Develop Reference

How to Order By or Sort an integer List and select the Nth element

I have a list, and I want to select the fifth highest element from it:
List<int> list = new List<int>();
list.Add(2);
list.Add(18);
list.Add(21);
list.Add(10);
list.Add(20);
list.Add(80);
list.Add(23);
list.Add(81);
list.Add(27);
list.Add(85);
But OrderbyDescending is not working for this int list...

int fifth = list.OrderByDescending(x => x).Skip(4).First();

Depending on the severity of the list not having more than 5 elements you have 2 options.
If the list never should be over 5 i would catch it as an exception:
int fifth;
try
{
fifth = list.OrderByDescending(x => x).ElementAt(4);
}
catch (ArgumentOutOfRangeException)
{
//Handle the exception
}
If you expect that it will be less than 5 elements then you could leave it as default and check it for that.
int fifth = list.OrderByDescending(x => x).ElementAtOrDefault(4);
if (fifth == 0)
{
//handle default
}
This is still some what flawed because you could end up having the fifth element being 0. This can be solved by typecasting the list into a list of nullable ints at before the linq:
var newList = list.Select(i => (int?)i).ToList();
int? fifth = newList.OrderByDescending(x => x).ElementAtOrDefault(4);
if (fifth == null)
{
//handle default
}

Without LINQ expressions:
int result;
if(list != null && list.Count >= 5)
{
list.Sort();
result = list[list.Count - 5];
}
else // define behavior when list is null OR has less than 5 elements
This has a better performance compared to LINQ expressions, although the LINQ solutions presented in my second answer are comfortable and reliable.
In case you need extreme performance for a huge List of integers, I'd recommend a more specialized algorithm, like in Matthew Watson's answer.
Attention: The List gets modified when the Sort() method is called. If you don't want that, you must work with a copy of your list, like this:
List<int> copy = new List<int>(original);
List<int> copy = original.ToList();

The easiest way to do this is to just sort the data and take N items from the front. This is the recommended way for small data sets - anything more complicated is just not worth it otherwise.
However, for large data sets it can be a lot quicker to do what's known as a Partial Sort.
There are two main ways to do this: Use a heap, or use a specialised quicksort.
The article I linked describes how to use a heap. I shall present a partial sort below:
public static IList<T> PartialSort<T>(IList<T> data, int k) where T : IComparable<T>
{
int start = 0;
int end = data.Count - 1;
while (end > start)
{
var index = partition(data, start, end);
var rank = index + 1;
if (rank >= k)
{
end = index - 1;
}
else if ((index - start) > (end - index))
{
quickSort(data, index + 1, end);
end = index - 1;
}
else
{
quickSort(data, start, index - 1);
start = index + 1;
}
}
return data;
}
static int partition<T>(IList<T> lst, int start, int end) where T : IComparable<T>
{
T x = lst[start];
int i = start;
for (int j = start + 1; j <= end; j++)
{
if (lst[j].CompareTo(x) < 0) // Or "> 0" to reverse sort order.
{
i = i + 1;
swap(lst, i, j);
}
}
swap(lst, start, i);
return i;
}
static void swap<T>(IList<T> lst, int p, int q)
{
T temp = lst[p];
lst[p] = lst[q];
lst[q] = temp;
}
static void quickSort<T>(IList<T> lst, int start, int end) where T : IComparable<T>
{
if (start >= end)
return;
int index = partition(lst, start, end);
quickSort(lst, start, index - 1);
quickSort(lst, index + 1, end);
}
Then to access the 5th largest element in a list you could do this:
PartialSort(list, 5);
Console.WriteLine(list[4]);
For large data sets, a partial sort can be significantly faster than a full sort.
Addendum
See here for another (probably better) solution that uses a QuickSelect algorithm.

