Stretch or resample 1D array





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This question might be something really simple and I might be missing something really basic, but how do you interpolate a 1D array in C#?



Lets say I have this array of n elements



int array1 = new int { 1, 3, 5, 7, 1 };


How to stretch or compress the array so that it has n values and interpolates the values, just like when you resize an image, thats it, not chopping or adding zeros or empty values to the array.



For example if I want to convert the array so it has n = 4 elements, get this



array1

>>[2, 4, 6, 4]


Strech example



what I'm trying to do is the same as the resample function from matlab does
https://mathworks.com/help/signal/ref/resample.html






c# arrays int







share|improve this question

























  • I don't understand how you got from the first array to the second one. Did you take the first n elements and add 1 to them? Something else?

    – mjwills
    Nov 17 '18 at 7:14













  • What is the calculation you are looking for? What is stretch supose to do?

    – Aldert
    Nov 17 '18 at 7:15











  • @mjwills It seems to be an average of 1 and 3, 3 and 5, etc.

    – John
    Nov 17 '18 at 7:15






  • 4





    Please show us at least 7 other sample inputs and the expected results for those inputs. Ensure that some of the inputs don't have numbers that are exactly 2 apart from each other. Also explain what n is for.

    – mjwills
    Nov 17 '18 at 7:23






  • 1





    There's a C++ version here if it helps.

    – John
    Nov 17 '18 at 9:22


















-2















This question might be something really simple and I might be missing something really basic, but how do you interpolate a 1D array in C#?



Lets say I have this array of n elements



int array1 = new int { 1, 3, 5, 7, 1 };


How to stretch or compress the array so that it has n values and interpolates the values, just like when you resize an image, thats it, not chopping or adding zeros or empty values to the array.



For example if I want to convert the array so it has n = 4 elements, get this



array1

>>[2, 4, 6, 4]


Strech example



what I'm trying to do is the same as the resample function from matlab does
https://mathworks.com/help/signal/ref/resample.html






c# arrays int







share|improve this question

























  • I don't understand how you got from the first array to the second one. Did you take the first n elements and add 1 to them? Something else?

    – mjwills
    Nov 17 '18 at 7:14













  • What is the calculation you are looking for? What is stretch supose to do?

    – Aldert
    Nov 17 '18 at 7:15











  • @mjwills It seems to be an average of 1 and 3, 3 and 5, etc.

    – John
    Nov 17 '18 at 7:15






  • 4





    Please show us at least 7 other sample inputs and the expected results for those inputs. Ensure that some of the inputs don't have numbers that are exactly 2 apart from each other. Also explain what n is for.

    – mjwills
    Nov 17 '18 at 7:23






  • 1





    There's a C++ version here if it helps.

    – John
    Nov 17 '18 at 9:22














-2












-2








-2


1






This question might be something really simple and I might be missing something really basic, but how do you interpolate a 1D array in C#?



Lets say I have this array of n elements



int array1 = new int { 1, 3, 5, 7, 1 };


How to stretch or compress the array so that it has n values and interpolates the values, just like when you resize an image, thats it, not chopping or adding zeros or empty values to the array.



For example if I want to convert the array so it has n = 4 elements, get this



array1

>>[2, 4, 6, 4]


Strech example



what I'm trying to do is the same as the resample function from matlab does
https://mathworks.com/help/signal/ref/resample.html






c# arrays int







share|improve this question
















This question might be something really simple and I might be missing something really basic, but how do you interpolate a 1D array in C#?



Lets say I have this array of n elements



int array1 = new int { 1, 3, 5, 7, 1 };


How to stretch or compress the array so that it has n values and interpolates the values, just like when you resize an image, thats it, not chopping or adding zeros or empty values to the array.



For example if I want to convert the array so it has n = 4 elements, get this



array1

>>[2, 4, 6, 4]


Strech example



what I'm trying to do is the same as the resample function from matlab does
https://mathworks.com/help/signal/ref/resample.html






c# arrays int




c# arrays int






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 17 '18 at 9:13







Boris

















asked Nov 17 '18 at 7:06









BorisBoris

166




166













  • I don't understand how you got from the first array to the second one. Did you take the first n elements and add 1 to them? Something else?

    – mjwills
    Nov 17 '18 at 7:14













  • What is the calculation you are looking for? What is stretch supose to do?

    – Aldert
    Nov 17 '18 at 7:15











  • @mjwills It seems to be an average of 1 and 3, 3 and 5, etc.

