MediumNLPnlp-351

Laplace Smoothed Unigram Probability

Problem

Laplace smoothing for a unigram adds one to every count: (count + 1) / (total_tokens + vocab_size).

Given the token count, the total tokens, and the vocabulary size, return the probability.

Hint

Every word in the vocabulary gets one pseudo-count.

case 1

in: [5,100,50]

expected: 0.04

case 2

in: [0,10,5]

expected: 0.06666666666666667

case 3

in: [30,1000,100]

expected: 0.028181818181818183

Starter codePython

def laplace_smoothed_unigram_probability(count, total_tokens, vocab_size):
    # Your code here
    pass

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