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I am planning to use data for a clustering problem that contains a column with a binary value BUY/SELL.

Should I be converting this attribute and assign it binary values (BUY=1, SELL=0), and keep it on the same column, thus reducing the number dimensions

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Hot encode the attribute (adding two columns BUY and SELL and putting 1 on the appropriate column)?

How do these two methods of nominal to numeric conversion affect the final model for popular clustering algorithms (K means, Hierarchical, etc...)

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2 Answers 2

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Not much of difference in your case. The difference is in just 1 dimension which does not affect much. The only point I can add is that if the number of BUY and SELL values are not the same, you can replace them with their frequencies i.e. if 40% BUY and 60% SELL, then replace BUY with 0.4 and SELL with 0.6

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  • $\begingroup$ Exactly this. Also don't forget to scale it to your data (if you have are working on a TV shop, and have a variable 'price', varrying between 200 and 800, having a binary variable varrying between 0 and 1 will not have any impact on your clustering. Think about normalizing.) $\endgroup$
    – Adept
    Jul 29, 2021 at 12:03
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If one value has more priority than the other, then you can go with binary encoding. e.x) If the values are based on education level, you can assign 0 to school-level education and 1 to college-level education.

If the values do not have any arithmetical dependency, then you need to go for one-hot encoding.

In your case, hot encoding is better.

Edit: If we have only two values, either binary encoding or hot encoding will work. This edit is based on the comment from @beamsadept.

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  • $\begingroup$ This would work with 3 or + values, it's not a problem if they're only 2. The problem is, when you have 1, 2 and 3, that distance between 1 and 2 is not the same that distance between 1 and 3, so it defines an order. Here, since you have 2 possibilities, the distance will always be the same if you don't have the same value $\endgroup$
    – Adept
    Jul 29, 2021 at 11:59
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    $\begingroup$ @BeamsAdept Yes, after giving some thought, I had to agree with you. If we have only two values, binary coding will work fine. $\endgroup$
    – Venkat
    Jul 29, 2021 at 12:27

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