As part of problem C, we're going to randomly select words to create sentences. However, we don't want it to be uniformly random. We will want to influence the random selection by having some words be more likely to appear than others. This is called weighted random selection. In this problem, we will build towards that goal by creating a function that does a weighted random selection from the keys of a dictionary, using the values as the weights. Create a function weighted_choice that takes as a parameter a dictionary of words and their corresponding counts. The function should randomly choose one of the words, using the counts as weights in the selection. For example, if I pass the dictionary {'green': 1, 'eggs': 1} To this function, it should return "green" half the time and "eggs" half the time. If I pass the dictionary {'green': 1, 'eggs': 3, 'ham': 2} It should return "green" one-sixth of the time, "eggs" half (3/6) of the time, and “ham” one-third (2/6) of the time. There are a number of ways to accomplish this task. Many solutions require you to iterate over the given dictionary to add up the counts to get the total of the counts before using a function from the random module to get a random number in the range from 1 to the total. Then, you can use that number to figure out which word it corresponds to and return the appropriate word. There's also a simple method that utilizes making a list with x copies of each key (where x is the value for said key), and then using random.choice. Examples (random, so won't necessarily match, but make sure that both come up about 50% of the time): >>> weighted_choice({'green': 1, 'eggs': 1}) 'green' >>> weighted_choice({'green': 1, 'eggs': 1}) 'eggs' >>> weighted_choice({ 'green': 1, 'eggs': 1}) 'eggs' 1
As part of problem C, we're going to randomly select words to create sentences. However, we don't want it to be uniformly random. We will want to influence the random selection by having some words be more likely to appear than others. This is called weighted random selection. In this problem, we will build towards that goal by creating a function that does a weighted random selection from the keys of a dictionary, using the values as the weights. Create a function weighted_choice that takes as a parameter a dictionary of words and their corresponding counts. The function should randomly choose one of the words, using the counts as weights in the selection. For example, if I pass the dictionary {'green': 1, 'eggs': 1} To this function, it should return "green" half the time and "eggs" half the time. If I pass the dictionary {'green': 1, 'eggs': 3, 'ham': 2} It should return "green" one-sixth of the time, "eggs" half (3/6) of the time, and “ham” one-third (2/6) of the time. There are a number of ways to accomplish this task. Many solutions require you to iterate over the given dictionary to add up the counts to get the total of the counts before using a function from the random module to get a random number in the range from 1 to the total. Then, you can use that number to figure out which word it corresponds to and return the appropriate word. There's also a simple method that utilizes making a list with x copies of each key (where x is the value for said key), and then using random.choice. Examples (random, so won't necessarily match, but make sure that both come up about 50% of the time): >>> weighted_choice({'green': 1, 'eggs': 1}) 'green' >>> weighted_choice({'green': 1, 'eggs': 1}) 'eggs' >>> weighted_choice({ 'green': 1, 'eggs': 1}) 'eggs' 1
Computer Networking: A Top-Down Approach (7th Edition)
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Chapter1: Computer Networks And The Internet
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In python
Transcribed Image Text:As part of problem C, we're going to randomly select
words to create sentences. However, we don't want it to
be uniformly random. We will want to influence the
random selection by having some words be more likely to
appear than others. This is called weighted random
selection. In this problem, we will build towards that goal
by creating a function that does a weighted random
selection from the keys of a dictionary, using the values
as the weights.
Create a function weighted_choice that takes as a
parameter a dictionary of words and their corresponding
counts. The function should randomly choose one of the
words, using the counts as weights in the selection. For
example, if I pass the dictionary
{'green': 1, 'eggs': 1}
To this function, it should return "green" half the time and
"eggs" half the time. If I pass the dictionary
{'green': 1, 'eggs': 3, 'ham': 2}
It should return "green" one-sixth of the time, "eggs" half
(3/6) of the time, and “ham” one-third (2/6) of the time.
There are a number of ways to accomplish this task.
Many solutions require you to iterate over the given
dictionary to add up the counts to get the total of the
counts before using a function from the random module
to get a random number in the range from 1 to the total.
hen, you can use that number to figure out which word
it corresponds to and return the appropriate word.
There's also a simple method that utilizes making a list
with x copies of each key (where x is the value for said
key), and then using random.choice.
Examples (random, so won't necessarily match, but
make sure that both come up about 50% of the time):
>>> weighted_choice({'green': 1, 'eggs':
1})
'green'
>>> weighted_choice({'green': 1, 'eggs':
1})
'eggs'
>>> weighted_choice({'green': 1, 'eggs':
1})
'eggs'
![Note that testing this function is a little tricky because it
relies on randomness. You will not see the same output
every time! We can control for this a bit by running the
function many times, to ensure that on average, the
probabilities roughly match the outcome. For example, if
we run
weighted_choice({'green': 1, 'eggs': 3,
'ham': 2})
600 times, then we would expect to get 'green' about 100
times, 'eggs' about 300 times, and 'ham' about 200 times.
[weighted_choice({'green':
>>> results =
1, 'eggs': 3, 'ham':2}) for i in
range (600)]
>>> results.count('green')
99
>>> results.count('eggs')
296
>>> results.count('ham')
205
If your results are consistently off by more than 20 or so,
you should probably check your algorithm.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F77fc5231-144e-4658-a4e4-843a776de72b%2F3d1b2c3f-494a-449d-be55-1e42216d1dd5%2F1k8rcsg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Note that testing this function is a little tricky because it
relies on randomness. You will not see the same output
every time! We can control for this a bit by running the
function many times, to ensure that on average, the
probabilities roughly match the outcome. For example, if
we run
weighted_choice({'green': 1, 'eggs': 3,
'ham': 2})
600 times, then we would expect to get 'green' about 100
times, 'eggs' about 300 times, and 'ham' about 200 times.
[weighted_choice({'green':
>>> results =
1, 'eggs': 3, 'ham':2}) for i in
range (600)]
>>> results.count('green')
99
>>> results.count('eggs')
296
>>> results.count('ham')
205
If your results are consistently off by more than 20 or so,
you should probably check your algorithm.
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