Practical advice, please.

[Deleted] 2020.06.06 15:18

Question: How do I evaluate the results correctly?

The error of each module is given as a percentage. 0% is the ideal result.

________________ PARAMETERS ________________	Mod 1	Mod 2	Mod 3	Mode 4	Mode 5	(vi) mod 6	(vi) Mode 7	(vi) Mode 8	(vi) Mod 9	(vi) Mod 10	(vi) Mod 11	(vi) Mod 12	(viii) mod 13	(vi) 14	Mode 15	Average error	From the attempts
2_48_24_2160_12_VECTOR_UP_HIDDEN_LAYERS_HAND	4,43	17,09	15,82	2,53	0,63	17,72	28,48	5,70	13,29	5,70	8,23	6,33	0,63	3,16	6,96	9,11	158,00
2_48_24_2160_12_VECTOR_UP_HIDDEN_LAYERS_MT1	5,06	17,72	12,66	3,80	0,63	19,62	29,11	4,43	9,49	5,06	6,33	6,33	1,90	1,90	6,33	8,69	158,00
2_48_24_2160_12_VECTOR_UP_HIDDEN_LAYERS_MT2	4,43	20,25	16,46	4,43	0,63	17,72	29,75	6,33	5,06	8,23	10,13	5,06	0,63	1,27	4,43	8,99	158,00

I would like the error of each module to be minimal, but I would also like the spread to be minimal.

Alexandr Andreev 2020.06.06 15:50 #1

Сергей Таболин:

Question: How do I evaluate the results correctly?

The error of each module is given as a percentage. 0% is the ideal result.

________________ PARAMETERS ________________	Mod 1	Mod 2	Mod 3	Mode 4	Mode 5	(vi) Mode 6	(vi) Mode 7	(vi) Mode 8	(vi) Mod 9	(vi) Mod 10	(vi) Mod 11	(vi) Mod 12	(viii) mod 13	(vi) 14	Mode 15	Average error	From the attempts
2_48_24_2160_12_VECTOR_UP_HIDDEN_LAYERS_HAND	4,43	17,09	15,82	2,53	0,63	17,72	28,48	5,70	13,29	5,70	8,23	6,33	0,63	3,16	6,96	9,11	158,00
2_48_24_2160_12_VECTOR_UP_HIDDEN_LAYERS_MT1	5,06	17,72	12,66	3,80	0,63	19,62	29,11	4,43	9,49	5,06	6,33	6,33	1,90	1,90	6,33	8,69	158,00
2_48_24_2160_12_VECTOR_UP_HIDDEN_LAYERS_MT2	4,43	20,25	16,46	4,43	0,63	17,72	29,75	6,33	5,06	8,23	10,13	5,06	0,63	1,27	4,43	8,99	158,00

I want the error of each module to be minimal, but I also want the spread to be minimal.

product of logarithms

ZS: not really sure what is required, but the logarithm will allow errors to be treated progressively, it will give better results in single cases (individual modules). And multiplication is an attempt to reduce dispersion

Zero sample correlation does Hearst index Why have subscriptions been

[Deleted] 2020.06.06 16:05 #2

Alexandr Andreev:

product of logarithms

ZS: I don't really understand what exactly is required, but logarithm will allow errors to be treated progressively, it will give better results in single cases (individual modules). And multiplication is an attempt to reduce the spread

Thank you. Andpractically how is it?

Alexandr Andreev 2020.06.06 16:22 #3

Сергей Таболин:

Thank you. Andpractically how is that?

Probably just the product of

Option1 ) Translate each module into the style (1-x%) and multiply them..... the answer is also subtracted from the unit.

x% is the cell value

Option2

With logarithms we simply take value of a cell and count logarithm from it)))) the closer to zero value is to progressive evaluation i.e. with some base setting 0.1 is better than 0.01 and also 0.1 is better than 1. There will only be a base parameter for the logarithm which is worth playing around with.

From theory to practice I'm getting a bit Questions on OOP (Object

[Deleted] 2020.06.06 18:51 #4

Alexandr Andreev:

Perhaps a simple product would do.

Option1 ) Translate each module into the style (1-x%) and multiply them..... The answer is also subtracted from the unit.

x% is the cell value

Option2

With logarithms we simply take value of a cell and count logarithm from it)))) the closer to zero value is to progressive evaluation i.e. with some base setting 0.1 is better than 0.01 and also 0.1 is better than 1. There will only be a basis parameter of the logarithm which is worth playing around with.

