I would like to compare 2 lists of arrays. Let's take for instance this example:
List<int[]> list1 = new List<int[]>() { new int[4] { 1, 2, 3, 4 }, new int[4] { 1, 2, 3, 5 } };
List<int[]> list2 = new List<int[]>() { new int[2] { 1, 2 }, new int[2] { 3, 4 }, new int[2] { 3, 5 } };
I would like to know for each element in list1 to calculate for every element in list 2 how many common elements they have.
Ex. 1,2,3,4 compared with 1,2 will result to 2 matching elements.
1,2,3,4 compared with 3,5 will result to 1 matching element.
This is not duplicated as I don't wish to compare regular lists. I wish to see for each record in list1 how many items from list2 contain how many common items.
List<int[]> list1 = new List<int[]>() { new int[4] { 1, 2, 3, 4 }, new int[4] { 1, 2, 3, 5 } };
List<int[]> list2 = new List<int[]>() { new int[2] { 1, 2 }, new int[2] { 3, 4 }, new int[2] { 3, 5 } };
var results = list1.Select(x => list2.Select(y => y.Intersect(x).Count()).ToList()).ToList();
Result contains following data: [ [ 2, 2, 1 ], [ 2, 1, 2 ] ]
You can use Enumerable.Intersect to find out common items of first found in second list.
var commonList = list1.Intersect(list2);
The intersection of two sets A and B is defined as the set that
contains all the elements of A that also appear in B, but no other
elements
Edit Since you have array as element of list you have to go through each list item.
List<int[]> list1 = new List<int[]>() { new int[4] { 1, 2, 3, 4 }, new int[4] { 1, 2, 3, 5 } };
List<int[]> list2 = new List<int[]>() { new int[2] { 1, 2 }, new int[2] { 3, 4 }, new int[2] { 3, 5 } };
List<int[]> list3 = new List<int[]>();
for (int i = 0; i < list1.Count; i++)
{
list3.Add(list1[i].Intersect(list2[i]).ToArray());
}
You'd do something like this:
var results =
from x in list1.Select((array, index) => new { array, index })
from y in list2.Select((array, index) => new { array, index })
select new
{
list1_index = x.index,
list2_index = y.index,
count = x.array.Intersect(y.array).Count()
};
foreach(var r in results)
{
Console.WriteLine("({0}, {1}) have {2} item(s) in common.", r.list1_index, r.list2_index, r.count);
}
// (0, 0) have 2 item(s) in common.
// (0, 1) have 2 item(s) in common.
// (0, 2) have 1 item(s) in common.
// (1, 0) have 2 item(s) in common.
// (1, 1) have 1 item(s) in common.
// (1, 2) have 2 item(s) in common.
var commons = list1.Select(x => list2.Select(x.Intersect).ToArray()).ToArray();
Console.WriteLine(commons[0][0]); // Commons between list1[0] and list2[0]
Console.WriteLine(commons[0][1]); // Commons between list1[0] and list2[1]
Console.WriteLine(commons[3][0]); // Commons between list1[3] and list2[0]
Console.WriteLine(commons[3][0].Length); // Number of commons between [3] and [0]
Accourding to ur requirement, I think in C# there isnt such inbuild function in list which do exactly what you want but following function return exactly what you required in resultList. Hope this help you.