    – John
    Nov 17 '18 at 7:15






  • 4





    Please show us at least 7 other sample inputs and the expected results for those inputs. Ensure that some of the inputs don't have numbers that are exactly 2 apart from each other. Also explain what n is for.

    – mjwills
    Nov 17 '18 at 7:23






  • 1





    There's a C++ version here if it helps.

    – John
    Nov 17 '18 at 9:22



















  • I don't understand how you got from the first array to the second one. Did you take the first n elements and add 1 to them? Something else?

    – mjwills
    Nov 17 '18 at 7:14













  • What is the calculation you are looking for? What is stretch supose to do?

    – Aldert
    Nov 17 '18 at 7:15











  • @mjwills It seems to be an average of 1 and 3, 3 and 5, etc.

    – John
    Nov 17 '18 at 7:15






  • 4





    Please show us at least 7 other sample inputs and the expected results for those inputs. Ensure that some of the inputs don't have numbers that are exactly 2 apart from each other. Also explain what n is for.

    – mjwills
    Nov 17 '18 at 7:23






  • 1





    There's a C++ version here if it helps.

    – John
    Nov 17 '18 at 9:22

















I don't understand how you got from the first array to the second one. Did you take the first n elements and add 1 to them? Something else?

– mjwills
Nov 17 '18 at 7:14







I don't understand how you got from the first array to the second one. Did you take the first n elements and add 1 to them? Something else?

– mjwills
Nov 17 '18 at 7:14















What is the calculation you are looking for? What is stretch supose to do?

– Aldert
Nov 17 '18 at 7:15





What is the calculation you are looking for? What is stretch supose to do?

– Aldert
Nov 17 '18 at 7:15













@mjwills It seems to be an average of 1 and 3, 3 and 5, etc.

– John
Nov 17 '18 at 7:15





@mjwills It seems to be an average of 1 and 3, 3 and 5, etc.

– John
Nov 17 '18 at 7:15




4




4





Please show us at least 7 other sample inputs and the expected results for those inputs. Ensure that some of the inputs don't have numbers that are exactly 2 apart from each other. Also explain what n is for.

– mjwills
Nov 17 '18 at 7:23





Please show us at least 7 other sample inputs and the expected results for those inputs. Ensure that some of the inputs don't have numbers that are exactly 2 apart from each other. Also explain what n is for.

– mjwills
Nov 17 '18 at 7:23




1




1





There's a C++ version here if it helps.

– John
Nov 17 '18 at 9:22





There's a C++ version here if it helps.

– John
Nov 17 '18 at 9:22












1 Answer
1






active

oldest

votes


















2














I suggest this solution for the case that the new array is shorter than the old one:



int array1 = new int { 1, 3, 5, 7, 9 };

int array2 = new int[4];



for (var i = 0; i < array2.Length; i++)

{

    var doubleIndex1 = (double)i * array1.Length / array2.Length;

    var index1 = (int)Math.Floor(doubleIndex1);

    var rel = doubleIndex1 - index1;



    array2[i] = (int)Math.Round((1.0 - rel) * array1[index1] + rel * array1[index1 + 1]);

}





share|improve this answer
























  • yeah, it works that way but only when the values in the array are linearly growing or decreasing, thanks anyway

    – Boris
    Nov 17 '18 at 7:47













  • Well yes, this is linear interpolation. Do you need someting more sophisticated? Then have a look at en.wikipedia.org/wiki/Interpolation

    – Klaus Gütter
    Nov 17 '18 at 8:11






  • 2





    @Boris that's what you get for not providing a good set of examples when asked to.

    – John
    Nov 17 '18 at 8:12











  • just a resize dude, as when you stretch or shrink an image, I really dont get whats so hard to understand, I feel like I'm speaking a totally different language from anyone else, dont know how else to explain it, sorry

    – Boris
    Nov 17 '18 at 8:48






  • 1





    @Boris as a programmer and an amateur photographer, I'll tell you that when it comes to resizing an image Photoshop gives me a lot of resampling options, including: bicubic, nearest neighbour, and bilinear. There are a vast number of interpolation algorithms, etc. With a single example, it's impossible to tell which one you want. Klaus believed that linear was the method you wanted, now you're saying it's not.