Option 1

-3,43	-16,09	-14,82	-1,53	0,37	-16,72	-27,48	-4,70	-12,29	-4,70	-7,23	-5,33	0,37	-2,16	-5,96	10601305851,38
-4,06	-16,72	-11,66	-2,80	0,37	-18,62	-28,11	-3,43	-8,49	-4,06	-5,33	-5,33	-0,90	-0,90	-5,33	-6223799946,09
-3,43	-19,25	-15,46	-3,43	0,37	-16,72	-28,75	-5,33	-4,06	-7,23	-9,13	-4,06	0,37	-0,27	-3,43	1237520122,21

What does this tell me?

Option 2

-0,64640373	-1,23274206	-1,19920648	-0,40312052	0,200659451	-1,24846372	-1,45453998	-0,75587486	-1,12352498	-0,75587486	-0,91539984	-0,80140371	0,200659451	-0,49968708	-0,84260924
-0,70415052	-1,24846372	-1,10243371	-0,5797836	0,200659451	-1,292699	-1,46404221	-0,64640373	-0,97726621	-0,70415052	-0,80140371	-0,80140371	-0,2787536	-0,2787536	-0,80140371
-0,64640373	-1,30642503	-1,21642983	-0,64640373	0,200659451	-1,24846372	-1,47348697	-0,80140371	-0,70415052	-0,91539984	-1,00560945	-0,70415052	0,200659451	-0,10380372	-0,64640373

This is the logarithm with base 0.1

What should I do with it?

I have tried other functions. Only how do I understand them too? ....

STANDOTCLONE	7,8208
	7,9133
	8,4150

DISP.B	61,1650
	62,6198
	70,8128

QUADROTKL	856,3093
	876,6772
	991,3799

MEDIANA	6,3300
	6,3300
	5,0600

SCOS	1,1805
	1,5197
	1,3018

EXCESS	1,1322
	1,9702
	1,1832

Scoreboard Result of the previous PREDICT time period

Dmitry Fedoseev 2020.06.06 18:58 #5

Find the maximum in each row, then select the row with the minimum maximum. Pardon the pun))

[Deleted] 2020.06.06 19:12 #6

Dmitry Fedoseev:
Find the maximum in each row, then choose the row with the minimum maximum. Pardon the pun))

Maximum maximum in line 3, minimum maximum in line 1. И? )))

Dmitry Fedoseev 2020.06.06 19:18 #7

Сергей Таболин:

Maximum maximum in line 3, minimum maximum in line 1. И? )))

Choose the first line.

Alexandr Andreev 2020.06.06 21:15 #8

Сергей Таболин:

Option 1

-3,43	-16,09	-14,82	-1,53	0,37	-16,72	-27,48	-4,70	-12,29	-4,70	-7,23	-5,33	0,37	-2,16	-5,96	10601305851,38
-4,06	-16,72	-11,66	-2,80	0,37	-18,62	-28,11	-3,43	-8,49	-4,06	-5,33	-5,33	-0,90	-0,90	-5,33	-6223799946,09
-3,43	-19,25	-15,46	-3,43	0,37	-16,72	-28,75	-5,33	-4,06	-7,23	-9,13	-4,06	0,37	-0,27	-3,43	1237520122,21

What does this tell me?

Option 2

-0,64640373	-1,23274206	-1,19920648	-0,40312052	0,200659451	-1,24846372	-1,45453998	-0,75587486	-1,12352498	-0,75587486	-0,91539984	-0,80140371	0,200659451	-0,49968708	-0,84260924
-0,70415052	-1,24846372	-1,10243371	-0,5797836	0,200659451	-1,292699	-1,46404221	-0,64640373	-0,97726621	-0,70415052	-0,80140371	-0,80140371	-0,2787536	-0,2787536	-0,80140371
-0,64640373	-1,30642503	-1,21642983	-0,64640373	0,200659451	-1,24846372	-1,47348697	-0,80140371	-0,70415052	-0,91539984	-1,00560945	-0,70415052	0,200659451	-0,10380372	-0,64640373

This is the logarithm with base 0.1

What should I do with it?

I have tried other functions. But how do I understand them too?

