CODEWITHSUNDEEP
c#linqpivot-tabletranspose
B.Balamanigandan
B.Balamanigandan

Reputation: 4875

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

I'm having a List<List<string>>, which is return from the remote data source (i.e., WCF). So, I need to modify the following data into a user-friendly list using LINQ

The C# Code is

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

Appropriate Screen Shot: Existing

enter image description here

I need to rotate the data as like the below Screenshot: Proposed

enter image description here

Kindly assist me how to rotate the data using LINQ C#

Upvotes: 16

Views: 9889

Answers (7)

Masaki Ohashi
Masaki Ohashi

Reputation: 491

Just do something like:

var persons = Enumerable.Range(0, PersonInfo.First().Count()).Select(i => PersonInfo.Select(e => e[i]).ToList()).ToList();

or

var persons = Enumerable.Range(0, PersonInfo[0].Count()).Select(i => {
    return PersonInfo.Select(e => {
        return e[i];
    }).ToList();
}).ToList();

and check the result like below:

persons.ForEach(p => Console.WriteLine("{0} {1}", p[0], p[1]));

Upvotes: 0

fubo
fubo

Reputation: 45947

Here is a generic 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();

Upvotes: 10

andredking
andredking

Reputation: 1

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
);

Upvotes: 0

Tim Schmelter
Tim Schmelter

Reputation: 460058

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

Upvotes: 2

Oliver
Oliver

Reputation: 9002

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();

Upvotes: 32

Dennis_E
Dennis_E

Reputation: 8894

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();

Upvotes: 2

sachin
sachin

Reputation: 2361

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());
}

Upvotes: -1

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