    – John
    Nov 17 '18 at 9:08














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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














I suggest this solution for the case that the new array is shorter than the old one:



int array1 = new int { 1, 3, 5, 7, 9 };

int array2 = new int[4];



for (var i = 0; i < array2.Length; i++)

{

    var doubleIndex1 = (double)i * array1.Length / array2.Length;

    var index1 = (int)Math.Floor(doubleIndex1);

    var rel = doubleIndex1 - index1;



    array2[i] = (int)Math.Round((1.0 - rel) * array1[index1] + rel * array1[index1 + 1]);

}





share|improve this answer
























  • yeah, it works that way but only when the values in the array are linearly growing or decreasing, thanks anyway

    – Boris
    Nov 17 '18 at 7:47













  • Well yes, this is linear interpolation. Do you need someting more sophisticated? Then have a look at en.wikipedia.org/wiki/Interpolation

    – Klaus Gütter
    Nov 17 '18 at 8:11






  • 2





    @Boris that's what you get for not providing a good set of examples when asked to.

    – John
    Nov 17 '18 at 8:12











  • just a resize dude, as when you stretch or shrink an image, I really dont get whats so hard to understand, I feel like I'm speaking a totally different language from anyone else, dont know how else to explain it, sorry

    – Boris
    Nov 17 '18 at 8:48






  • 1





    @Boris as a programmer and an amateur photographer, I'll tell you that when it comes to resizing an image Photoshop gives me a lot of resampling options, including: bicubic, nearest neighbour, and bilinear. There are a vast number of interpolation algorithms, etc. With a single example, it's impossible to tell which one you want. Klaus believed that linear was the method you wanted, now you're saying it's not.

    – John
    Nov 17 '18 at 9:08


















2














I suggest this solution for the case that the new array is shorter than the old one:



int array1 = new int { 1, 3, 5, 7, 9 };

int array2 = new int[4];



for (var i = 0; i < array2.Length; i++)

{

    var doubleIndex1 = (double)i * array1.Length / array2.Length;

    var index1 = (int)Math.Floor(doubleIndex1);

    var rel = doubleIndex1 - index1;



    array2[i] = (int)Math.Round((1.0 - rel) * array1[index1] + rel * array1[index1 + 1]);

}





share|improve this answer
























  • yeah, it works that way but only when the values in the array are linearly growing or decreasing, thanks anyway

    – Boris
    Nov 17 '18 at 7:47













  • Well yes, this is linear interpolation. Do you need someting more sophisticated? Then have a look at en.wikipedia.org/wiki/Interpolation

    – Klaus Gütter
    Nov 17 '18 at 8:11






  • 2





    @Boris that's what you get for not providing a good set of examples when asked to.

    – John
    Nov 17 '18 at 8:12











  • just a resize dude, as when you stretch or shrink an image, I really dont get whats so hard to understand, I feel like I'm speaking a totally different language from anyone else, dont know how else to explain it, sorry

    – Boris
    Nov 17 '18 at 8:48






  • 1





    @Boris as a programmer and an amateur photographer, I'll tell you that when it comes to resizing an image Photoshop gives me a lot of resampling options, including: bicubic, nearest neighbour, and bilinear. There are a vast number of interpolation algorithms, etc. With a single example, it's impossible to tell which one you want. Klaus believed that linear was the method you wanted, now you're saying it's not.

    – John
    Nov 17 '18 at 9:08
















2












2








2







I suggest this solution for the case that the new array is shorter than the old one:



int array1 = new int { 1, 3, 5, 7, 9 };

int array2 = new int[4];



for (var i = 0; i < array2.Length; i++)

{

    var doubleIndex1 = (double)i * array1.Length / array2.Length;

    var index1 = (int)Math.Floor(doubleIndex1);

    var rel = doubleIndex1 - index1;



    array2[i] = (int)Math.Round((1.0 - rel) * array1[index1] + rel * array1[index1 + 1]);

}





share|improve this answer













I suggest this solution for the case that the new array is shorter than the old one:



int array1 = new int { 1, 3, 5, 7, 9 };

int array2 = new int[4];



for (var i = 0; i < array2.Length; i++)

{

    var doubleIndex1 = (double)i * array1.Length / array2.Length;

    var index1 = (int)Math.Floor(doubleIndex1);

    var rel = doubleIndex1 - index1;



    array2[i] = (int)Math.Round((1.0 - rel) * array1[index1] + rel * array1[index1 + 1]);

}






share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 17 '18 at 7:41









Klaus GütterKlaus Gütter

2,60231422




2,60231422













  • yeah, it works that way but only when the values in the array are linearly growing or decreasing, thanks anyway

    – Boris
    Nov 17 '18 at 7:47













  • Well yes, this is linear interpolation. Do you need someting more sophisticated? Then have a look at en.wikipedia.org/wiki/Interpolation

    – Klaus Gütter
    Nov 17 '18 at 8:11






  • 2





    @Boris that's what you get for not providing a good set of examples when asked to.

