Double [] test

Question

I am struggling I am trying to pass in double [] array and added to my ArrayDictionary  can someone assist me. enum ArrayType { Array1 Array2 Array3 } public Dictionary > ArrayDictionary { get; set; } public void ConstructArray(double[] array1 double[] array2 double[] array3) { foreach (ArrayType arrayItem in Enum.GetValues(typeof(ArrayType))) { ArrayDictionary.Add(arrayItem .ToString() new List ()); switch (opticalArrayItem) { case ArrayType.Array1: foreach (double arrayItem in array1) { // Add to ArrayDictionary } break; case ArrayType.Array2: foreach (double arrayItem in array1) { // Add to ArrayDictionary b } break; case ArrayType.Array3: foreach (double arrayItem in array1) { // Add to ArrayDictionary } break; } } }

Vulpes · Answer

There are 401 values in the series 1.30 1.31 ....  5.30 So if 'i' is running from 0 to (n - 1) for each GivenArray there will be (401 * n) elements in each NewArray. The following is off the top of my head and so is untested: int len = GivenArray1.Length; int ix = 0; double temp = 0.0; double[] NewArray1 = new double[401 * len]; double[] NewArray2 = new double[401 * len]; double[] NewArray3 = new double[401 * len]; for(int i = 0; i { for(int w = 130; w { ix = 401 * i + w - 130; temp = GivenArray1[i] * GivenArray2[i] * GivenArray3[i] * (GivenArray2[i] - GivenArray2[i] * Math.Pow(w/100.0 5)); NewArray1[ix] = temp + GivenArray2[i]; NewArray2[ix] = temp + NewArray1[ix]; NewArray3[ix] = temp + NewArray2[ix]; } } double total123 = 0.0 total1 = 0.0; for(int i = 0; i for(int i = 0; i double finalOutput = total123/total1;

David Smith · Answer

thanks vuples you have truly been a help to me. thanks at least i can say i learn alot about link and the first time really playing around with an array of double like i am doing now. this was an interesting project and fun.

David Smith · Answer

Sorry about that copy paste error NewArray1 [] = ((GivenArray1[] * GivenArray2[] * GivenArray3[] ( GivenArray2[] - GivenArray2[] (Math.Pow( waveLength 5) )) + GivenArray2[] ); NewArray2 [] = ((GivenArray1[] * GivenArray2[] * GivenArray3[] ( GivenArray2[] - GivenArray2[] (Math.Pow( waveLength 5) )) + NewArray1 []); NewArray3 [] = ((GivenArray1[] * GivenArray2[] * GivenArray3[] ( GivenArray2[] - GivenArray2[] (Math.Pow( waveLength 5) )) + NewArray3 [] );

Vulpes · Answer

Can you check those formulae as GivenArray3 doesn't appear in them and presumably should do?

David Smith · Answer

Its iterating by 10 nano meters. I guess i give you the whole picture i was trying to do piece by piece basically all array are calculater by 10 nano meters that I am given from 1.30 to 5.30 Example : 1.30 1.31 1.33 1.34 1.35 ........to 5.28 5.29 5.30 etc........ I cant go into detail but  I am going to use 3 to 4 arrays only like we have been. We are on track just got to re think the logic to get to the final output. I am still using the enum ArrayType { Array1 Array2 Array3 } Dictionary >> ArrayTypeDictionary { get; set; } double[] GivenArray1{ get; set; } double[] GivenArray2{ get; set; } double[] GivenArray3{ get; set; } 1. Construct all ways from 1.30 to 5.30 by 10 nano meters by given formula below Note: wavelength is the values changing from 1.30 to 5.30 by 10 nano meters NewArray1 [] = ((GivenArray1[] * GivenArray2[] * GivenArray2[] ( GivenArray2[] - GivenArray2[] (Math.Pow( waveLength 5) )) + GivenArray2[] ); NewArray2 [] = ((GivenArray1[] * GivenArray2[] * GivenArray2[] ( GivenArray2[] - GivenArray2[] (Math.Pow( waveLength 5) )) + NewArray1 []); NewArray3 [] = ((GivenArray1[] * GivenArray2[] * GivenArray2[] ( GivenArray2[] - GivenArray2[] (Math.Pow( waveLength 5) )) + NewArray3 [] ); 2. After constructing each given array  Each array will construct another output Array from 1.30 to 5.30 by 10 nano meters by formula given. 3. There will be a total of 7 arrays need to get to the final single output double value. double finalOutput = (Summation (NewArray1 [] * NewArray2 [] * NewArray3 [])) / (Summation (NewArray1 []))

