I can't figure out how to do this, if even possible.
An example:
int[][] myArrays = {
new int[] {1, 2, 3},
new int[] {4, 5, 3},
new int[] {1, 2, 3}
};
int[] avgArray = myArrays.//Some LINQ statement to average every Nth element in the second dimension (essentially "flatten" the array)
//avgArray == {2, 3, 3}
To do this so far, I can only think of:
int ndDimLen = myArrays.GetLength(1);
int[] avgArray = new int[ndDimLen];
myArrays.Select(
arr => arr.Select( (n, i) => avgArray[i] += n )
);
avgArray = avgArray.Select( n => n / ndDimLen ).ToArray();
But this defeats the purpose, and isn't a particularly good idea on jagged arrays...
Also, I'd definitely like to avoid transposition, as it's quite a slow operation when operating on large arrays!
Thank you for your time!
You could iterate throught the [Columns] index while a [Row].Length reports that it contains a [Column] in the dimension whose values you need to average.
(Using the terms Column and Row for simplicity, as a visual aid)
An example, using Linq's .Average() to compute the average value of the sequence:
int[][] myArrays = {
new int[] {1, 2, 3},
new int[] {4, 5, 3},
new int[] {1, 2, 3},
};
int column = 2;
double avg = myArrays.Select((arr, i) => myArrays[i][column]).Average();
Result: 3
With a more complex structure, we need to verify whether the current [Column] contains a value:
int[][] myArrays = {
new int[] {1, 2, 3},
new int[] {3, 4, 5},
new int[] {3, 4},
new int[] {1, 2, 3, 4},
new int[] {1},
new int[] {4, 5, 3}
};
int column= 2;
double? avg = myArrays.Select((arr, i) =>
((arr.Length > column) ? myArrays?[i][column] : null)).Average();
Result Column 1: { 1, 3, 3, 1, 1, 4 } => 2.1666666...
Result Column 2: { 2, 4, 4, 2, 5 } => 3.4
Result Column 3: { 3, 5, 3, 3 } => 3.5
Result Column 4: { 4 } => 4
Same method, applied to all the [Columns]:
var Averages = Enumerable.Range(0, myArrays.Max(Arr => Arr.Length))
.Select(col => myArrays
.Select((arr, idx) =>
((arr.Length > col) ? myArrays?[idx][col] : null))
.Average()).ToList();
Enumerable.Range gives some flexibility.
The code above generates a series of int elements starting from 0 and incrementing the value to the number of Colums in the Array (Max(Arr => Arr.Length) selects the Array's Row containing the higher number of elements).
So, you could average the numbers in the first Column only (Index = 0) with:
var Averages = Enumerable.Range(0, 1).Select( ... )
or from Columns 1 to 3 (Indexes 1 to 3):
var Averages = Enumerable.Range(1, 3).Select( ... )
Yes, it is possible, but not on this object.
Basically, myArrays is an array of arrays, so LINQ only sees one row at a time, you cannot make it to see columns, because it's just not how it works.
What you could do, is to transpose this "table" first, that is change places of columns and rows. How to do it has already been discussed here so I will just refer you to it.
Using knowledge how to transpose it, you can make a method that will do it, and use LINQ on it, like:
Transpose(myArray).Select(predicate);
You didn't specify what you want if the arrays have unequal length:
int[][] myArrays =
{
new int[] {1, 2},
new int[] {4, 5, 3, 7, 5, 3, 4, 5, 1},
new int[] {1, 2, 3}
};
Let's assume your arrays all have the same length.
If you plan to use this functionality regularly, consider writing an extension function for two dimensional arrays. See Extension Methods Demystified
public static IEnumerable<int> ToVerticalAverage(this int[][] array2D)
{
if (array2D == null) throw new ArgumentNullException(nameof(array2D);
// if input empty: return empty
if (array2D.Any())
{ // there is at least one row
// check number of columns:
int nrOfColumns = array2D.First().Length;
if (!array2D.All(row => row.Length == nrOfColumns)
{
// TODO: decide what to do if arrays have unequal length
}
// if here, at least one row, all rows have same number of columns
for(int columNr = 0; columNr < nrOfColumns; ++columNr)
{
yield return array2D.Select(row => row[i]).Average();
}
}
// else: no rows, returns empty sequence
}
Usage:
int[][] myInputValues = ...