This LINQ approach retrieves the 5th biggest element OR throws an exception WHEN the list is null or contains less than 5 elements:
int fifth = list?.Count >= 5 ?
list.OrderByDescending(x => x).Take(5).Last() :
throw new Exception("list is null OR has not enough elements");
This one retrieves the 5th biggest element OR null WHEN the list is null or contains less than 5 elements:
int? fifth = list?.Count >= 5 ?
list.OrderByDescending(x => x).Take(5).Last() :
default(int?);
if(fifth == null) // define behavior
This one retrieves the 5th biggest element OR the smallest element WHEN the list contains less than 5 elements:
if(list == null || list.Count <= 0)
throw new Exception("Unable to retrieve Nth biggest element");
int fifth = list.OrderByDescending(x => x).Take(5).Last();
All these solutions are reliable, they should NEVER throw "unexpected" exceptions.
PS: I'm using .NET 4.7 in this answer.

Here there is a C# implementation of the QuickSelect algorithm to select the nth element in an unordered IList<>.
You have to put all the code contained in that page in a static class, like:
public static class QuickHelpers
{
// Put the code here
}
Given that "library" (in truth a big fat block of code), then you can:
int resA = list.QuickSelect(2, (x, y) => Comparer<int>.Default.Compare(y, x));
int resB = list.QuickSelect(list.Count - 1 - 2);
Now... Normally the QuickSelect would select the nth lowest element. We reverse it in two ways:
For resA we create a reverse comparer based on the default int comparer. We do this by reversing the parameters of the Compare method. Note that the index is 0 based. So there is a 0th, 1th, 2th and so on.
For resB we use the fact that the 0th element is the list-1 th element in the reverse order. So we count from the back. The highest element would be the list.Count - 1 in an ordered list, the next one list.Count - 1 - 1, then list.Count - 1 - 2 and so on
Theorically using Quicksort should be better than ordering the list and then picking the nth element, because ordering a list is on average a O(NlogN) operation and picking the nth element is then a O(1) operation, so the composite is O(NlogN) operation, while QuickSelect is on average a O(N) operation. Clearly there is a but. The O notation doesn't show the k factor... So a O(k1 * NlogN) with a small k1 could be better than a O(k2 * N) with a big k2. Only multiple real life benchmarks can tell us (you) what is better, and it depends on the size of the collection.
A small note about the algorithm:
As with quicksort, quickselect is generally implemented as an in-place algorithm, and beyond selecting the k'th element, it also partially sorts the data. See selection algorithm for further discussion of the connection with sorting.
So it modifies the ordering of the original list.

How would i find a range of numbers inside of an array that suit a condition?

I'm trying to understand arrays better and I've found this piece of code on this website that fills an array with a load of random numbers. I was wondering how you would go about, say, extracting a range of numbers. So if i wanted to find out how many of these random numbers inside of the array, were between 25 and 50, how would i go about doing this? I've heard about the Array.FindAll<> however i have no clue on how to use it.
Thank you in advance.
Random r = new Random();
int count = 100;
// Create an array with count elements.
int[] numbers = new int[count];
// Loop over each index
for (int i = 0; i < count; i++)
{
// Generate and store a random number at current index
numbers[i] = r.Next(1, 100);
}

Using LINQ you can try the following:
var numbersInRange = numbers.Where(number => number > 25 && number < 60)
.ToArray();
The above will filter out all the numbers in the range (25,60). If you want to get also the numbers that are equal to either 25 or 60, you just have to use >= and <= repsectively.
If you don't want to use LINQ (which I couldn't explain), you could try the following:
var numbersInRange = new List<int>();
foreach(var number in numbers)
{
if(number > 25 && number < 60)
{
numbersInRange.Add(number);
}
}
So at the end you would have a list that would contains your numbers.
Comparing this to the LINQ version, I think that LINQ is the winner in simplicity and readability.

Using Linq it is a single line
int[] inRange = numbers.Where(x => x >= 25 && x <= 50).ToArray();
The Enumerable methods are a collection of methods that operates on sequences of values and apply a predicate expression (x => 25 && x <= 50) to each member (x) of the sequence (numbers). In your particular case the Where method filters the sequence based on the result of the boolean expression (the return includes the members where the expression returns true).

