简体   繁体   中英

How to use a LINQ : If there are several min and max values, just remove only one min and max value in List

I asked below question. I got an answer.

How to use a LINQ in order to remove Min and Max value in List

However, I have a problem in such scenario there are several min and max values in List. I want to use a LINQ expression in order to remove only one min and Max value in List.

Code snippet :

namespace ConsoleApplication_lamdaTest
{
    public struct TValue
    {
        public double x, y;
        public double value { get { return Math.Sqrt(x * x + y * y); } }
    }

    class Program
    {
        static void Main(string[] args)
        {
            List<TValue> temp = new List<TValue> { 
                new TValue { x = 1, y =2 },
                new TValue { x = 3, y =4 },
                new TValue { x = 4, y =3 },
                new TValue { x = 3, y =1 },
                new TValue { x = 2, y =3 },
                new TValue { x = 1, y =4 },
                new TValue { x = 1, y =2 },
                new TValue { x = 1, y =2 }
            };

            foreach(TValue item in temp)
               Console.WriteLine(item.value.ToString());

            var newValue = from pair in temp
                           where pair.value < temp.Max(m => m.value) && pair.value > temp.Min(m => m.value)
                           select pair;

            foreach (TValue item in newValue)
                Console.WriteLine(item.value.ToString());

            Console.ReadKey();
        }
    }
}

output is ;

2.23606797749979
5
5
3.16227766016838
3.60555127546399
4.12310562561766
2.23606797749979
2.23606797749979
-------------------
3.16227766016838
3.60555127546399
4.12310562561766

but, I want to get output like below ;

2.23606797749979
5
5
3.16227766016838
3.60555127546399
4.12310562561766
2.23606797749979
2.23606797749979
-------------------
5
3.16227766016838
3.60555127546399
4.12310562561766
2.23606797749979
2.23606797749979

I'm thinking about several steps to solve this issue. Using LINQ: Is it Possible to use LINQ ?

1. Sorting
2. Remove First and Last index 

Code Snippet : Any Help ?

var newSortedValue = from pair in temp
                                 orderby pair.value descending
                                 where pair = temp.RemoveAt(0) && pair = temp.RemoveAt(temp.Count()-1)
                                 select pair;

Sorting and removing first and last value should work just fine. You can do it like that:

var tempWithoutMinAndMax = temp.OrderBy(m => m.value).Skip(1).Take(temp.Count-2);

//Edit: I was curious about Rashid Ali's solution (below) so I decided to perform a quick test. I created a list with 10 000 elements:

var list = new List<KeyValuePair<int, int>>();
for(int i=0;i<10000;i++)
{
    list.Add(new KeyValuePair<int, int>(i,random.Next()));
}

Then i removed single Min and Max element from the list using both methods and measured the time. My test code:

Stopwatch watch = new Stopwatch();
watch.Start();
var ver1 = list.OrderBy(m => m.Value).Skip(1).Take(list.Count - 2).ToList();
watch.Stop();
var ver1time = watch.ElapsedMilliseconds;
watch.Reset();
watch.Start();
list.Remove(list.Where(x => x.Value == list.Max(y => y.Value)).First());
list.Remove(list.Where(x => x.Value == list.Min(y => y.Value)).First());
watch.Stop();
var ver2time = watch.ElapsedMilliseconds;
Console.WriteLine("First method (order by): {0}ms\nSecond method (remove): {1}ms",
ver1time,ver2time);

Result: First method (order by): 11ms , Second method (remove): 3424ms I ran this test a couple of times and all results were similar. I didn't check what IL code earch method produces, but clearly using OrderBy outperforms combining Remove, Where and Min/Max

Here is another way to get the result set you want without performing orderby operation:

temp.Remove(temp.Where(x => x.value == temp.Max(y => y.value)).First());
temp.Remove(temp.Where(x => x.value == temp.Min(y => y.value)).First());

The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM