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    <p>My immediate reaction would be something like <code>(number_up - number_down) / (number_up + number_down)</code>. This basically gives up or down as a percentage of the whole. The obvious shortcoming is that for a really short list, the percentage can be pretty high from a fairly small absolute difference (e.g., 3 up, 1 down).</p> <p>Edit: One possible way to keep small lists from excessively impacting overall scores is to add a couple of constants into the equation:</p> <pre><code>min_denom = 20; factor = 10; result = ((number_up - number_down) / (number_up + number_down + min_denom)) * factor; </code></pre> <p>This lets you take both relative and absolute differences into account to some degree. For example, with 3 up/1 down, it'll give 0.833. With 6 up/2 down (same ratio, but twice as many of each) it'll give 1.4. At the same time, relative differences are still taken into account, so (for example) 10 up/1 down will give 2.9.</p> <p>In effect, this retains the same general idea, but allows you to pick some degree (adjustable by changing the min_denom) to which you give extra weight to larger samples. Strictly speaking, <code>factor</code> isn't entirely necessary -- it just helps keep the results in a convenient range.</p> <p>Of course, this may not be appropriate -- for what you're dealing with, a sample of four may carry the same weight as a sample of 100. Another possible shortcoming is that the result values become more open-ended, instead of a nice, neat -1..1.</p>
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    1. POEvaluating a score that indicates a state of a group of elements based on the counts of specific properties of the elements
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    1. USJerry Coffin
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    1. POEvaluating a score that indicates a state of a group of elements based on the counts of specific properties of the elements
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    1. COThat's also one of the issue I have. I'm trying to take into account the size of the elements.
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      1. USEinar
    2. COthis method returns a number between -1 and 1. You can scale it at your will, for example by the logarithm of the count.
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    3. CO@Einar: I guessed that might be the case, and was editing in a possible solution as you commented.
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