It's a bitch."Error of each module is given as a percentage. 0% is the ideal result." Number 1 means 100% - x% or 1-X*0.01

Alexandr Andreev 2020.06.06 21:19 #9

@Sergey Tabolin @Sergey Tabolin

															TOTAL
0,9557	0,8291	0,8418	0,9747	0,9937	0,8228	0,7152	0,943	0,8671	0,943	0,9177	0,9367	0,9937	0,9684	0,9304	0,77439
0,9494	0,8228	0,8734	0,962	0,9937	0,8038	0,7089	0,9557	0,9051	0,9494	0,9367	0,9367	0,981	0,981	0,9367	0,758606
0,9557	0,7975	0,8354	0,9557	0,9937	0,8228	0,7025	0,9367	0,9494	0,9177	0,8987	0,9494	0,9937	0,9873	0,9557	0,771806

the second line is the best, and the first and the third are very similar

the total subtracted from the unit, i.e. the closer to 0 the total, the better the results..... in other words, the results so far are not very good because 0.75 is your 75, although it depends on what to compare it with..... the worst score would be 1 (100%) the best score 0

You have to understand that a score of 90 is ten times better than a score of 99.... a score of 99 is ten times better than a score of 99.9... 100 is in fact possible only when all modules have an error score of 100... i.e. a score of 0.1 is ten times worse than a score of 0.01. At the same time, a score of 10 is ten times worse than a score of 1.

with the logarithm over think..... the answer should be exclusively positive values... usually a logarithm of 1.1... in the range of 1 to 2, not 0 to 1.... if they want to increase the number and from 2 if they want to decrease it progressively

The quadratic deviation method is definitely out of the question. So are all the others that count deviations. Because ideally the quadratic deviation from a linear regression would be used in order to understand the variance. But then we get an estimate of these deviations without any average of the numbers themselves.....

A single quality indicator Any rookie question, so Indicator multi color -

Roman 2020.06.07 01:48 #10

Сергей Таболин:

Question: How do I evaluate the results correctly?

The error of each module is given as a percentage. 0% is the ideal result.

________________ PARAMETERS ________________	Mod 1	Mod 2	Mod 3	Mode 4	Mode 5	(vi) Mode 6	(vi) Mode 7	(vi) Mode 8	(vi) Mod 9	(vi) Mod 10	(vi) Mod 11	(vi) Mod 12	(viii) mod 13	(vi) 14	Mode 15	Average error	From the attempts
2_48_24_2160_12_VECTOR_UP_HIDDEN_LAYERS_HAND	4,43	17,09	15,82	2,53	0,63	17,72	28,48	5,70	13,29	5,70	8,23	6,33	0,63	3,16	6,96	9,11	158,00
2_48_24_2160_12_VECTOR_UP_HIDDEN_LAYERS_MT1	5,06	17,72	12,66	3,80	0,63	19,62	29,11	4,43	9,49	5,06	6,33	6,33	1,90	1,90	6,33	8,69	158,00
2_48_24_2160_12_VECTOR_UP_HIDDEN_LAYERS_MT2	4,43	20,25	16,46	4,43	0,63	17,72	29,75	6,33	5,06	8,23	10,13	5,06	0,63	1,27	4,43	8,99	158,00

I want the error of each module to be minimal, but I also want the scatter to be minimal.

It is probably better to get the sum of squares for the module errors, and extract the root.
Thus, we will get the total module error estimate.
The closer to zero this value is, the better.
So it goes like this.

//+------------------------------------------------------------------+
//|                                                EstimateError.mq5 |
//|                        Copyright 2020, MetaQuotes Software Corp. |
//|                                             https://www.mql5.com |
//+------------------------------------------------------------------+
#property copyright "Copyright 2020, MetaQuotes Software Corp."
#property link      "https://www.mql5.com"
#property version   "1.00"


double ModN[15][3] = {{4.43, 5.06, 4.43},
                      {17.09, 17.72, 20.25},
                      {15.82, 12.66, 16.46},
                      {2.53, 3.80, 4.43},
                      {0.63, 0.63, 0.63},
                      {17.72, 19.62, 17.72},
                      {28.48, 29.11, 29.75},
                      {5.70, 4.43, 6.33},
                      {13.29, 9.49, 5.06},
                      {5.70, 5.06, 8.23},
                      {8.23, 6.33, 10.13},
                      {6.33, 6.33, 5.06},
                      {0.63, 6.33, 0.63},
                      {3.16, 1.90, 1.27},
                      {6.96, 6.33, 4.43}};