We can use Linq. First have using System.Linq in your namespaces so the extension methods are available.
To find the numbers:
numbers.Where(x => x >= 25 & x <= 50)
This takes a lambda expression (in this case x => x >= 25 & x <= 50) that takes on object of the type in the sequence (in this case int) returns bool and then evaluates for that expression for all contained items, filtering out those that do not.
You can use ToArray() on the result to get it back into an array, or just work on the results directly.
To get which numbers, that is to say, which indices the matches where at, you can use:
numbers.Select((x, i) => new {El = x, Idx = i}).Where(x => x.EL >= 25 & x.El <= 50).Select(x => x.Idx)
This first creates a new anonymous object for each item where El is the value held and Idx is the index it was at. Then it does the same Where as before, but on the El property of the object we created in the first step. Finally it extracts out the Idx property of the survivors so we have all of the matching indices.

Although "Linq" can do the job as shown by other answers ; there is also, as you heard about, Array.FindAll which behaves the same but already returns an array (and so doesn't need a call to ToArray)
int[] matchedNumbers = Array.FindAll (currentNumber => 25 <= currentNumber && currentNumber <= 50);
As for the "strange" arrow notation it's a lambda expression

Get all possible distinct triples using LINQ

I have a List contains these values: {1, 2, 3, 4, 5, 6, 7}. And I want to be able to retrieve unique combination of three. The result should be like this:
{1,2,3}
{1,2,4}
{1,2,5}
{1,2,6}
{1,2,7}
{2,3,4}
{2,3,5}
{2,3,6}
{2,3,7}
{3,4,5}
{3,4,6}
{3,4,7}
{3,4,1}
{4,5,6}
{4,5,7}
{4,5,1}
{4,5,2}
{5,6,7}
{5,6,1}
{5,6,2}
{5,6,3}
I already have 2 for loops that able to do this:
for (int first = 0; first < test.Count - 2; first++)
{
int second = first + 1;
for (int offset = 1; offset < test.Count; offset++)
{
int third = (second + offset)%test.Count;
if(Math.Abs(first - third) < 2)
continue;
List<int> temp = new List<int>();
temp .Add(test[first]);
temp .Add(test[second]);
temp .Add(test[third]);
result.Add(temp );
}
}
But since I'm learning LINQ, I wonder if there is a smarter way to do this?

