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Reciprocal Rank Fusion Score
Problem
Reciprocal rank fusion contributes 1 / (k + rank) for a document at the given rank, where rank starts at one and k is the smoothing constant.
Given the rank and k, return the contribution.
Hint
The constant damps the influence of very high ranks.
Test cases
case 1
in: [1,60]
expected: 0.01639344262295082
case 2
in: [10,60]
expected: 0.014285714285714285
case 3
in: [3,0]
expected: 0.3333333333333333
Starter codePython
def reciprocal_rank_fusion_score(rank, k):
# Your code here
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