HardNLPnlp-030

Interpolation Probability

Problem

Return the linearly interpolated trigram probability lambdas[0] * P(w3 | w1 w2) + lambdas[1] * P(w3 | w2) + lambdas[2] * P(w3).

The three dictionaries are keyed by 'w1 w2 w3', 'w2 w3', and w3 respectively. Missing entries contribute 0.0.

Hint

Look up each order with dict.get and blend the three values with the weights.

case 1

in: [["a","b"],"c",{"a b c":0.2},{"b c":0.4},{"c":0.1},[0.5,0.3,0.2]]

expected: 0.24

case 2

in: [["a","b"],"c",{},{},{"c":0.5},[0.2,0.3,0.5]]

expected: 0.25

case 3

in: [["a","b"],"c",{},{"b c":1},{},[0,1,0]]

expected: 1

case 4

in: [["x","y"],"z",{},{},{},[0.33,0.33,0.34]]

expected: 0

Starter codePython

def interpolation_prob(context, token, tri_probs, bi_probs, uni_probs, lambdas):
    # Your code here
    pass

4 test cases · run your code against them in the editor