UPDATE: I used this question as the subject of a series of articles starting here; I'll go through two slightly different algorithms in that series. Thanks for the great question!
The two solutions posted so far are correct but inefficient for the cases where the numbers get large. The solutions posted so far use the algorithm: first enumerate all the possibilities:
{1, 1, 1 }
{1, 1, 2 },
{1, 1, 3 },
...
{7, 7, 7}
And while doing so, filter out any where the second is not larger than the first, and the third is not larger than the second. This performs 7 x 7 x 7 filtering operations, which is not that many, but if you were trying to get, say, permutations of ten elements from thirty, that's 30 x 30 x 30 x 30 x 30 x 30 x 30 x 30 x 30 x 30, which is rather a lot. You can do better than that.
I would solve this problem as follows. First, produce a data structure which is an efficient immutable set. Let me be very clear what an immutable set is, because you are likely not familiar with them. You normally think of a set as something you add items and remove items from. An immutable set has an Add operation but it does not change the set; it gives you back a new set which has the added item. The same for removal.
Here is an implementation of an immutable set where the elements are integers from 0 to 31:
using System.Collections;
using System.Collections.Generic;
using System.Diagnostics;
using System.Linq;
using System;
// A super-cheap immutable set of integers from 0 to 31 ;
// just a convenient wrapper around bit operations on an int.
internal struct BitSet : IEnumerable<int>
{
public static BitSet Empty { get { return default(BitSet); } }
private readonly int bits;
private BitSet(int bits) { this.bits = bits; }
public bool Contains(int item)
{
Debug.Assert(0 <= item && item <= 31);
return (bits & (1 << item)) != 0;
}
public BitSet Add(int item)
{
Debug.Assert(0 <= item && item <= 31);
return new BitSet(this.bits | (1 << item));
}
public BitSet Remove(int item)
{
Debug.Assert(0 <= item && item <= 31);
return new BitSet(this.bits & ~(1 << item));
}
IEnumerator IEnumerable.GetEnumerator() { return this.GetEnumerator(); }
public IEnumerator<int> GetEnumerator()
{
for(int item = 0; item < 32; ++item)
if (this.Contains(item))
yield return item;
}
public override string ToString()
{
return string.Join(",", this);
}
}
Read this code carefully to understand how it works. Again, always remember that adding an element to this set does not change the set. It produces a new set that has the added item.
OK, now that we've got that, let's consider a more efficient algorithm for producing your permutations.
We will solve the problem recursively. A recursive solution always has the same structure:
Can we solve a trivial problem? If so, solve it.
If not, break the problem down into a number of smaller problems and solve each one.
Let's start with the trivial problems.
Suppose you have a set and you wish to choose zero items from it. The answer is clear: there is only one possible permutation with zero elements, and that is the empty set.
Suppose you have a set with n elements in it and you want to choose more than n elements. Clearly there is no solution, not even the empty set.
We have now taken care of the cases where the set is empty or the number of elements chosen is more than the number of elements total, so we must be choosing at least one thing from a set that has at least one thing.
Of the possible permutations, some of them have the first element in them and some of them do not. Find all the ones that have the first element in them and yield them. We do this by recursing to choose one fewer elements on the set that is missing the first element.
The ones that do not have the first element in them we find by enumerating the permutations of the set without the first element.
static class Extensions
{
public static IEnumerable<BitSet> Choose(this BitSet b, int choose)
{
if (choose < 0) throw new InvalidOperationException();
if (choose == 0)
{
// Choosing zero elements from any set gives the empty set.
yield return BitSet.Empty;
}
else if (b.Count() >= choose)
{
// We are choosing at least one element from a set that has
// a first element. Get the first element, and the set
// lacking the first element.
int first = b.First();
BitSet rest = b.Remove(first);
// These are the permutations that contain the first element:
foreach(BitSet r in rest.Choose(choose-1))
yield return r.Add(first);
// These are the permutations that do not contain the first element:
foreach(BitSet r in rest.Choose(choose))
yield return r;
}
}
}
Now we can ask the question that you need the answer to:
class Program
{
static void Main()
{
BitSet b = BitSet.Empty.Add(1).Add(2).Add(3).Add(4).Add(5).Add(6).Add(7);
foreach(BitSet result in b.Choose(3))
Console.WriteLine(result);
}
}
And we're done. We have generated only as many sequences as we actually need. (Though we have done a lot of set operations to get there, but set operations are cheap.) The point here is that understanding how this algorithm works is extremely instructive. Recursive programming on immutable structures is a powerful tool that many professional programmers do not have in their toolbox.

You can do it like this:
var data = Enumerable.Range(1, 7);
var r = from a in data
from b in data
from c in data
where a < b && b < c
select new {a, b, c};
foreach (var x in r) {
Console.WriteLine("{0} {1} {2}", x.a, x.b, x.c);
}
Demo.
Edit: Thanks Eric Lippert for simplifying the answer!

var ints = new int[] { 1, 2, 3, 4, 5, 6, 7 };
var permutations = ints.SelectMany(a => ints.Where(b => (b > a)).
SelectMany(b => ints.Where(c => (c > b)).
Select(c => new { a = a, b = b, c = c })));

