Sort a list from another list IDs

Question

I have a list with some identifiers like this:

List<long> docIds = new List<long>() { 6, 1, 4, 7, 2 };

Morover, I have another list of <T> items, which are represented by the ids described above.

List<T> docs = GetDocsFromDb(...)

I need to keep the same order in both collections, so that the items in List<T> must be in the same position than in the first one (due to search engine scoring reasons). And this process cannot be done in the GetDocsFromDb() function.

If necessary, it's possible to change the second list into some other structure (Dictionary<long, T> for example), but I'd prefer not to change it.

Is there any simple and efficient way to do this "ordenation depending on some IDs" with LINQ?

are you assured that every docId occurs exactly once in docs, what property will hold the Id or will a selector Func<T, long> be required? — Jodrell
– Jodrell, Commented Mar 7, 2013 at 15:43
Does the first list represent a "master list"? Another words, will the second list be a subset representing a portion (or the entirety) of the first list? — code4life
– code4life, Commented Mar 7, 2013 at 16:00

Denys Denysenko · Accepted Answer · 2013-03-07 15:42:22Z

502

docs = docs.OrderBy(d => docsIds.IndexOf(d.Id)).ToList();

answered Mar 7, 2013 at 15:42

Denys Denysenko

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7 Comments

Jodrell Over a year ago

@BorjaLópez, a quick note. You mention efficiency in your question. IndexOf is perfectly acceptable for your example and nice and simple. If you had a lot of data my answer might be better suited. stackoverflow.com/questions/3663014/…

silkfire Over a year ago

@DenysDenysenko Fantastic. Thanks so much; exactly what I was looking for.

Dan Hunex Over a year ago

does not work if you have elements in docs that do not have ids in the ordering list

sdds Over a year ago

Quite inefficient - IndexOf is being called for every element in the source collection and OrderBy has to order the elements. The solution by @Jodrell is much faster.

nicolascolman Over a year ago

@DanHunex if you need the non existing ids to appear at the end of the list you can do: .OrderBy(d=> { var index = docIds.IndexOf(d.Id); if (index == -1) index = docIds.Count; return index; })

|

Jodrell · Accepted Answer · 2022-09-27 08:54:54Z

55

Since you don't specify T,

public static IEnumerable<T> OrderBySequence<T, TId>(
       this IEnumerable<T> source,
       IEnumerable<TId> order,
       Func<T, TId> idSelector)
{
    var lookup = source.ToDictionary(idSelector, t => t);
    foreach (var id in order)
    {
        yield return lookup[id];
    }
}

Is a generic extension for what you want.

You could use the extension like this perhaps,

var orderDocs = docs.OrderBySequence(docIds, doc => doc.Id);

A safer version might be

public static IEnumerable<T> OrderBySequence<T, TId>(
       this IEnumerable<T> source,
       IEnumerable<TId> order,
       Func<T, TId> idSelector)
{
    var lookup = source.ToLookup(idSelector, t => t);
    foreach (var id in order)
    {
        foreach (var t in lookup[id])
        {
           yield return t;
        }
    }
}

which will work if source does not zip exactly with order.

edited Sep 27, 2022 at 8:54

answered Mar 7, 2013 at 15:53

Jodrell

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4 Comments

vinmm Over a year ago

I used this solution and it worked. Just that, i had to make the method static and the class static.

jim tollan Over a year ago

10 years old (almost) and still a wonderful solution. tnx (likewise had to embed in a static helper class but does offer great reuse)

Jodrell Over a year ago

@jimtollan, Nice of you to say so. I guess it is also a good reflection on Stack Overflow. You should bear in mind that Join works too. Sort a list from another list IDs

Flou Over a year ago

+1 more versatile, e.g : EmployeeList.OrderBySequence(HospitalList.OrderBy(h => h.Position).Select(h => h.Id), e => e.HospitalId) gives : employees order by their hospital itself order by its position

Kladzey · Accepted Answer · 2019-07-10 17:56:48Z

37

Jodrell's answer is best, but actually he reimplemented System.Linq.Enumerable.Join. Join also uses Lookup and keeps ordering of source.

    docIds.Join(
      docs,
      i => i,
      d => d.Id,
      (i, d) => d);

edited Jul 10, 2019 at 17:56

answered Apr 12, 2019 at 11:11

Kladzey

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2 Comments

Orace Over a year ago

That is the answer we are looking for

PRMan Over a year ago

This just proves that Join is too difficult to understand since everyone agreed that rewriting it was easier.

endeffects · Accepted Answer · 2024-04-04 08:56:31Z

I think the given solution is incorrect because it loses all items that are not included in the order collection. The correct solution would be:

public static IEnumerable<T> OrderBySequence<T, TId>(this IEnumerable<T> source, IEnumerable<TId> order, Func<T, TId> idSelector)
{
    var lookup = source.ToLookup(idSelector, t => t);
    var included = new HashSet<TId>();

    foreach (var id in order)
    {
        if (!lookup.Contains(id)) continue;

        foreach (var t in lookup[id])
        {
            yield return t;
            included.Add(id);
        }
    }

    foreach (var item in source)
    {
        if (!included.Contains(idSelector(item)))
        {
            yield return item;
        }
    }
}

Albin Sunnanbo · Accepted Answer · 2013-03-07 15:44:00Z

-4

One simple approach is to zip with the ordering sequence:

List<T> docs = GetDocsFromDb(...).Zip(docIds, Tuple.Create)
               .OrderBy(x => x.Item2).Select(x => x.Item1).ToList();

answered Mar 7, 2013 at 15:44

Albin Sunnanbo

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3 Comments

Albin Sunnanbo Over a year ago

Because Zip combines each index (into a Tuple) with the document in the same position in the corresponding list. Then the OrderBy sorts the Tuples by the index part and then the select digs our just the docs from the now ordered list.

Jodrell Over a year ago

but, the result of GetDocsFromDb is unordered so you'll be creating Tuples where Item1 is unrelated to Item2.

Michael Logutov Over a year ago

I think this will produce incorrect results since ordering performing uppon id and not the index.

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