N-gram language models in NLP - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create bigrams from a list of words.

NLP

bigrams = [(words[i], words[i+[1]]) for i in range(len(words)-1)]

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A0

B2

C1

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 or 2 as the offset causes index errors or wrong pairs.

The bigram pairs each word with the next word, so the index offset is 1.

2fill in blank

medium

Complete the code to count the frequency of each trigram in the text.

NLP

from collections import Counter
trigrams = [(words[i], words[i+1], words[i+[1]]) for i in range(len(words)-2)]
trigram_counts = Counter(trigrams)

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A1

B0

C3

D2

Attempts:

3 left

💡 Hint

Common Mistakes

Using 1 or 3 causes incorrect trigram formation or index errors.

For trigrams, the third word is two positions ahead of the first word, so the offset is 2.

3fill in blank

hard

Fix the error in the code that calculates the probability of a bigram using counts.

NLP

bigram_prob = bigram_counts[bigram] / [1]

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Asum(bigram_counts.values())

Blen(bigram_counts)

Clen(bigram_counts[bigram])

Dbigram_counts[bigram[0]]

Attempts:

3 left

💡 Hint

Common Mistakes

Using length of bigram_counts or counts of a single word causes wrong probabilities.

The probability of a bigram is its count divided by the total count of all bigrams.

4fill in blank

hard

Fill both blanks to create a dictionary of bigram probabilities from counts.

NLP

bigram_probs = {bg: bigram_counts[bg] / [1] for bg in [2]

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Asum(bigram_counts.values())

Bbigram_counts.keys()

Cbigram_counts.items()

Dlen(bigram_counts)

Attempts:

3 left

💡 Hint

Common Mistakes

Using items() instead of keys() causes errors in dictionary comprehension.

We divide each bigram count by the total count of all bigrams (sum of values) and iterate over all bigram keys.

5fill in blank

hard

Fill the two blanks to build a conditional probability dictionary for bigrams.

NLP

cond_probs = {}
for (w1, w2), count in bigram_counts.items():
    w1_total = sum(c for (word1, _), c in bigram_counts.items() if word1 == [1])
    cond_probs.setdefault(w1, {})[[2]] = count / w1_total

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Aw2

Bw1

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up w1 and w2 keys causes wrong dictionary structure or division errors.

The conditional probability P(w2|w1) is count divided by total counts of bigrams starting with w1. We use w2 as the key in the nested dict, divide by w1_total, and sum counts where word1 equals w1.