//+------------------------------------------------------------------+
//| Script program start function                                    |
//+------------------------------------------------------------------+
void OnStart()
{
   double ModX[];
   ArrayResize(ModX, 3);
   ZeroMemory(ModX);   
   
   int    num = 1;
   double est = 0.0;
   
   for(int i=0; i<15; i++)
   {
      for(int j=0; j<3; j++)
      {
         ModX[j] = ModN[i][j];         
      }
         
      est = EstimateError(ModX);  
      
      PrintFormat("Mod"+(string)num+" EstimateError: %.3f", est); 
      num++;
   }         
}// End OnStart

//+------------------------------------------------------------------+
double EstimateError(double & arr[])
{
   int size = ArraySize(arr);
   if(size == 0 || size < 3) 
      return(0.0);
   
   //double avg       = ArrayMean(arr); 
   double max       = ArrayMax(arr);   
   double min       = ArrayMin(arr); 
   double sum_sqr_e = 0.0;
   double est_e     = 0.0;      
   
   
   for(int i=0; i<size; i++)   
      sum_sqr_e += MathPow(arr[i] - (max-min)/* или avg*/, 2.0) / (size - 2.0);
      
   est_e = MathSqrt(sum_sqr_e);   

   return(est_e);
}

//+-------------------------------------------------------------------
//Возвращает максимальное значение элементов массива
double ArrayMax(double & arrIn[])
{
   uint size = ArraySize(arrIn);
   if(size == 0) 
      return(0.0);          
   
   double max = arrIn[0];
   for(uint i=1; i<size; i++)
      if(arrIn[i] > max) 
         max = arrIn[i];

   return(max);
}

//--------------------------------------------------------------------
//Возвращает минимальное значение элементов массива
double ArrayMin(double & arrIn[])
{
   uint size = ArraySize(arrIn);
   if(size == 0) 
      return(0.0);   
   
   double min = arrIn[0];  
   for(uint i=1; i<size; i++)
      if(arrIn[i] < min) 
         min = arrIn[i];

   return(min);
}

//--------------------------------------------------------------------
//Возвращает средне арефметическое значение элементов массива
double ArrayMean(double & arrIn[])
{
   uint size = ArraySize(arrIn);
   if(size == 0) 
      return(0.0);         

   double sum = 0.0;       
   for(uint i=0; i<size; i++) 
      sum += arrIn[i];     
   
   return(sum/size);
}

//--------------------------------------------------------------------

2020.06.07 04:59:23.227 EstimateError (AUDUSD,M5)       Mod1 EstimateError: 6.965
2020.06.07 04:59:23.227 EstimateError (AUDUSD,M5)       Mod2 EstimateError: 26.422
2020.06.07 04:59:23.227 EstimateError (AUDUSD,M5)       Mod3 EstimateError: 19.577
2020.06.07 04:59:23.227 EstimateError (AUDUSD,M5)       Mod4 EstimateError: 3.226
2020.06.07 04:59:23.227 EstimateError (AUDUSD,M5)       Mod5 EstimateError: 1.091
2020.06.07 04:59:23.227 EstimateError (AUDUSD,M5)       Mod6 EstimateError: 28.540
2020.06.07 04:59:23.227 EstimateError (AUDUSD,M5)       Mod7 EstimateError: 48.234
2020.06.07 04:59:23.227 EstimateError (AUDUSD,M5)       Mod8 EstimateError: 6.361
2020.06.07 04:59:23.227 EstimateError (AUDUSD,M5)       Mod9 EstimateError: 6.102
2020.06.07 04:59:23.227 EstimateError (AUDUSD,M5)       Mod10 EstimateError: 5.965
2020.06.07 04:59:23.227 EstimateError (AUDUSD,M5)       Mod11 EstimateError: 8.130
2020.06.07 04:59:23.227 EstimateError (AUDUSD,M5)       Mod12 EstimateError: 8.098
2020.06.07 04:59:23.227 EstimateError (AUDUSD,M5)       Mod13 EstimateError: 7.198
2020.06.07 04:59:23.227 EstimateError (AUDUSD,M5)       Mod14 EstimateError: 1.413
2020.06.07 04:59:23.227 EstimateError (AUDUSD,M5)       Mod15 EstimateError: 6.138

The estimate shows that Mod5 has the smallest error.

"The 'perfect' trading system Dr.Fx corner What's the root cause

1 2 3 4 5 6 7

New comment