private List<int> MatchList()
{
List<int[]> list1 = new List<int[]>() { new int[4] { 1, 2, 3, 4 }, new int[4] { 1, 2, 3, 5 } };
List<int[]> list2 = new List<int[]>() { new int[2] { 1, 2 }, new int[2] { 3, 4 }, new int[2] { 3, 5 } };
List<int> resultList = new List<int>();
for (int i = 0; i < list1.Count; i++)
{
for (int j = 0; j < list2.Count; j++)
{
if (i == j)
{
int result = 0;
foreach (int list1Element in list1[i])
{
foreach (int list2Element in list2[j])
{
if (list1Element == list2Element)
{
result +=1;
}
}
}
resultList.Add(result);
}
}
}
return resultList;
}
Related
Suppose that I have a list of integer or whatever
List<int> motherlist = { 1, 1, 2, 5, 7, 2, 2, 2, 6, 1 }
Console.WriteLine(children.Count); // 10
I would like to find all duplicates and not remove them from the list but to distribute them across other lists so the final count of all childrens should be the same as motherlist:
List<List<int>> children = { { 1, 2, 5, 7, 6 }, { 1, 2 }, { 1, 2 }, { 2 }}
Console.WriteLine(children.Sum(l => l.Count())); // 10 same as mother
I tried so far a brute force approach by looping through all elements of mother, comparing the elements with all other elements and to check for duplicates, If duplicate found I add it to a list of buckets (List of Lists) and so forth until the last elements.
But the brute force approach takes 7 CPU seconds for only a mother list of 300 items.
I imagine that if I had 1000 items this would take forever.
Is there a faster way to do this in C# .NET ?
I suggest grouping duplicates and then loop taking into account size of the groups:
public static IEnumerable<List<T>> MyDo<T>(IEnumerable<T> source,
IEqualityComparer<T> comparer = null) {
if (null == source)
throw new ArgumentNullException(nameof(source));
var groups = new Dictionary<T, List<T>>(comparer ?? EqualityComparer<T>.Default);
int maxLength = 0;
foreach (T item in source) {
if (!groups.TryGetValue(item, out var list))
groups.Add(item, list = new List<T>());
list.Add(item);
maxLength = Math.Max(maxLength, list.Count);
}
for (int i = 0; i < maxLength; ++i) {
List<T> result = new List<T>();
foreach (var value in groups.Values)
if (i < value.Count)
result.Add(value[i]);
yield return result;
}
}
Demo:
int[] source = new int[] { 1, 1, 2, 5, 7, 2, 2, 2, 6, 1 };
var result = MyDo(source).ToList();
string report = string.Join(Environment.NewLine, result
.Select(line => $"[{string.Join(", ", line)}]"));
Console.Write(report);
Outcome:
[1, 2, 5, 7, 6]
[1, 2]
[1, 2]
[2]
Stress Demo:
Random random = new Random(1234); // seed, the results to be reproducible
// We don't want 1000 items be forever; let's try 1_000_000 items
int[] source = Enumerable
.Range(1, 1_000_000)
.Select(x => random.Next(1, 1000))
.ToArray();
Stopwatch sw = new Stopwatch();
sw.Start();
var result = MyDo(source).ToList();
sw.Stop();
Console.WriteLine($"Time: {sw.ElapsedMilliseconds} ms");
Outcome: (may vary from workstation to workstation)
Time: 50 ms
I would GroupBy the elements of the list, and then use the count of elements to know the number of sublists an element has to be added in
List<int> motherlist = new List<int> { 1, 1, 2, 5, 7, 2, 2, 2, 6, 1 };
var childrens = motherlist.GroupBy(x => x).OrderByDescending(x => x.Count());
var result = new List<List<int>>();
foreach (var children in childrens)
{
for (var i = 0; i < children.Count(); i++)
{
if (result.Count() <= i) result.Add(new List<int>());
result[i].Add(children.Key);
}
}
Console.WriteLine("{");
foreach (var res in result)
{
Console.WriteLine($"\t{{ { string.Join(", ", res) } }}");
}
Console.WriteLine("}");
This outputs :
{
{ 2, 1, 5, 7, 6 }
{ 2, 1 }
{ 2, 1 }
{ 2 }
}
Just a quick shot, but it seems to work quite well...
