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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)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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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'
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.
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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