How to Sort a List in Desending Order by the "Value" and Sort the Duplicates by the "Key" ?

Joe Caggiano

I have looked All over the Internet to try and find an example of How to fix the following.

How to Sort a List in Desending Order by the "Value" and Sort the Duplicates by the "Key" ?
Then Print out the Results in the format below.

I have enclosed my code and it works, but the problem happens when there are duplicate values, which occurs when you use SortedList(). I would GREATLY APPRECIATE it if someone could PLEASE Modify this Code or Show me EXACTLY how to do this another way, that is just as quick and efficent.

THANK YOU VERY MUCH in Advance.

BEFORE SORT:

VALUE's, KEY's

sL.Add(1269.63,"white");

sL.Add(1270.36,"orange");

sL.Add(1272.06,"yellow");

sL.Add(1271.50,"cyan");

sL.Add(1272.06,"black");

sL.Add(1274.12,"dodBlue");

sL.Add(1276.02,"blue");

sL.Add(1273.21,"green");

sL.Add(1275.52,"red");

AFTER SORT:

VALUE's, KEY's

sL.Add(1276.02,"blue");

sL.Add(1275.52,"red");

sL.Add(1274.12,"dodBlue");

sL.Add(1273.21,"green");

sL.Add(1272.06,"black");

sL.Add(1272.06,"yellow");

sL.Add(1271.50,"cyan");

sL.Add(1270.36,"orange");

sL.Add(1269.63,"white");

CODE:

SortedList sL = new SortedList();

sL.Add(SMA(8)[0], "white");

sL.Add(SMA(10)[0], "orange");

sL.Add(SMA(13)[0], "yellow");

sL.Add(SMA(20)[0], "cyan");

sL.Add(SMA(30)[0], "black");

sL.Add(SMA(40)[0], "dodBlue");

sL.Add(SMA(50)[0], "blue");

sL.Add(SMA(100)[0], "green");

sL.Add(SMA(200)[0], "red");

Print(" " + " " + sL.GetByIndex(8) + " " + ">=" + " " + sL.GetByIndex(7));

Print("&&" + " " + sL.GetByIndex(7) + " " + ">=" + " " + sL.GetByIndex(6));

Print("&&" + " " + sL.GetByIndex(6) + " " + ">=" + " " + sL.GetByIndex(5));

Print("&&" + " " + sL.GetByIndex(5) + " " + ">=" + " " + sL.GetByIndex(4));

Print("&&" + " " + sL.GetByIndex(4) + " " + ">=" + " " + sL.GetByIndex(3));

Print("&&" + " " + sL.GetByIndex(3) + " " + ">=" + " " + sL.GetByIndex(2));

Print("&&" + " " + sL.GetByIndex(2) + " " + ">=" + " " + sL.GetByIndex(1));

Print("&&" + " " + sL.GetByIndex(1) + " " + ">=" + " " + sL.GetByIndex(0));

PRINT OUT RESULTS:

blue >= red ;

&& red >= dodBlue ;

&& dodBlue >= green ;

&& green >= yellow ;

&& yellow >= black ;

&& black >= cyan ;

&& cyan >= orange ;

&& orange >= white ;

9 Replies

Know the answer? Post it — somebody with the same question will find it here.

Vulpes15y ago
Accepted answer
Whenever you wrap some code in a method and then call that method, there is a tiny reduction in performance. This is because the system has to create a 'stack frame' for the method on the current thread's stack which contains (amongst other things) memory slots for parameters, local variables and the method's return address.

In nearly all applications, this performance difference is negligible and I'd simply ignore what I've just said and create as many methods as you like:)

Although there wasn't much difference in the amount of code on this occasion because the array only had 9 members, it would as you say have been different if it had 99 members!

This is why I tend to generalize code as far as I can so it will scale better for large(r) amounts of data.
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Joe Caggiano15y ago
Hi Vulpes,

Truely, THANK YOU for all the Help and your Time !!!!!!

Joe
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Joe Caggiano15y ago
Hi Vulpes,

THANK YOU for your through explanation, It realy helped me to understand what's going on.

You were wright it was my error that caused the "&&" to Print.

I would like to ask you one more question.

I created an alternative way to Print out the Results.

I was wondering if Performance (Time) wise, structural wise, etc... is one method to Print out the Results more efficient than the other ???

The only benefit I can see using the for(i = 0 to i = x) --> Print method, is that it would be easier to create a Print out, if your "var list" was very large (less code).