IEnumerable<int> averageColumnValues = myInputValues.ToVerticalAverage();
If you have several functions where you need the values of the columns, write an extension function to fetch the columns (= transpose the matrix)
public static IEnumerable<IEnumerable<int>> Transpose(this int[][] array2D)
{
// TODO: check input
int nrOfColumns = array2D.First().Length;
for(int columNr = 0; columNr < nrOfColumns; ++columNr)
{
yield return array2D.Select(row => row[columnNr];
}
}
public static IEnumerable<int> ToVerticalAverage(this int[][] array2D)
{
// TODO: check input
foreach (IEnumerable<int> column in array2D.Transpose())
{
yield return column.Average();
}
Related
I need to find a way to return the longest match found in number of sets/lists (values returns only once) when the order of items is important.
the list is not cyclic.
A match is a sequence of values that exists in all the lists and maintains the same order of elements in all the lists.
e.g. 1:
List<int> list1 = new List<int> { 1, 2, 3, 4, 7, 9 };
List<int> list2 = new List<int> { 1, 2, 5, 6, 3, 4, 7, 9 };
List<int> list3 = new List<int> { 1, 2, 3, 6, 8, 9 };
List<int> list4 = new List<int> { 1, 2, 5, 6, 8, 9 };
result { 1, 2 }
e.g. 2:
List<int> list1 = new List<int> { 2, 3, 6, 8, 1, 18 };
List<int> list2 = new List<int> { 2, 3, 4, 6, 8, 1, 18, 19, 17, 14 };
List<int> list3 = new List<int> { 2, 5, 6, 8, 1, 18, 16, 13, 14 };
List<int> list4 = new List<int> { 2, 6, 8, 1, 18, 19, 17, 14 };
result { 6, 8, 1, 18 }
The match doesn't have to be found at the beginning or at the end and can be on any part of any list.
I hope that I explained my problem good enough :)
Thanks!
You can build a map from pairs of ints to a count of how many of the lists they appear adjacent in.
Pseudo-code:
For each list L {
For each adjacent pair (x, y) in L {
Counts[x, y] += 1
}
}
Now you can iterate through the first list (or the shortest list), and find the longest run such that each adjacent pair (x, y) in the run with Counts[x, y] showing that the pair appears in every list.
Pseudo-code:
run = []
best_run = []
For x in L[0] {
if len(run) is zero or Counts[run[len(run)-1], x] == number of lists {
run = run + x
} else {
run = [x]
}
if run is longer than best_run {
best_run = run
}
}
This works given the assumption in the question that no integer appears twice in the same list.
This algorithm runs in O(N) time, where N is the sum of the lengths of all the lists.
Here's my approach.
First I need a way to compare lists:
public class ListCompare<T> : IEqualityComparer<List<T>>
{
public bool Equals(List<T> left, List<T> right)
{
return left.SequenceEqual(right);
}
public int GetHashCode(List<T> list)
{
return list.Aggregate(0, (a, t) => a ^ t.GetHashCode());
}
}
Next a method to produce all subsequences of a source list:
Func<List<int>, IEnumerable<List<int>>> subsequences = xs =>
from s in Enumerable.Range(0, xs.Count)
from t in Enumerable.Range(1, xs.Count - s)
select xs.Skip(s).Take(t).ToList();
Now I can create a list of lists:
var lists = new [] { list1, list2, list3, list4, };
Finally a query that pulls it all together:
var answer =
lists
.Skip(1)
.Aggregate(
subsequences(lists.First()),
(a, l) => a.Intersect(subsequences(l), new ListCompare<int>()))
.OrderByDescending(x => x.Count)
.FirstOrDefault();
Given the sample data provided in the question this produces the expected results.
First generate an ordered combination of int from the shortest list
Compare the lists other than shortest list with the combination. For easy comparison of lists I just convert to string and use string.Contains()
Return immediately if find the match as the items left are next order or the shorter one.
public static List<int> GetLongestMatch(params List<int>[] all)
{
var shortest = all.Where(i => i.Count == all.Select(j => j.Count).Min()).First();
var permutations = (from length in Enumerable.Range(1, shortest.Count)
orderby length descending
from count in Enumerable.Range(1, shortest.Count - length + 1)
select shortest.Skip(count - 1).Take(length).ToList())
.ToList();
Func<List<int>, string> stringfy = (list) => { return string.Join(",", list.Select(i => i.ToString()).ToArray()); };
foreach (var item in permutations)
{
Debug.WriteLine(string.Join(", ", item.Select(i => i.ToString()).ToArray()));
if (all.All(list => stringfy(list).Contains(stringfy(item))))
{
Debug.WriteLine("Matched, skip process and return");
return item;
}
}
return new List<int>();
}
Usage
var result = GetLongestMatch(list1, list2, list3, list4);
Result
2, 3, 6, 8, 1, 18
2, 3, 6, 8, 1
3, 6, 8, 1, 18
2, 3, 6, 8
3, 6, 8, 1
6, 8, 1, 18
Matched, skip process and return
I want to find the top 3 maximum repeated numbers in a Integer array?
Below is the piece of code which I have tried but I couldn't find the desired result:
static void Main(string[] args)
{
int[,] numbers = {
{1, 2, 0, 6 },
{5, 6, 7, 0 },
{9, 3, 6, 2 },
{6, 4, 8, 1 }
};
int count = 0;
List<int> checkedNumbers = new List<int>();
foreach (int t in numbers)
{
if (!checkedNumbers.Contains(t))
{
foreach (int m in numbers)
{
if (m == t)
{
count++;
}
}
Console.WriteLine("Number {0} is Repeated {1} Times ", t, count);
count = 0;
checkedNumbers.Add(t);
}
}
Console.ReadLine();
}
You can use GroupBy from LINQ then OrderByDescending based on count in each group:
var result = list.GroupBy(i => i)
.OrderByDescending(g => g.Count())
.Select(g => g.Key)
.Take(3);
Edit: With your code, you can use OfType to flatten your matrix then use the code above:
int[,] numbers = {
{1, 2, 0, 6 },
{5, 6, 7, 0 },
{9, 3, 6, 2 },
{6, 4, 8, 1 }
};
var list = numbers.OfType<int>();
int[] numbers = {1, 2, 3, 5, 6, 32, 2, 4, 42, 2, 4, 4, 5, 6, 3, 4};
var counts = new Dictionary<int, int>();
foreach (var number in numbers)
{
counts[number] = counts[number] + 1;
}
var top3 = counts.OrderByDescending(x => x.Value).Select(x => x.Key).Take(3);
Hint:
You can do this with the help of LINQ.
This is the code to find most frequest occuring element:-
List<int> list = new List<int>() { 1,1,2,2,3,4,5 };
// group by value and count frequency
var query = from i in list
group i by i into g
select new {g.Key, Count = g.Count()};
// compute the maximum frequency
int frequency = query.Max(g => g.Count);
// find the values with that frequency
IEnumerable<int> modes = query
.Where(g => g.Count == frequency)
.Select(g => g.Key);
// dump to console
foreach(var mode in modes) {
Console.WriteLine(mode);
}
In the same manner you can find the other two also.
I see that none of the existing answers provide an explanation, so I will try to explain.
What you need to do is to count how many times each item appears in the array. To do that, there are various methods (dictionaries, linq etc). Probably it would be easiest to use a dictionary which contains the number, and how may times it appeared:
int numbers[] = {1, 3, 6, 10, 9, 3, 3, 1, 10} ;
Dictionary<int, int> dic = new Dictionary<int, int>();
Now iterate through every element in numbers, and add it to the dictionary. If it was already added, simply increase the count value.
foreach (var i in numbers)
{
dic[i]++; // Same as dic[i] = dic[i]+1;
}
The dictionary will automatically adds a new item if it doesn't exist, so we can simply do dic[i]++;
Next, we need to get the highest 3 values. Again, there are many ways to do this, but the easiest one would be to sort it.
var sorted_dic = dic.OrderByDescending(x => x.Value);
Now the first 3 items in sorted_dic are going to be the 3 values you are looking for.
There are various methods to get only these 3, for example using the Take method:
var first_3 = sorted_dic.Take(3);
Now you can iterate through these 3 values, and for example print them on the screen:
foreach (var i in first_3)
{
Console.Write("{0} appeared {1} times.", i.Key, i.Value);
}
public class Unicorn
{
public List<int> Numbers { get; set; }
}
unicorns.Add(new Unicorn() { Numbers = {1, 2, 3} } );
unicorns.Add(new Unicorn() { Numbers = {4, 5, 6} } );
unicorns.Add(new Unicorn() { Numbers = {7, 8, 9} } );
What's the most efficient way in c# 4 to concatenate all the lists into one list of {1, 2, 3, 4, 5, 6, 7, 8, 9 } ?
Preferably (ideally; by preference; if one had a choice) no loops and Linq-less. I tinkered around with .FindAll, but it's not digging it.
List<int> theInts = unicorns.SelectMany(unicorn => unicorn.Numbers).ToList();
Without measuring, the only thing in here that gives me pause from a performance perspective is the .ToList
If there are a TON of numbers, it's possible that ToList may repeatedly re-allocate its backing array. In order to escape this behavior, you have to have some idea about how many Numbers to expect.
