More efficient way to plot x-axis points?

Question

I have been working on a project that requires a bar graph to be populated with price results. The chart displays the number of items within a given price range. For instance, if on amazon there are 9 items within the price range of $0-$10 the x-axis would display $0-$10 and the y-axis would be populated with a value of 9.

My bar graph has 8 bars, all with similar price ranges. $0-$10, $10-$20, $20-$30... etc.

My question is this: What is the best way to define those individual points? There is no common price range between these items, so the x-axis cannot be static numbers. They must be dynamically calculated within the range of results.

As such, currently I am creating the x-axis points as follows:

I take the lowest result: @numbers[0] = results[0];

And I take the highest result: @numbers[8] = results[-1];

Then I find the median of the two: @numbers[4] = (@numbers[0]+@numbers[8])/2;

I then repeat the process 6 more times

@numbers[2] = (@numbers[0]+@numbers[4])/2; @numbers[6] = (@numbers[4]+@numbers[8])/2; @numbers[1] = (@numbers[0]+@numbers[2])/2; @numbers[3] = (@numbers[2]+@numbers[4])/2; @numbers[5] = (@numbers[4]+@numbers[6])/2; @numbers[7] = (@numbers[6]+@numbers[8])/2;

This gives me the results I need, but it seems awfully repetitive and I would imagine there is a better way.

I tried creating a loop, but I could not write it in a less verbose manner.

Is there a quicker way to do this, or perhaps something more along the lines of DRY?

Answer 1

Are your bins always of equal size? In your example, all share range=10. If so, then you could do:

binspacing = overall range / (numberofbins-1);

and the position of bin n would be the x-axis position of numbers[0] plus n times the binspacing.

Answer 2

Perhaps my way would have been superior if the goal was to write as convoluted code as possible.... Thanks.