using System;
using System.Collections.Generic;
using System.Linq;
namespace ConsoleApp2
{
class Program
{
static void Main(string[] args)
{
List<int> motherlist = new List<int> { 1, 1, 2, 5, 7, 2, 2, 2, 6, 1 };
var rnd = new Random(1);
for (int i = 0; i < 1000; i++)
{
motherlist.Add(rnd.Next(1, 200));
}
var resultLists = new List<IEnumerable<int>>();
while (motherlist.Any())
{
var subList = motherlist.Distinct().OrderBy(x => x).ToList();
subList.ForEach(x => motherlist.Remove(x));
resultLists.Add(subList);
}
}
}
}
You can use a Dictionary<int, int> to keep track of the number of occurrences of each element and build the child lists in a single iteration with O(n) time complexity(most of the time) and without any LINQ:
var motherlist = new List<int>() { 1, 1, 2, 5, 7, 2, 2, 2, 6, 1 };
var counts = new Dictionary<int, int>();
var children = new List<List<int>>();
foreach(var element in motherlist)
{
counts.TryGetValue(element, out int count);
counts[element] = ++count;
if (children.Count < count)
{
children.Add(new List<int>() { element });
}
else
{
children[count - 1].Add(element);
}
}
OUTPUT
{ 1, 2, 5, 7, 6 }
{ 1, 2 }
{ 2, 1 }
{ 2 }
I have two lists:
var list1 = new List<int> { 0, 1, 2 };
var list2 = new List<int> { 1, 2, 3 };
I want to be able to check if the ending chunk of list1 is present at the start of list2. After that I want to delete one of the chunks from any of the lists, merging both into a third list (sequentially, list1 + list2).
var list3 = list1.Something(list2);
//Returns 0,1,2,3 instead of 0,1,2,1,2,3
There's another problem, one list can be smaller than the other, such as:
0,1,2,3 <-- 2,3,4 = 0,1,2,3,4
5,6 <-- 6,7,8 = 5,6,7,8
And of course, both lists can be different:
0,1,2 <-- 5,6,7 = 0,1,2,5,6,7
[empty] <-- 1,2 = 1,2
Is there any method provided by .Net Framework that allows me to do that?
If not, could you help me create one?
The end and start can only "kill" each other if they are sequentially equal.
Example, if list1 ends in 1,2 and list2 starts with 2,1 they are not equal.
So, Distinct() is not helpful.
My use case:
private List<int> Cut(this List<int> first, List<int> second)
{
//Code
return new List<int>();
}
internal List<int> MergeKeyList()
{
var keyList = new List<int>() {0, 1, 2};
var newList = new List<int>() {1, 2, 3};
return keyList.InsertRange(keyList.Count, keyList.Cut(newList));
}
Would be much more efficient with for loops .. but whatever:
keyList.TakeWhile((_, i) => !keyList.Skip(i).SequenceEqual(newList.Take(keyList.Count - i)))
.Concat(newList)
Try this:
void Main()
{
var keyList = new List<int>() {0, 1, 2};
var newList = new List<int>() {1, 2, 3};
var result = keyList.Cut(newList);
}
public static class Ex
{
public static List<int> Cut(this List<int> first, List<int> second)
{
var skip =
second
.Select((x, n) => new { x, n })
.Where(xn => xn.x == first.Last())
.Where(xn =>
first
.Skip(first.Count - xn.n - 1)
.SequenceEqual(second.Take(xn.n + 1)))
.Reverse()
.Select(xn => xn.n + 1)
.FirstOrDefault();
return first.Concat(second.Skip(skip)).ToList();
}
}
result becomes:
Also:
{ 0, 1, 2 } & { 1, 2, 1, 2, 3 } => { 0, 1, 2, 1, 2, 3 }
{ 0, 1, 2, 1 } & { 1, 2, 1, 2, 3 } => { 0, 1, 2, 1, 2, 3 }
I have a list of lists like : List<List<int>> a ;
How can I initialize it like this:
<2 1> <3 0> <5 1>
I mean the list "a" has three lists that each one has 2 elements
And then how can i access each element or change the value of them?
