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I am trying to look for methods to aggregate and find the average of qualitative data that are outputted from a simulation.

There are 20 qualitative measures, each divided unevenly into 4 cycles labeled 1-4. I am trying to find which bucket would be the mean? I cannot simply take the average, as that would cause problems if most were labeled 1 and 4. The average could not be 2.5.

I will say that 1 is closely related to the numbers it's next to, 4 and 2(because its a cycle). So, in my previous example, the answer of 2.5 is further incorrect because its saying the mean cycle is one that's most opposite of where most qualitative data is.

I was looking into cell-cycle prediction and thought this may be something similar.

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If the features are considered categorical, think of the values as A,B,C,D. There's no possible mean value in this case, the most common way to aggregate would be to pick the mode, i.e. the value which obtains the maximum frequency in the data.

Apparently the values might be ordinal, i.e. they are not continuous but they have an order. Sometimes these are treated as numerical, it depends on the application.

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  • $\begingroup$ Thank you for the answer! I guess i could explain more, but even so, maybe the answer still is to have the mode. The features ARE considered categorical, though I wanted to try to quantify it more, to have a more granular view of the cycle. If there are 4 stages, and the mode is telling me I am in, say, stage 1, am I at the beginning or end of that stage? What if Stage 4 had the second highest count? That could mean I am at the beginning of Stage 1. If Stage 2 had the second highest count, that could mean I am at the end of Stage 1. $\endgroup$ Commented Aug 30, 2022 at 19:02
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    $\begingroup$ @worldCurrencies what you're talking about is the problem of ordinal values, for which there is no standard solution in ML (to my knowledge). As far as I know this is usually addressed with custom solutions specific to the application. $\endgroup$
    – Erwan
    Commented Aug 31, 2022 at 8:59

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