = 2; int valueToAppend = 0; int numberOfItemsToAppend = itemsInGroup - nums.Count() % itemsInGroup; var pairs = nums. Concat(Enumerable.Repeat(valueToAppend, numExtraItems)). Select((n, i) => new { GroupNumber = i / itemsInGroup, Number = n }). GroupBy(n => n.GroupNumber). Select(g => g.Select(n => n.Number).ToList()). ToList();

Answer 4

int[] numbers = new int[] { 1, 2, 3, 4, 5, 6, 7 };
var result = numbers.Zip(numbers.Skip(1).Concat(new int[] { 0 }), (x, y) => new
        {
            First = x,
            Second = y
        }).Where((item, index) => index % 2 == 0);

Answer 5

（警告：看起来很难看）

var pairs = x.Where((i, val) => i % 2 == 1)
            .Zip(
            x.Where((i, val) => i % 2 == 0),
                (first, second) =>
                new
                {
                    First = first,
                    Second = second
                })
            .Concat(x.Count() % 2 == 1 ? new[]{
                new
                {
                    First = x.Last(),
                    Second = default(int)
                }} : null);

Answer 6

public static IEnumerable<List<T>> InSetsOf<T>(this IEnumerable<T> source, int max)
{
    return InSetsOf(source, max, false, default(T));
}

public static IEnumerable<List<T>> InSetsOf<T>(this IEnumerable<T> source, int max, bool fill, T fillValue)
{
    var toReturn = new List<T>(max);
    foreach (var item in source)
    {
        toReturn.Add(item);
        if (toReturn.Count == max)
        {
            yield return toReturn;
            toReturn = new List<T>(max);
        }
    }
    if (toReturn.Any())
    {
        if (fill)
        {
            toReturn.AddRange(Enumerable.Repeat(fillValue, max-toReturn.Count));
        }
        yield return toReturn;
    }
}

用法：

var pairs = nums.InSetsOf(2, true, 0).ToArray();

Answer 7

IList<int> numbers = new List<int> {1, 2, 3, 4, 5, 6, 7};
var grouped = numbers.GroupBy(num =>
{
   if (numbers.IndexOf(num) % 2 == 0)
   {
      return numbers.IndexOf(num) + 1;
   }
   return numbers.IndexOf(num);
});

如果你需要填充零的最后一对，你可以在进行分组之前添加它，如果listcount是奇数。

if (numbers.Count() % 2 == 1)
{
   numbers.Add(0);
}

另一种方法可能是：

var groupedIt = numbers
   .Zip(numbers.Skip(1).Concat(new[]{0}), Tuple.Create)
   .Where((x,i) => i % 2 == 0);

或者您使用具有许多有用扩展名的MoreLinq：

IList<int> numbers = new List<int> {1, 2, 3, 4, 5, 6, 7};
var batched = numbers.Batch(2);

Answer 8

 var nums = new float[] { 1, 2, 3, 4, 5, 6, 7 };
 var enumerable = 
        Enumerable
          .Range(0, nums.Length)
          .Where(i => i % 2 == 0)
          .Select(i => 
             new { F = nums[i], S = i == nums.Length - 1 ? 0 : nums[i + 1] });

Answer 9

    var w =
        from ei in nums.Select((e, i) => new { e, i })
        group ei.e by ei.i / 2 into g
        select new { f = g.First(), s = g.Skip(1).FirstOrDefault() };

Answer 10

另一个选择是使用SelectMany LINQ方法。对于希望遍历项目列表的人员来说，这是更多的选择，并且对于每个项目，返回其2个或多个属性。无需再次遍历每个属性的列表，只需一次。

var list = new [] {//Some list of objects with multiple properties};

//Select as many properties from each Item as required. 
IEnumerable<string> flatList = list.SelectMany(i=> new[]{i.NameA,i.NameB,i.NameC});

Answer 11

使用index和index + 1的另一种简单解决方案。

var nums = Enumerable.Range(1, 10);
var pairs = nums.Select((item, index) =>
  new { First = item, Second = nums.ElementAtOrDefault(index + 1) })
    .SkipLastN(1);

pairs.ToList().ForEach(p => Console.WriteLine($"({p.First}, {p.Second}) "));

最后一项无效，必须用SkipLastN()删除。

Answer 12

这给出了所有可能的对（vb.net）：

Dim nums() = {1, 2, 3, 4, 5, 6, 7}
Dim pairs = From a In nums, b In nums Where a <> b Select a, b

编辑：

 Dim allpairs = From a In nums, b In nums Where b - a = 1 Select a, b
 Dim uniquePairs = From p In allpairs Where p.a Mod 2 <> 0 Select p

注意：最后一对丢失，正在处理

编辑：

uniquePairs {nums.Last,0}对{{1}}

Linq to Objects - 从数字列表中返回数字对

12 个答案: