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source: branches/2994-AutoDiffForIntervals/HeuristicLab.Common/3.3/EnumerableExtensions.cs @ 18078

Last change on this file since 18078 was 17209, checked in by gkronber, 5 years ago

#2994: merged r17132:17198 from trunk to branch

File size: 6.7 KB
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1#region License Information
2/* HeuristicLab
3 * Copyright (C) Heuristic and Evolutionary Algorithms Laboratory (HEAL)
4 *
5 * This file is part of HeuristicLab.
6 *
7 * HeuristicLab is free software: you can redistribute it and/or modify
8 * it under the terms of the GNU General Public License as published by
9 * the Free Software Foundation, either version 3 of the License, or
10 * (at your option) any later version.
11 *
12 * HeuristicLab is distributed in the hope that it will be useful,
13 * but WITHOUT ANY WARRANTY; without even the implied warranty of
14 * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
15 * GNU General Public License for more details.
16 *
17 * You should have received a copy of the GNU General Public License
18 * along with HeuristicLab. If not, see <http://www.gnu.org/licenses/>.
19 */
20#endregion
21
22using System;
23using System.Collections.Generic;
24using System.Linq;
25
26namespace HeuristicLab.Common {
27  public static class EnumerableExtensions {
28    /// <summary>
29    /// Selects all elements in the sequence that are maximal with respect to the given value.
30    /// </summary>
31    /// <remarks>
32    /// Runtime complexity of the operation is O(N).
33    /// </remarks>
34    /// <typeparam name="T">The type of the elements.</typeparam>
35    /// <param name="source">The enumeration in which the items with a maximal value should be found.</param>
36    /// <param name="valueSelector">The function that selects the value to compare.</param>
37    /// <returns>All elements in the enumeration where the selected value is the maximum.</returns>
38    public static IEnumerable<T> MaxItems<T>(this IEnumerable<T> source, Func<T, IComparable> valueSelector) {
39      IEnumerator<T> enumerator = source.GetEnumerator();
40      if (!enumerator.MoveNext()) return Enumerable.Empty<T>();
41      IComparable max = valueSelector(enumerator.Current);
42      var result = new List<T>();
43      result.Add(enumerator.Current);
44
45      while (enumerator.MoveNext()) {
46        T item = enumerator.Current;
47        IComparable comparison = valueSelector(item);
48        if (comparison.CompareTo(max) > 0) {
49          result.Clear();
50          result.Add(item);
51          max = comparison;
52        } else if (comparison.CompareTo(max) == 0) {
53          result.Add(item);
54        }
55      }
56      return result;
57    }
58
59    /// <summary>
60    /// Selects all elements in the sequence that are minimal with respect to the given value.
61    /// </summary>
62    /// <remarks>
63    /// Runtime complexity of the operation is O(N).
64    /// </remarks>
65    /// <typeparam name="T">The type of the elements.</typeparam>
66    /// <param name="source">The enumeration in which items with a minimal value should be found.</param>
67    /// <param name="valueSelector">The function that selects the value.</param>
68    /// <returns>All elements in the enumeration where the selected value is the minimum.</returns>
69    public static IEnumerable<T> MinItems<T>(this IEnumerable<T> source, Func<T, IComparable> valueSelector) {
70      IEnumerator<T> enumerator = source.GetEnumerator();
71      if (!enumerator.MoveNext()) return Enumerable.Empty<T>();
72      IComparable min = valueSelector(enumerator.Current);
73      var result = new List<T>();
74      result.Add(enumerator.Current);
75
76      while (enumerator.MoveNext()) {
77        T item = enumerator.Current;
78        IComparable comparison = valueSelector(item);
79        if (comparison.CompareTo(min) < 0) {
80          result.Clear();
81          result.Add(item);
82          min = comparison;
83        } else if (comparison.CompareTo(min) == 0) {
84          result.Add(item);
85        }
86      }
87      return result;
88    }
89
90    /// <summary>
91    /// Compute the n-ary cartesian product of arbitrarily many sequences: http://blogs.msdn.com/b/ericlippert/archive/2010/06/28/computing-a-cartesian-product-with-linq.aspx
92    /// </summary>
93    /// <typeparam name="T">The type of the elements inside each sequence</typeparam>
94    /// <param name="sequences">The collection of sequences</param>
95    /// <returns>An enumerable sequence of all the possible combinations of elements</returns>
96    public static IEnumerable<IEnumerable<T>> CartesianProduct<T>(this IEnumerable<IEnumerable<T>> sequences) {
97      IEnumerable<IEnumerable<T>> result = new[] { Enumerable.Empty<T>() };
98      return sequences.Where(s => s.Any()).Aggregate(result, (current, s) => (from seq in current from item in s select seq.Concat(new[] { item })));
99    }
100
101    /// <summary>
102    /// Compute all k-combinations of elements from the provided collection.
103    /// <param name="elements">The collection of elements</param>
104    /// <param name="k">The combination group size</param>
105    /// <returns>An enumerable sequence of all the possible k-combinations of elements</returns>
106    /// </summary>
107    public static IEnumerable<IEnumerable<T>> Combinations<T>(this IList<T> elements, int k) {
108      if (k > elements.Count)
109        throw new ArgumentException();
110
111      if (k == 1) {
112        foreach (var element in elements)
113          yield return new[] { element };
114        yield break;
115      }
116
117      int n = elements.Count;
118      var range = Enumerable.Range(0, k).ToArray();
119      var length = BinomialCoefficient(n, k);
120
121      for (int i = 0; i < length; ++i) {
122        yield return range.Select(x => elements[x]).ToArray();
123
124        if (i == length - 1) break;
125        var m = k - 1;
126        var max = n - 1;
127
128        while (range[m] == max) { --m; --max; }
129        range[m]++;
130        for (int j = m + 1; j < k; ++j) {
131          range[j] = range[j - 1] + 1;
132        }
133      }
134    }
135
136    /// <summary>
137    /// This function gets the total number of unique combinations based upon N and K,
138    /// where N is the total number of items and K is the size of the group.
139    /// It calculates the total number of unique combinations C(N, K) = N! / ( K! (N - K)! )
140    /// using the  recursion C(N+1, K+1) = (N+1 / K+1) * C(N, K).
141    /// <remarks>http://blog.plover.com/math/choose.html</remarks>
142    /// <remark>https://en.wikipedia.org/wiki/Binomial_coefficient#Multiplicative_formula</remark>
143    /// <param name="n">The number of elements</param>
144    /// <param name="k">The size of the group</param>
145    /// <returns>The binomial coefficient C(N, K)</returns>
146    /// </summary>
147    public static long BinomialCoefficient(long n, long k) {
148      if (k > n) return 0;
149      if (k == n) return 1;
150      if (k > n - k)
151        k = n - k;
152
153      // enable explicit overflow checking for very large coefficients
154      checked {
155        long r = 1;
156        for (long d = 1; d <= k; d++) {
157          r *= n--;
158          r /= d;
159        }
160        return r;
161      }
162    }
163  }
164}
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