You can initialise your list like this:
var list = new List<List<int>>()
{
new List<int>() {2, 1 },
new List<int>() {3, 0 },
new List<int>() {5, 1 }
};
And then you can access each element like this:
var x = list[0][1]; // 1
var y = list[1][0]; // 3
And you can access each inner list like this:
var inner = list[0];// List<int> (2, 1)
And you can update the list like this:
list[0][1] = 42;
or
list[0] = new List<int>() { 10, 11 };
EDIT: How to initialise the list with 10 lists of 1, 1
var list = new List<List<int>>();
for(var i=0;i<10;i++)
{
list.Add(new List<int>() {1, 1});
};
he needs a list of 3 lists of 2 elements each.
List<List<int>> a = new List<List<int>> { new List<int>() { 2, 1 }, new List<int>() { 3, 0 }, new List<int>() { 5, 1 } };
I'd like to sort multiple lists (variable number of them) into single list, but keeping the specific order. For example:
List A: { 1,2,3,4,5 }
List B: { 6,7,8 }
List C: { 9,10,11,12 }
Result List: { 1,6,9,2,7,10,3,8,11,4,12,5 }
The only idea I got was to remove the first element from each list and put it into resulting set (and repeat until all lists are empty), but maybe there is a better way that doesn't require to create copy of each list and doesn't affect the original lists as well?
I suggest using IEnumerator<T> to enumerate lists while they have items:
private static IEnumerable<T> Merge<T>(params IEnumerable<T>[] sources) {
List<IEnumerator<T>> enums = sources
.Select(source => source.GetEnumerator())
.ToList();
try {
while (enums.Any()) {
for (int i = 0; i < enums.Count;)
if (enums[i].MoveNext()) {
yield return enums[i].Current;
i += 1;
}
else {
// exhausted, let's remove enumerator
enums[i].Dispose();
enums.RemoveAt(i);
}
}
}
finally {
foreach (var en in enums)
en.Dispose();
}
}
Test
List<int> A = new List<int>() { 1, 2, 3, 4, 5 };
List<int> B = new List<int>() { 6, 7, 8 };
List<int> C = new List<int>() { 9, 10, 11, 12 };
var result = Merge(A, B, C)
.ToList();
Console.Write(string.Join(", ", result));
The outcome is
1, 6, 9, 2, 7, 10, 3, 8, 11, 4, 12, 5
For more flexible use
public static string MergeArrays(params IList<int>[] items)
{
var result = new List<int>();
for (var i = 0; i < items.Max(x => x.Count); i++)
result.AddRange(from rowList in items where rowList.Count > i select rowList[i]);
return string.Join(",", result);
}
.
var a = new List<int>() { 1, 2, 3, 4, 5 };
var b = new List<int>() { 6, 7, 8 };
var c = new List<int>() { 9, 10, 11, 12, 0, 2, 1 };
var r = MergeArrays(a, b, c);
There is no sense in over complicating this in my opinion, why not use a simple for loop to accomplish what you need?
List<int> list1 = new List<int> { 1, 2, 3, 4, 5 };
List<int> list2 = new List<int> { 6, 7, 8 };
List<int> list3 = new List<int> { 9, 10, 11, 12 };
List<int> resultList = new List<int>();
for (int i = 0; i < list1.Count || i < list2.Count || i < list3.Count; i++)
{
if (i < list1.Count) resultList.Add(list1[i]);
if (i < list2.Count) resultList.Add(list2[i]);
if (i < list3.Count) resultList.Add(list3[i]);
}
Result: 1,6,9,2,7,10,3,8,11,4,12,5
Here's a fairly simple way. It was fun to write up anyway.
No, it isn't the best, but it works and you could expand it to suit your needs of using a List<List<int>> very easily.
