How to get the most common value in an Int array using C#
如何使用C#获取Int数组中最常见的值
eg: Array has the following values: 1, 1, 1, 2
例如:Array具有以下值:1,1,1,2
Ans should be 1
Ans应为1
5 个解决方案
#1
27
var query = (from item in array
group item by item into g
orderby g.Count() descending
select new { Item = g.Key, Count = g.Count() }).First();
For just the value and not the count, you can do
对于价值而不是数量,你可以做到
var query = (from item in array
group item by item into g
orderby g.Count() descending
select g.Key).First();
Lambda version on the second:
第二个Lambda版本:
var query = array.GroupBy(item => item).OrderByDescending(g => g.Count()).Select(g => g.Key).First();
#2
14
Some old fashioned efficient looping:
一些老式的高效循环:
var cnt = new Dictionary<int, int>();
foreach (int value in theArray) {
if (cnt.ContainsKey(value)) {
cnt[value]++;
} else {
cnt.Add(value, 1);
}
}
int mostCommonValue = 0;
int highestCount = 0;
foreach (KeyValuePair<int, int> pair in cnt) {
if (pair.Value > highestCount) {
mostCommonValue = pair.Key;
highestCount = pair.Value;
}
}
Now mostCommonValue
contains the most common value, and highestCount
contains how many times it occured.
现在,mostCommonValue包含最常见的值,而highestCount包含它发生的次数。
#3
1
Maybe O(n log n), but fast:
也许是O(n log n),但很快:
sort the array a[n]
// assuming n > 0
int iBest = -1; // index of first number in most popular subset
int nBest = -1; // popularity of most popular number
// for each subset of numbers
for(int i = 0; i < n; ){
int ii = i; // ii = index of first number in subset
int nn = 0; // nn = count of numbers in subset
// for each number in subset, count it
for (; i < n && a[i]==a[ii]; i++, nn++ ){}
// if the subset has more numbers than the best so far
// remember it as the new best
if (nBest < nn){nBest = nn; iBest = ii;}
}
// print the most popular value and how popular it is
print a[iBest], nBest
#4
1
public static int get_occure(int[] a)
{
int[] arr = a;
int c = 1, maxcount = 1, maxvalue = 0;
int result = 0;
for (int i = 0; i < arr.Length; i++)
{
maxvalue = arr[i];
for (int j = 0; j <arr.Length; j++)
{
if (maxvalue == arr[j] && j != i)
{
c++;
if (c > maxcount)
{
maxcount = c;
result = arr[i];
}
}
else
{
c=1;
}
}
}
return result;
}
#5
1
I know this post is old, but someone asked me the inverse of this question today.
我知道这篇文章很老了,但今天有人问我这个问题的反面。
LINQ Grouping
LINQ分组
sourceArray.GroupBy(value => value).OrderByDescending(group => group.Count()).First().First();
Temp Collection, similar to Guffa's:
Temp Collection,类似于Guffa的:
var counts = new Dictionary<int, int>();
foreach (var i in sourceArray)
{
if (!counts.ContainsKey(i)) { counts.Add(i, 0); }
counts[i]++;
}
return counts.OrderByDescending(kv => kv.Value).First().Key;
#1
27
var query = (from item in array
group item by item into g
orderby g.Count() descending
select new { Item = g.Key, Count = g.Count() }).First();
For just the value and not the count, you can do
对于价值而不是数量,你可以做到
var query = (from item in array
group item by item into g
orderby g.Count() descending
select g.Key).First();
Lambda version on the second:
第二个Lambda版本:
var query = array.GroupBy(item => item).OrderByDescending(g => g.Count()).Select(g => g.Key).First();
#2
14
Some old fashioned efficient looping:
一些老式的高效循环:
var cnt = new Dictionary<int, int>();
foreach (int value in theArray) {
if (cnt.ContainsKey(value)) {
cnt[value]++;
} else {
cnt.Add(value, 1);
}
}
int mostCommonValue = 0;
int highestCount = 0;
foreach (KeyValuePair<int, int> pair in cnt) {
if (pair.Value > highestCount) {
mostCommonValue = pair.Key;
highestCount = pair.Value;
}
}
Now mostCommonValue
contains the most common value, and highestCount
contains how many times it occured.
现在,mostCommonValue包含最常见的值,而highestCount包含它发生的次数。
#3
1
Maybe O(n log n), but fast:
也许是O(n log n),但很快:
sort the array a[n]
// assuming n > 0
int iBest = -1; // index of first number in most popular subset
int nBest = -1; // popularity of most popular number
// for each subset of numbers
for(int i = 0; i < n; ){
int ii = i; // ii = index of first number in subset
int nn = 0; // nn = count of numbers in subset
// for each number in subset, count it
for (; i < n && a[i]==a[ii]; i++, nn++ ){}
// if the subset has more numbers than the best so far
// remember it as the new best
if (nBest < nn){nBest = nn; iBest = ii;}
}
// print the most popular value and how popular it is
print a[iBest], nBest
#4
1
public static int get_occure(int[] a)
{
int[] arr = a;
int c = 1, maxcount = 1, maxvalue = 0;
int result = 0;
for (int i = 0; i < arr.Length; i++)
{
maxvalue = arr[i];
for (int j = 0; j <arr.Length; j++)
{
if (maxvalue == arr[j] && j != i)
{
c++;
if (c > maxcount)
{
maxcount = c;
result = arr[i];
}
}
else
{
c=1;
}
}
}
return result;
}
#5
1
I know this post is old, but someone asked me the inverse of this question today.
我知道这篇文章很老了,但今天有人问我这个问题的反面。
LINQ Grouping
LINQ分组
sourceArray.GroupBy(value => value).OrderByDescending(group => group.Count()).First().First();
Temp Collection, similar to Guffa's:
Temp Collection,类似于Guffa的:
var counts = new Dictionary<int, int>();
foreach (var i in sourceArray)
{
if (!counts.ContainsKey(i)) { counts.Add(i, 0); }
counts[i]++;
}
return counts.OrderByDescending(kv => kv.Value).First().Key;