Print( list[0].Value + " >= " + list[1].Value + ";");
Print("&& " + list[1].Value + " >= " + list[2].Value + ";");
Print("&& " + list[2].Value + " >= " + list[3].Value + ";");
Print("&& " + list[3].Value + " >= " + list[4].Value + ";");
Print("&& " + list[4].Value + " >= " + list[5].Value + ";");
Print("&& " + list[5].Value + " >= " + list[6].Value + ";");
Print("&& " + list[6].Value + " >= " + list[7].Value + ";");
Print("&& " + list[7].Value + " >= " + list[8].Value + ";");
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Suthish Nair15y ago
If the query got resolved, then mark the post that helped you as Answer so will be helpful for others. Otherwise if you got the solution please share with us.
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Vulpes15y ago
It worked OK when I tested it. Are you sure that you've copied this line correctly?

string and = (i > 0) ? "&& " : "";

The idea is that if 'i' is greater than 0 (i.e. we're not on the first line) then the variable 'and' is set to "&& ", otherwise it's just an empty string.

"\r\n" means a carriage return, line feed combination.

So, if you have two of them together, it means move to the next line and then move to the next line after that.

The first argument to the String.Format method: "{0}{1} >= {2};{3}" contains placeholders for the remaining arguments to that method. So {0} is the placeholder for 'and', {1} is the placeholder for list[i].Value and so on. In other words, it's like a template for a string where you successively fill in the gaps.
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Joe Caggiano15y ago
Hi Vulpes,

THANKs AGAIN!!!

This also Works!!!

There's Only one Problem the First Line also Prints Out with the "&&".
&& blue >= red ; <------- Should Be: blue >= red; ???
&& red >= dodBlue ;
&& dodBlue >= green ;
&& green >= yellow ;
&& yellow >= black ;
&& black >= cyan ;
&& cyan >= orange ;
&& orange >= white ;
Could you PLEASE elaborate on what "\r\n\r\n" means and also "{0}{1} >= {2};{3}" what is it saying and doing ???
THANK You Once Again for you Time and Support !!!
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Vulpes15y ago
It looks like Print is a custom method which prints a string to some output window.

However, assuming it recognizes new lines (normally "\r\n" in Windows), this should work:

string output = "";

for(int i = 0; i < list.Count - 1; i++)
{
   string and = (i > 0) ? "&& " : "";
   string newline = (i < list.Count - 2) ? "\r\n\r\n" : "";
   output += String.Format("{0}{1} >= {2};{3}", and, list[i].Value, list[i+1].Value, newline);
}

Print(output);
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Joe Caggiano15y ago
Hi Vulpes,

Thanks for your Help, it Sorts the Duplicate Items like a Charm.

I still have a problem with Printing out the Sort Results.

My software does not recognize the Console.WriteLine("{0:F2} : {1}", pair.Key, pair.Value).

It uses the Print() command to Print to the OutPut Window of the software.

Could you Please tell me how I can access the Values in the sorted List so I can print them out accordingly?

Print(pair.Value(0) + " " + ">=" + " " + pair.Value(1) + ";")   <------------------- ???
Print("&&" + " " + pair.Value(1) + " " + ">=" + " " + pair.Value(2) + ";")   <------- ???
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
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vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
Print("&&" + " " + pair.Value(6) + " " + ">=" + " " + pair.Value(7) + ";")   <-------- ???
Print("&&" + " " + pair.Value(7) + " " + ">=" + " " + pair.Value(8) + ";")   <-------- ???
So the Print out looks Like the Following:
blue >= red ;
&& red >= dodBlue ;
&& dodBlue >= green ;
&& green >= yellow ;
&& yellow >= black ;
&& black >= cyan ;
&& cyan >= orange ;
&& orange >= white ;
THANK YOU VERY MUCH IN ADVANCE.
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Vulpes15y ago
I'm not completely clear which is your key and which is your value here but one way of doing this would be to pass the SortedList a custom IComparer to give you the sort order you need.

However, a better approach nowadays is to use a LINQ query on an unsorted collection. As duplicates are present, I've used a List of KeyValue pairs rather than a Dictionary in the following example code:

using System;
using System.Collections.Generic;
using System.Linq;

class Test
{
   static void Main()
   {
   var list = new List>();

   list.Add(new KeyValuePair(1269.63,"white"));

   list.Add(new KeyValuePair(1270.36,"orange"));

   list.Add(new KeyValuePair(1272.06,"yellow"));

   list.Add(new KeyValuePair(1271.50,"cyan"));

   list.Add(new KeyValuePair(1272.06,"black"));

   list.Add(new KeyValuePair(1274.12,"dodBlue"));

   list.Add(new KeyValuePair(1276.02,"blue"));

   list.Add(new KeyValuePair(1273.21,"green"));

   list.Add(new KeyValuePair(1275.52,"red"));

   var query = from kvp in list orderby kvp.Key descending, kvp.Value
   select new KeyValuePair(kvp.Key, kvp.Value);

   list = query.ToList();

   foreach(var pair in list) Console.WriteLine("{0:F2} : {1}", pair.Key, pair.Value);

   Console.ReadKey();
   }
}
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