//Using arrays for simplicity, you get the idea.
int[] A = { 1, 2, 3, 4, 5 };
int[] B = { 6, 7, 8 };
int[] C = { 9, 10, 11, 12 };
List<int> ResultSet = new List<int>();
//Determine this somehow. I'm doing this for simplicity.
int longest = 5;
for (int i = 0; i < longest; i++)
{
if (i < A.Length)
ResultSet.Add(A[i]);
if (i < B.Length)
ResultSet.Add(B[i]);
if (i < C.Length)
ResultSet.Add(C[i]);
}
//ResultSet contains: { 1, 6, 9, 2, 7, 10, 3, 8, 11, 4, 12, 5 }
As you can see, just pop this out into a method and loop through your lists of lists, properly determining the max length of all lists.
I'd go with:
static void Main(string[] args)
{
var a = new List<int>() { 1, 2, 3, 4, 5 };
var b = new List<int>() { 6, 7, 8 };
var c = new List<int>() { 9, 10, 11, 12 };
var abc = XYZ<int>(new[] { a, b, c }).ToList();
}
static IEnumerable<T> XYZ<T>(IEnumerable<IList<T>> lists)
{
if (lists == null)
throw new ArgumentNullException();
var finished = false;
for (int index = 0; !finished; index++)
{
finished = true;
foreach (var list in lists)
if (list.Count > index) // list != null (prior checking for count)
{
finished = false;
yield return list[index];
}
}
}
I had to use use IList to have indexer and Count. It doesn't creates anything (no enumerators, no lists, etc.), purely yield return.
For your problem I create static method, which can merge any collections as you want:
public static class CollectionsHandling
{
/// <summary>
/// Merge collections to one by index
/// </summary>
/// <typeparam name="T">Type of collection elements</typeparam>
/// <param name="collections">Merging Collections</param>
/// <returns>New collection {firsts items, second items...}</returns>
public static IEnumerable<T> Merge<T>(params IEnumerable<T>[] collections)
{
// Max length of sent collections
var maxLength = 0;
// Enumerators of all collections
var enumerators = new List<IEnumerator<T>>();
foreach (var item in collections)
{
maxLength = Math.Max(item.Count(), maxLength);
if(collections.Any())
enumerators.Add(item.GetEnumerator());
}
// Set enumerators to first item
enumerators.ForEach(e => e.MoveNext());
var result = new List<T>();
for (int i = 0; i < maxLength; i++)
{
// Add elements to result collection
enumerators.ForEach(e => result.Add(e.Current));
// Remobve enumerators, in which no longer have elements
enumerators = enumerators.Where(e => e.MoveNext()).ToList();
}
return result;
}
}
Example of using:
static void Main(string[] args)
{
var a = new List<int> { 1, 2, 3, 4, 5 };
var b = new List<int> { 6, 7, 8 };
var c = new List<int> { 9, 10, 11, 12 };
var result= CollectionsHandling.Merge(a, b, c);
}
When you understand how it works, it will be possible to reduce the method of smaller.
Shortest and probably slowest solution
int[] A = { 1, 2, 3, 4, 5 };
int[] B = { 6, 7, 8 };
int[] C = { 9, 10, 11, 12 };
var arrs = new[] { A, B, C };
var merged = Enumerable.Range(0, arrs.Max(a => a.Length))
.Select(x => arrs.Where(a=>a.Length>x).Select(a=>a[x]))
.SelectMany(x=>x)
.ToArray();
upd.
Another way to solve - I just refactored #Sinatr answer.
static IEnumerable<T> XYZ<T>(IEnumerable<IList<T>> lists)
{
if (lists == null)
throw new ArgumentNullException();
var index = 0;
while (lists.Any(l => l.Count > index))
{
foreach (var list in lists)
if (list.Count > index)
yield return list[index];
index++;
}
}
This question already has answers here:
Combination of List<List<int>>
(9 answers)
Closed 9 years ago.
Differs from sugested solution above in that a list item can only appear once for each row.
This is for a booking system for my spa. Different employees can perform different treatments.