Transform sequence in Linq while being aware of each Select/SelectMany result

Here's my problem. I have one specific list, which I'll present as a int[] for simplicity's sake.
int[] a = {1,2,3,4,5};
Suppose I need to transform each item on this list, but depending on the situation, I may return an int or an array of ints.
As an example, suppose I need to return {v} if the value is odd, and {v,v+1} if the value is even. I've done this:
int[] b = a.SelectMany(v => v % 2 == 0 ? new int[] { v, v+1 } : new int[] { v })
.ToArray();
So if I run this, I'll get the expected response:
{1,2,3,3,4,5,5}
See that I have repeating numbers, right? 3 and 5. I don't want those repeating numbers. Now, you may tell me that I can just call .Distinct() after processing the array.
This is the problem. The SelectMany clause is fairly complex (I just made up a simpler example), and I definitely don't want to process 3 if it's already present in the list.
I could check if 3 is present in the original list. But if I got 3 in the SelectMany clause, I don't want to get it again. For instance, if I had this list:
int[] a = {1,2,3,4,5,2};
I would get this:
{1,2,3,3,4,5,5,2,3}
Thus returning v (my original value) and v+1 again at the end. Just so you can understand it better v+1 represents some processing I want to avoid.
Summarizing, this is what I want:
I have a list of objects. (Check)
I need to filter them, and depending on the result, I may need to return more than one object. (Check, used SelectMany)
I need them to be distinct, but I can't do that at the end of the process. I should be able to return just {v} if {v+1} already exists. (Clueless...)
One thing I thought about is writing a custom SelectMany which may suit my needs, but I want to be sure there's no built-in way to do this.
EDIT: I believe I may have mislead you guys with my example. I know how to figure out if v+1 is in a list. To be clear, I have one object which has 2 int properties, Id and IdParent. I need to "yield return" all the objects and their parents. But I just have the ParentId, which comes from the objects themselves. I'm able to know if v+1 is in the list because I can check if any object there has the same Id as the ParentId I'm checking.
ANSWER: I ended up using Aggregate, which can be used to do exactly what I'm looking for.

Does this simple loop with the HashSet<int> help?
int[] a = {1,2,3,4,5,2};
var aLookupList = new HashSet<int>();
foreach (int i in a)
{
bool isEven = i % 2 == 0;
if (isEven)
{
aLookupList.Add(i);
aLookupList.Add(i + 1);
}
else
{
aLookupList.Add(i);
}
}
var result = aLookupList.ToArray();

What about this using Aggregate method. You won't be processing numbers that are already in the list, wheather they were in the original list or as a result of applying (v + 1)
int[] v = { 1, 2, 3, 4, 5, 2 };
var result = v.Aggregate(new List<int>(),
(acc, next) =>
{
if (!acc.Contains(next))
return (next % 2 == 0) ? acc.Concat(new int[] { next, next + 1 }).ToList()
: acc.Concat(new int[] { next }).ToList();
else
return acc;
}).ToArray();

var existing = new HashSet<int>(a);
var result = existing
.Where(v => v % 2 == 0 && !existing.Contains(v + 1))
.Select(v => v + 1)
.Concat(existing)
.ToArray();

As I understand you have this input:
int[] a = {1,2,3,4,5};
And the output should also be {1,2,3,4,5} because you don't want duplicated numbers as you describe.
Because you use an array as input, you can try this code:
var output = a.SelectMany((x,i)=> x % 2 == 0 ? new []{x,x+1} :
i > 0 && a[i-1]==x-1 ? new int[]{} : new []{x});
//if the input is {1,2,4,5}
//The output is also {1,2,3,4,5}

How to extract values from arrays using upper and lower limits?