    – John
    Nov 17 '18 at 8:12











  • just a resize dude, as when you stretch or shrink an image, I really dont get whats so hard to understand, I feel like I'm speaking a totally different language from anyone else, dont know how else to explain it, sorry

    – Boris
    Nov 17 '18 at 8:48






  • 1





    @Boris as a programmer and an amateur photographer, I'll tell you that when it comes to resizing an image Photoshop gives me a lot of resampling options, including: bicubic, nearest neighbour, and bilinear. There are a vast number of interpolation algorithms, etc. With a single example, it's impossible to tell which one you want. Klaus believed that linear was the method you wanted, now you're saying it's not.

    – John
    Nov 17 '18 at 9:08





















  • yeah, it works that way but only when the values in the array are linearly growing or decreasing, thanks anyway

    – Boris
    Nov 17 '18 at 7:47













  • Well yes, this is linear interpolation. Do you need someting more sophisticated? Then have a look at en.wikipedia.org/wiki/Interpolation

    – Klaus Gütter
    Nov 17 '18 at 8:11






  • 2





    @Boris that's what you get for not providing a good set of examples when asked to.

    – John
    Nov 17 '18 at 8:12











  • just a resize dude, as when you stretch or shrink an image, I really dont get whats so hard to understand, I feel like I'm speaking a totally different language from anyone else, dont know how else to explain it, sorry

    – Boris
    Nov 17 '18 at 8:48






  • 1





    @Boris as a programmer and an amateur photographer, I'll tell you that when it comes to resizing an image Photoshop gives me a lot of resampling options, including: bicubic, nearest neighbour, and bilinear. There are a vast number of interpolation algorithms, etc. With a single example, it's impossible to tell which one you want. Klaus believed that linear was the method you wanted, now you're saying it's not.

    – John
    Nov 17 '18 at 9:08



















yeah, it works that way but only when the values in the array are linearly growing or decreasing, thanks anyway

– Boris
Nov 17 '18 at 7:47







yeah, it works that way but only when the values in the array are linearly growing or decreasing, thanks anyway

– Boris
Nov 17 '18 at 7:47















Well yes, this is linear interpolation. Do you need someting more sophisticated? Then have a look at en.wikipedia.org/wiki/Interpolation

– Klaus Gütter
Nov 17 '18 at 8:11





Well yes, this is linear interpolation. Do you need someting more sophisticated? Then have a look at en.wikipedia.org/wiki/Interpolation

– Klaus Gütter
Nov 17 '18 at 8:11




2




2





@Boris that's what you get for not providing a good set of examples when asked to.

– John
Nov 17 '18 at 8:12





@Boris that's what you get for not providing a good set of examples when asked to.

– John
Nov 17 '18 at 8:12













just a resize dude, as when you stretch or shrink an image, I really dont get whats so hard to understand, I feel like I'm speaking a totally different language from anyone else, dont know how else to explain it, sorry

– Boris
Nov 17 '18 at 8:48





just a resize dude, as when you stretch or shrink an image, I really dont get whats so hard to understand, I feel like I'm speaking a totally different language from anyone else, dont know how else to explain it, sorry

– Boris
Nov 17 '18 at 8:48




1




1





@Boris as a programmer and an amateur photographer, I'll tell you that when it comes to resizing an image Photoshop gives me a lot of resampling options, including: bicubic, nearest neighbour, and bilinear. There are a vast number of interpolation algorithms, etc. With a single example, it's impossible to tell which one you want. Klaus believed that linear was the method you wanted, now you're saying it's not.

– John
Nov 17 '18 at 9:08







@Boris as a programmer and an amateur photographer, I'll tell you that when it comes to resizing an image Photoshop gives me a lot of resampling options, including: bicubic, nearest neighbour, and bilinear. There are a vast number of interpolation algorithms, etc. With a single example, it's impossible to tell which one you want. Klaus believed that linear was the method you wanted, now you're saying it's not.

– John
Nov 17 '18 at 9:08






















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