# Demo: conditional probability

# Word counts in the Brown corpus

# Demo for conditional probability

#

import nltk

from nltk.corpus import brown

# number of words in the Brown corpus

brown_numwords = len(brown.words())

# number of word pairs in the Brown corpus

# word pairs = "bigrams"

brown_num_bigrams = len(list(nltk.bigrams(brown.words())))

# quick look at typical word pairs: via pre-given collocations function

brown_nltk = nltk.Text(brown.words())

brown_nltk.collocations()

# Now we look at one of those collocations in more detail:

# How likely is it that the next word will be "ago",

# GIVEN THAT the word we just read was "years"?

# First we determine the

# frequencies for "years" and "ago"

count_years = brown.words().count("years")

count_ago = brown.words().count("ago")

# Probability for "years": relative frequency

prob_years = count_years / brown_numwords

# Now we determine the frequency of "years ago"

count_years_ago = 0

for word1, word2 in nltk.bigrams(brown.words()):

if word1 == "years" and word2 == "ago":

count_years_ago += 1

# Again, we estimate the probability as relative frequency

prob_years_ago = count_years_ago / brown_num_bigrams

# now check the probability of "ago" given "years"

prob_years_ago / prob_years

# This will come out the same as when we compute:

# out of all occurrences of "years _", what percentage is "years ago"?

count_years_something = 0

for word1, word2, in nltk.bigrams(brown.words()):

if word1 == "years":

count_years_something += 1

count_years_ago / count_years_something