So, wouldn't the denominator of precision @ k just be equal to k? And if not, why?

Precision@k = (# of recommended items @k that are relevant) / (# of recommended items @k)
Recall and Precision at k for Recommender Systems
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Maher Malaeb
Corduroy Bologna
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Dec 30, 2022
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So, wouldn't the denominator of precision @ k just be equal to k? And if not, why?
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Written by Corduroy Bologna186 Followers
·345 Following
Writing in blissful ignorance as the world as we know it collapses all around us :)
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