I have a List<List<int>>. These are therapists that can perform the treatment that is booked.
Each list (booking) contain a number of integers like this (these are therapists that can perform the booking):
{1, 3, 6}, //Booking 1
{1, 2, 6}, //Booking 2
{1}, //Booking 3
{2,3} //Booking 4
I'd like to see all possible combinations where the number can only appear in one Place. For the above list the two possible ombinations would be:
6,2,1,3 or
3,6,1,2
That is for the first combination:
Booking 1: Therapist 6
Booking 2: Therapist 2
Booking 3: Therapist 1
Booking 4: Therapist 3
Hope this makes the question a Little bit clearer.
Solve by recursion:
static IEnumerable<List<int>> GetCombinations(IEnumerable<List<int>> lists, IEnumerable<int> selected)
{
if (lists.Any())
{
var remainingLists = lists.Skip(1);
foreach (var item in lists.First().Where(x => !selected.Contains(x)))
foreach (var combo in GetCombinations(remainingLists, selected.Concat(new int[] { item })))
yield return combo;
}
else
{
yield return selected.ToList();
}
}
static void Main(string[] args)
{
List<List<int>> lists = new List<List<int>>
{
new List<int> { 1, 3, 6 },
new List<int> { 1, 2, 6 },
new List<int> { 1 },
new List<int> { 2, 3 }
};
var combos = GetCombinations(lists, new List<int>()).Distinct();
foreach (var combo in combos)
Console.WriteLine("{ " + string.Join(", ", combo.Select(x => x.ToString())) + " }");
return;
}
Output:
{ 3, 6, 1, 2 }
{ 6, 2, 1, 3 }
This solution is far from efficient:
private static void Main()
{
List<List<int>> list = new List<List<int>>
{
new List<int>() {1, 3, 6}, //Booking 1
new List<int>() {1, 2, 6}, //Booking 2
new List<int>() {1}, //Booking 3
new List<int>() {2, 3}
};
List<int[]> solutions = new List<int[]>();
int[] solution = new int[list.Count];
Solve(list, solutions, solution);
}
private static void Solve(List<List<int>> list, List<int[]> solutions, int[] solution)
{
if (solution.All(i => i != 0) && !solutions.Any(s => s.SequenceEqual(solution)))
solutions.Add(solution);
for (int i = 0; i < list.Count; i++)
{
if (solution[i] != 0)
continue; // a caller up the hierarchy set this index to be a number
for (int j = 0; j < list[i].Count; j++)
{
if (solution.Contains(list[i][j]))
continue;
var solutionCopy = solution.ToArray();
solutionCopy[i] = list[i][j];
Solve(list, solutions, solutionCopy);
}
}
}
It sounds like this can be solved more efficiently with Dynamic programming, but it's been a while since I took the relevant course.
A simple way to look at this problem would be to choose from all combinations of the list of values, where every value in the combination is unique.
First figure out what all the combinations of values are.
public static IEnumerable<IList<T>> Combinations<T>(IEnumerable<IList<T>> collections)
{
if (collections.Count() == 1)
{
foreach (var item in collections.Single())
yield return new List<T> { item };
}
else if (collections.Count() > 1)
{
foreach (var item in collections.First())
foreach (var tail in Combinations(collections.Skip(1)))
yield return new[] { item }.Concat(tail).ToList();
}
}
Then you need a way to determine if all the values are unique. A simple way to figure that out would be to check if the count of distinct values equals the count of all values.
public static bool AllUnique<T>(IEnumerable<T> collection)
{
return collection.Distinct().Count() == collection.Count();
}
Once you have all that, put it all together.
var collections = new[]
{
new List<int> { 1, 3, 6 },
new List<int> { 1, 2, 6 },
new List<int> { 1 },
new List<int> { 2, 3 },
};
var results =
from combination in Combinations(collections)
where AllUnique(combination)
select combination;
// results:
// 3,6,1,2
// 6,2,1,3