In python Note that testing this function is a little tricky because itrelies on randomness. You will not see the same outputevery time! We can control for this a bit by running thefunction many times, to ensure that on average, theprobabilities roughly match the outcome. For example, ifwe runweighted_choice({'green': 1, 'eggs': 3,'ham': 2})600 times, then we would expect to get 'green' about 100times, 'eggs' about 300 times, and 'ham' about 200 times.>>> results =1, 'eggs': 3, 'ham':2}) for i inrange (600)]>>> results.count('green')99>>> results.count('eggs')296>>> results.count('ham')205[weighted_choice({ 'green':If your results are consistently off by more than 20 or so,you should probably check your algorithm. As part of problem C, we're going to randomly selectwords to create sentences. However, we don't want it tobe uniformly random. We will want to influence therandom selection by having some words be more likely toappear than others. This is called weighted randomselection. In this problem, we will build towards that goalby creating a function that does a weighted randomselection from the keys of a dictionary, using the valuesas the weights.Create a function weighted_choice that takes as aparameter a dictionary of words and their correspondingcounts. The function should randomly choose one of thewords, using the counts as weights in the selection. Forexample, 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 givendictionary to add up the counts to get the total of thecounts before using a function from the random moduleto get a random number in the range from 1 to the total.Then, you can use that number to figure out which wordit corresponds to and return the appropriate word.There's also a simple method that utilizes making a listwith x copies of each key (where x is the value for saidkey), and then using random.choice.Examples (random, so won't necessarily match, butmake sure that both come up about 50% of the time):>>> weighted_choice({'green': 1, 'eggs':1})'green'>>> weighted_choice({'green': 1,1})eggs':'eggs'>>> weighted_choice({'green': 1, 'eggs':1})'eggs'

1) Below is program that creates weighted_choice that takes as a parameter dictionary of words and…

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. >>> results = 1, 'eggs': 3, 'ham':2}) for i in range (600)] >>> results.count('green') 99 >>> results.count('eggs') 296 [weighted_choice({ 'green': >>> results.count('ham') 205 If your results are consistently off by more than 20 or so, you should probably check your algorithm.

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. >>> results = 1, 'eggs': 3, 'ham':2}) for i in range (600)] >>> results.count('green') 99 >>> results.count('eggs') 296 [weighted_choice({ 'green': >>> results.count('ham') 205 If your results are consistently off by more than 20 or so, you should probably check your algorithm.

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Question

In python

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.
>>> results =
1, 'eggs': 3, 'ham':2}) for i in
range (600)]
>>> results.count('green')
99
>>> results.count('eggs')
296
>>> results.count('ham')
205
[weighted_choice({ 'green':
If your results are consistently off by more than 20 or so,
you should probably check your algorithm.

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,
1})
eggs':
'eggs'
>>> weighted_choice({'green': 1, 'eggs':
1})
'eggs'

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