Rotate - Transposing a List<List<string>> using LINQ C#

允我心安 提交于 2019-11-27 15:48:49

This is a simple and flexible solution, it will handle multiple inner lists with any number of dimensions.

List<List<string>> PersonInfo = new List<List<string>>()
{
    new List<string>() {"John", "Peter", "Watson"},
    new List<string>() {"1000", "1001", "1002"}
};


var result = PersonInfo
    .SelectMany(inner => inner.Select((item, index) => new { item, index }))
    .GroupBy(i => i.index, i => i.item)
    .Select(g => g.ToList())
    .ToList();

Here is a extension method

public static IEnumerable<IEnumerable<T>> Pivot<T>(this IEnumerable<IEnumerable<T>> source)
{
    var enumerators = source.Select(e => e.GetEnumerator()).ToArray();
    try
    {
        while (enumerators.All(e => e.MoveNext()))
        {
            yield return enumerators.Select(e => e.Current).ToArray();
        }
    }
    finally
    {
        Array.ForEach(enumerators, e => e.Dispose());
    }
}

so you can

var result = PersonInfo.Pivot();

Assuming there are only ever 2 lists inside PersonInfo:

var rotated = PersonInfo[0]
    .Zip(PersonInfo[1], (a, b) => new List<string> { a, b }).ToList();

If there can be any number of Lists inside of PersonInfo:

Enumerable.Range(0, PersonInfo[0].Count)
    .Select(i => PersonInfo.Select(lst => lst[i]).ToList()).ToList();

You can use Enumerable.Range and Enumerable.ElementAtOrDefault:

List<List<string>> rotated = Enumerable.Range(0, PersonInfo.Max(list => list.Count))
 .Select(i => PersonInfo.Select(list => list.ElementAtOrDefault(i)).ToList())
 .ToList();

PersonInfo.Max(list => list.Count) returns the max-size of the lists. This will be the new size of the main list, in this case 3. Enumerable.Range is like a for-loop. For every list it will now select all strings at these indexes. If the sizes are different you'll get null(because of ElementAtOrDefault).

If the lists had the same size you can apply the same query to get the original list back:

PersonInfo = Enumerable.Range(0, rotated.Max(list => list.Count))
 .Select(i => rotated.Select(list => list.ElementAtOrDefault(i)).ToList())
 .ToList();

As extension:

public static IEnumerable<IList<T>> Rotate<T>(this IEnumerable<IList<T>> sequences)
{
    var list = sequences as IList<IList<T>> ?? sequences.ToList();
    int maxCount = list.Max(l => l.Count);
    return Enumerable.Range(0, maxCount)
        .Select(i => list.Select(l => l.ElementAtOrDefault(i)).ToList());
}

Usage:

IEnumerable<IList<string>> rotated = PersonInfo.Rotate();
IEnumerable<IList<string>> rotatedPersonInfo = rotated.Rotate(); // append ToList to get the original list

This extends the Zip idea above to any number of lists. Zip will truncate the row lists to the smallest rank.

List<List<string>> PersonInfo = new List<List<string>>()
{
    new List<string>() {"John", "Peter", "Watson"},
    new List<string>() {"1000", "1001", "1002"},
    new List<string>() {"2000", "2001", "2002"},
    new List<string>() {"3000", "3001", "3002"}
};

var seed = Enumerable.Empty<List<string>>();
var transformed = PersonInfo.Aggregate(seed, (acc, r) =>
   acc.Any()
 ? acc.Zip(r, (row, nextElement) => { row.Add(nextElement); return row; })
 : r.Select(e => new List<string> { e }) //initialize target list using first row
); 

Try this:

List<List<string>> PersonInfo = new List<List<string>>(){
new List<string>() {"John", "Peter", "Watson"},
new List<string>() {"1000", "1001", "1002"}};

List<List<string>> PivitedPersonInfo = new List<List<string>>();
for (int i = 0; i < PersonInfo.First().Count; i++)
{
    PivitedPersonInfo.Add(PersonInfo.Select(x => x.ElementAt(i)).ToList());
}
易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!