0
$\begingroup$

I am showing that applying a certain method leads to improved performance on a machine learning problem, and that this effect is consistent across different datasets (in this case, datasets collected across different days).

However the effect is small (on the scale of 0.01), and when plotted out, is dwarfed by the large variation across the different datasets. Are there any alternative plots/visualization methods I can use to demonstrate my effect?

This seems to be a common problem when the effect is smaller than the variation between subjects/days.

enter image description here

$\endgroup$
  • $\begingroup$ A graph taller than wide might work better for you than one with your present aspect ratio. $\endgroup$ – Nick Cox Jun 11 '19 at 7:32
  • $\begingroup$ @NickCox It's true! But changing the aspect ratio will only help you until a certain point, so I don't see it as a final solution $\endgroup$ – David Jun 11 '19 at 7:33
  • $\begingroup$ @David I don't know what would be a general solution (the phrase final solution is better avoided; Google it to see why) without knowing the range of possibilities. Usually I would suggest log scale but the graph given doesn't encourage that idea. $\endgroup$ – Nick Cox Jun 11 '19 at 7:38
  • $\begingroup$ Detail: showing 1.1(0.1)1.9 on the x axis is utterly pointless when data points are for 1 and 2 and intermediate values aren't defined. $\endgroup$ – Nick Cox Jun 11 '19 at 7:40
  • $\begingroup$ @NickCox What about the idea in my answer? $\endgroup$ – David Jun 11 '19 at 7:43
0
$\begingroup$

My first idea was doing a boxplot (or maybe a histogram) of the differences, but this is somehow incomplete as it loses information about the actual values (which I guess you do not want to happen)

However, what you may be interested in is a plot where you represent the variation (Y-axis) versus the value on the first day (X-axis). By representing it this way, you get rid of the "scale" problem

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.