Given two arrays, I need to extract values from arrayB based on where the range(actual values) falls in arrayA.
Index 0 1 2 3 4 5 6 7 8 9 10 11 12
-------------------------------------------------------------
ArrayA = {0, 0.5, 1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5, 5, 5.5, 6}
ArrayB = {1, 0.2, 3, 4, 5, 6,5.5, 8, 9,11.1, 11, 12, 3}
Given the following ranges, I need to extract the following results
RangeToExtract* IndexInArrayA Expected Values To Extract
-------------- ------------- --------------------------
0 -> 1 [0,2] 1,0.2,3
1 -> 3 [3,6] 4,5,6,5.5
3 -> 5 [7,10] 5.5,8,9,11.1,11
1 -> 5 [3,10] 4,5,6,5.5,8,9,11.1,11
3 -> 10 [7,12] 8,9,11.1,11,12,3
* Refers to the actual values in ArrayA
Note: Given the RangeToExtract (0->1), determine the indexes in ArrayA where these values are, the result being (0->1) maps to [0,2] (The value 1 is in position 2 in ArrayA)
I only figured that the following special cases exists (not sure if there are more)
the lower limit is equal to zero and
when the upper limit does not exist in ArrayA
Further info:
Both arrays will be the same size
ArrayA will always be sorted
Code:
private double[] GetRange(double lower, double upper)
{
var myList = new double[ArrayA.Length];
var lowerIndex = Array.IndexOf(ArrayA, lower);
var upperIndex = Array.IndexOf(ArrayA, upper);
// special case 1
if (lowerIndex != 0)
{
lowerIndex = lowerIndex + 1;
}
// special case 2
if (upperIndex == -1)
{
upperIndex = ArrayA.Length-1;
}
for (int i = lowerIndex; i <= upperIndex; i++)
{
myList[i] = ArrayB[i];
}
return myList;
}
Given the above code, have all the special cases been taken into account? Is there a better way to write the above code?

Yap! There is a quite better way, that comes with lovely LINQ. I put here in two forms. First looks complicated but not at ALL! Believe me ;)
At the first step you have to take out those A'indexes that their values fall into your range (I call it min...max), based on your example I got that your range is closed from the lower boundary and closed on upper side, I means when you mentioned 3 -> 5 actually It is [3, 5)! It does not contain 5. Anyway that is not the matter.
This can be done by following LINQ
int[] selectedIndexes = a.Select((value, index) =>
new { Value = value, Index = index }).
Where(aToken => aToken.Value > min && aToken.Value <= max).
Select(t => t.Index).ToArray<int>();
The first select, generates a collection of [Value, Index] pairs that the first one is the array element and the second one is the index of the element within the array. I think this is the main trick for your question. So It provides you with this ability to work with the indexes same as usual values.
Finally in the second Select I just wrap whole indexes into an integer array. Hence after this you have the whole indexes that their value fall in the given range.
Now second step!
When you got those indexes, you have to select whole elements within the B under the selected Indexes from the A. The same thing should be done over the B. It means again we select B element into a collection of [Value, Index] pairs and then we select those guys that their indexes exist within the selected indexes from the A. This can be done as follow:
double[] selectedValues = b.Select((item, index) =>
new { Item = item, Index = index }).
Where(bToken => selectedIndexes.Contains(bToken.Index)).
Select(d => d.Item).ToArray<double>();
Ok, so first select is the one I talked about it in the fist part and then look at the where section that check whether the index of the bToken which is an element of B exists in the selectedIndexes (from A) or not!
Finally I wrap both codes into one as below:
double[] answers = b.Select((item, index) =>
new { Item = item, Index = index }).
Where(bTokent =>
a.Select((value, index) =>
new { Value = value, Index = index }).
Where(aToken => aToken.Value > min && aToken.Value <= max).
Select(t => t.Index).
Contains(bTokent.Index)).Select(d => d.Item).ToArray<double>();
Buy a beer for me, if it would be useful :)

I don't know if you're still interested, but I saw this one and I liked the challenge. If you use .Net 4 (having the Enumberable.Zip method) there is a very concise way to do this (given the conditions under futher info):
arrayA.Zip(arrayB, (a,b) => new {a,b})
.Where(x => x.a > lower && x.a < upper)
.Select (x => x.b)
You may want to use >= and <= to make the range comparisons inclusive.