Vulpes · Answer

If the wavelength is also iterating by 1 viz: 1.30 2.30 3.30 4.30 5.30 then it's always going to be 1.30 more than the integer index i which is iterating from 0 to 4. So that would give: double[] Output2 = new double[Array1.Length]; double[] NewArray = new double[Array1.Length]; for(int i = 0; i { Output2[i] = Array1[i] * Array2[i] * Array3[i] * (Array2[i] - Array1[i]) + Array1[i]; NewArray[] = Array1[i] + Array2[i] + Math.Pow(i + 1.3 5) - Output2[i]; }

David Smith · Answer

Ok now i have came across a situation basically my my wavelength limit in the array is suppose to be from 1.30 to 5.30 which are doubles but we are indexing by int how to fix this situation. The wavelength value can't be the [i] it has to be a specfic value iterating from 1.30 to 5.30. We are almost there getting to the end. double[] Output2 = new double[Array1.Length]; for(int i = 0; i { Output2[i] = Array1[i] * Array2[i] * Array3[i] * (Array2[i] - Array1[i]) + Array1[i]; NewArray[] = (Array1[i] + Array2[i] ) + (Math.Pow( WaveLength  5)) - Output2[i] ) }

Vulpes · Answer

OK here's the code for that. I've also included an alternative approach using LINQ which is closer to the traditional approach and therefore more efficient than using Zip: double[] Array1 = t.ArrayDictionary[ArrayType.Array1.ToString()][0]; double[] Array2 = t.ArrayDictionary[ArrayType.Array2.ToString()][0]; double[] Array3 = t.ArrayDictionary[ArrayType.Array3.ToString()][0]; // using LINQ double[] Output = Array1.Zip(Array2 (a1 a2) => a1 * a2 * (a2 - a1)).ToArray() .Zip(Array3 (a12 a3) => a12 * a3).ToArray() .Zip(Array1 (a123 a1) => a123 + a1).ToArray(); foreach(double d in Output) Console.WriteLine(d); // using traditional approach double[] Output2 = new double[Array1.Length]; for(int i = 0; i Output2[i] = Array1[i] * Array2[i] * Array3[i] * (Array2[i] - Array1[i]) + Array1[i]; foreach(double d in Output2) Console.WriteLine(d); // using alternative LINQ approach double[] Output3 = Enumerable.Range(0 Array1.Length).Select(i => Array1[i] * Array2[i] * Array3[i] *(Array2[i] - Array1[i]) + Array1[i]).ToArray(); foreach(double d in Output3) Console.WriteLine(d);

David Smith · Answer

I had the formula wrong im glad you caught that. thanks my eyes getting weak. OutPut[i]=(( ArrayDictionary [Array1][i] * ArrayDictionary [Array2][i] * ArrayDictionary [Array3][i] * ( ArrayDictionary Array2][i] - ArrayDictionary [Array1][i])) + ArrayDictionary [Array1][i]);

Vulpes · Answer

Are you sure that there isn't a plus operator where I've marked? Output[] =( (Array1[] * Array2[] * Array3[] + ( Array1[] * Array2[] ) )+ Array1[] )

David Smith · Answer

show both ways the traditional way look cleaner linq is cool  show both i getting a better at the linq day by day

David Smith · Answer

This is the true equation how to Zip this up correctly using linq Output[] =( (Array1[] * Array2[] * Array3[] ( Array1[] * Array2[] ) )+ Array1[] )

David Smith · Answer

sorry i didnt ready clearly sorry about that vuples you did right. brb

David Smith · Answer

Sorry for the confusion im try to store Array1 Array2 and Array3 in to the output Array. So this time it will not be an a single output the out put will be an array itself. I am trying to add Array1 Array2 and Array3 together. Review equation below. Look at the array brackets. Output[] =( Array1[] * Array2[] * Array3[] ) Output [] will not be a single bracket it will be the array output of ( Array1[] * Array2[] * Array3[] )

Vulpes · Answer

It can be done in LINQ but not very elegantly or efficiently. Frankly I'd use the traditional method. Here's the code for both: // using LINQ double[] Output = Array1.Zip(Array2 (a1 a2) => a1 * a2).ToArray() .Zip(Array3 (a12 a3) => a12 * a3).ToArray(); foreach(double d in Output) Console.WriteLine(d); // using traditional approach double[] Output2 = new double[Array1.Length]; for(int i = 0; i Output2[i] = Array1[i] * Array2[i] * Array3[i]; foreach(double d in Output2) Console.WriteLine(d); The output in either case will be : 28 80 162

David Smith · Answer

ok we almost there now that we have all the arrays i need to multiply all the array into one array how to do this using linq review below. Output[] = Array1[] * Array2[] * Array3[]

Vulpes · Answer

It won't compile if you do that. You'll be informed that List.Add has some invalid arguments - argument 1 cannot convert from double to double[]. If it's any help to you here's a little program I did to test my own code and make sure the Dictionary could be iterated subsequently: using System; using System.Collections.Generic; enum ArrayType { Array1 Array2 Array3 } class Test { static void Main() { Test t = new Test(); var array1 = new double[]{1 2 3}; var array2 = new double[]{4 5 6}; var array3 = new double[]{7 8 9}; t.ConstructArray(array1 array2 array3); foreach(string arrayItem in t.ArrayDictionary.Keys) { foreach(double[] da in t.ArrayDictionary[arrayItem]) foreach(double d in da) Console.WriteLine(d); } Console.ReadKey(); } public Dictionary > ArrayDictionary { get; set; } public void ConstructArray(double[] array1 double[] array2 double[] array3) { ArrayDictionary = new Dictionary >(); // if not instantiated elsewhere foreach (ArrayType arrayItem in Enum.GetValues(typeof(ArrayType))) { ArrayDictionary.Add(arrayItem.ToString() new List ()); switch (arrayItem) { case ArrayType.Array1: ArrayDictionary[arrayItem.ToString()].Add(array1); break; case ArrayType.Array2: ArrayDictionary[arrayItem.ToString()].Add(array2); break; case ArrayType.Array3: ArrayDictionary[arrayItem.ToString()].Add(array3); break; } } } }

David Smith · Answer

I had to add each foreach to each switch case to get each item out of the passed double array public Dictionary > ArrayDictionary { get; set; } public void ConstructArray(double[] array1 double[] array2 double[] array3) { ArrayDictionary = new Dictionary >(); // if not instantiated elsewhere foreach (ArrayType arrayItem in Enum.GetValues(typeof(ArrayType))) { ArrayDictionary.Add(arrayItem.ToString() new List ()); switch (arrayItem) { case ArrayType.Array1: foreach( var item in array1) { ArrayDictionary[arrayItem.ToString()].Add(item ); } break; case ArrayType.Array2: foreach( var item in array2) { ArrayDictionary[arrayItem.ToString()].Add(item ); } break; case ArrayType.Array3: foreach( var item in array3) { ArrayDictionary[arrayItem.ToString()].Add(item ); } break; } } }

Vulpes · Answer

That's right. Because the Dictionary's value is a List of double arrays you can simply add a double array to the list for a given key.

David Smith · Answer

Question so are you saying there is no reason to do a foreach or iteration in each switch case?

Vulpes · Answer

I'm assuming that 'opticalArrayItem' should just be 'arrayItem' as the code would make little sense otherwise: public Dictionary > ArrayDictionary { get; set; } public void ConstructArray(double[] array1 double[] array2 double[] array3) { ArrayDictionary = new Dictionary >(); // if not instantiated elsewhere foreach (ArrayType arrayItem in Enum.GetValues(typeof(ArrayType))) { ArrayDictionary.Add(arrayItem.ToString() new List ()); switch (arrayItem) { case ArrayType.Array1: ArrayDictionary[arrayItem.ToString()].Add(array1); break; case ArrayType.Array2: ArrayDictionary[arrayItem.ToString()].Add(array2); break; case ArrayType.Array3: ArrayDictionary[arrayItem.ToString()].Add(array3); break; } } }