List<int> theInts = new List<int>(expectedSize);
theInts.AddRange(unicorns.SelectMany(unicorn => unicorn.Numbers));
Here's a solution that fully meets your requirements: No Linq and no loop - I'm pretty sure you do not want to use this code though:
List<Unicorn> unicorns = new List<Unicorn>();
unicorns.Add(new Unicorn() { Numbers = new List<int>{1, 2, 3} } );
unicorns.Add(new Unicorn() { Numbers = new List<int> { 4, 5, 6 } });
unicorns.Add(new Unicorn() { Numbers = new List<int> { 7, 8, 9 } });
List<int> numbers = new List<int>();
int count = 0;
AddUnicorn:
if(count < unicorns.Count)
{
numbers.AddRange(unicorns[count++].Numbers);
goto AddUnicorn;
}
Your requirements are quite out of the ordinary, but I guess you can always write:
var numbers = new List<int>();
unicorns.ForEach(unicorn => numbers.AddRange(unicorn.Numbers));
ForEach() arguably qualifies as "LINQ-less", since it's a genuine member of List<T>, not an extension method on IEnumerable<T>, and it actually predates LINQ itself.
Using .NET 4.0's Zip operator:
var sums = b.Zip(a, (x, y) => x + y)
.Concat(b.Skip(a.Count()));
If you want to generalize this, check which has more elements and use that as the "b" above
I have an array with elemnents in order 1,2,3,4,5 and I would need to reverse it so it will be 5,4,3,2,1.
What about the following pseudo code? Is here not an easier way
EDIT: I Am sorry I thought multidimensional array
someclass [,] temporaryArray=new someclass [ArrayLenght,ArrayLenght];
//for each dimension then
for(int I=0;I<ArrayLenghtOfDimension;I++)
{
temporaryArray[ArrayLenghtOfDimension-I]=Array[I];
}
Array=temporaryArray;
The array base class has a Reverse() extension method built in
int[] originalArray = new int[] { 1, 2, 3, 4, 5 };
int[] reversedArray = originalArray.Reverse().ToArray();
Note that the Reverse method returns IEnumerable, so you need to call ToArray() on the result.
And if you need to just iterate over the elements in the array, then all you need is
foreach (int element in originalArray.Reverse())
Console.WriteLine(element);
Oops - Reverse is on IEnumerable, not Array, so you can use that with any collection.
IEnumerable<int> IEnumerableInt = new List<int>() { 1, 2, 3 };
int[] reversedArray2 = IEnumerableInt.Reverse().ToArray();
Yes there is fast solution exists in .net
int[] values = new int[] { 1, 2, 3, 4, 5 };
Array.Reverse(values);
Your array is reversed. so you can iterate through it
foreach (int i in values)
{
Response.Write(i.ToString());
}
the above code will display
54321
It will also work for string[], char[] or other type of arrays
Event though the Array class has Reverse methods defined:
Array.Reverse(originalArray); // original array is now reversed
If all you need to do is iterate backwards over it do the following:
for(int I= ArrayLength - 1; I >= 0; I--)
{
}
This avoid re-allocating memory for the reversed array.
Array.Reverse is the best way to do this. Do you care about order of the elements at all? If so,then you can do the following.
int[] originalArray = new int[] { 10, 2, 13, 4, 5 };
int[] descOrderedArray = originalArray.OrderByDescending(i => i).ToArray();
int[] ascOrderedArray = originalArray.OrderBy(i => i).ToArray();
For a multi-dimensional array it's the same idea
int[][] multiDimArray = new int[][] { new int[] { 1, 2, 3 }, new int[] { 4, 5, 6 } };
int[][] reversedMultiArray = multiDimArray.Reverse().ToArray();
produces an array of two arrays that is: {4, 5, 6}, {1, 2, 3}
I have a small array of ints. I want to reorder the array from largest to smallest. Is there a method to do this?
You could use Array.Sort:
int[] array = new[] { 1, 3, 2 };
Array.Sort(array, (x, y) => y.CompareTo(x));
As far as complexity is concerned:
On average, this method is an O(n log
n) operation, where n is the Length of
array; in the worst case it is an O(n
^ 2) operation
You can do it using Array Sort & Reverse:
Array.Sort(array);
Array.Reverse(array);
Example:
[Test]
public void Test()
{
var array = new[] { 1, 3, 2 };
Array.Sort(array);
Array.Reverse(array);
CollectionAssert.AreEquivalent(new[] { 3, 2, 1 }, array);
}
You can try something like this
int[] ints = new int[] {1, 2, 3, 4, 1, 2, 3};
var sorted = ints.OrderBy(i => i);
Found at Sort array of items using OrderBy<>