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Week 08 Laboratory Sample Solutions
Objectives
- Proficiency at text processing in Python.
- Understanding multi-dimensional dicts.
- Explore a simple machine learning algorithm.
Preparation
Before the lab you should re-read the relevant lecture slides and their accompanying examples.
Getting Started
Set up for the lab by creating a new directory called lab08 and changing to this directory.
mkdir lab08
cd lab08There are some provided files for this lab which you can fetch with this command:
2041 fetch lab08If you're not working at CSE, you can download the provided files as a zip file or a tar file.
Exercise: How many words in standard input?
In these exercises you will work with a dataset containing sing lyrics.
This dataset contains the lyrics of the songs of 10 well-known artists.
unzip lyrics.zip
Archive: lyrics.zip
creating: lyrics/
inflating: lyrics/David_Bowie.txt
inflating: lyrics/Adele.txt
inflating: lyrics/Metallica.txt
inflating: lyrics/Rage_Against_The_Machine.txt
inflating: lyrics/Taylor_Swift.txt
inflating: lyrics/Keith_Urban.txt
inflating: lyrics/Ed_Sheeran.txt
inflating: lyrics/Justin_Bieber.txt
inflating: lyrics/Rihanna.txt
inflating: lyrics/Leonard_Cohen.txt
inflating: song0.txt
inflating: song1.txt
inflating: song2.txt
inflating: song3.txt
inflating: song4.txtThe lyrics for each song have been re-ordered to avoid copyright concerns.
The dataset also contains lyrics from 5 songs where we don't know the artists.
cat song0.txt
I've made up my mind, Don't need to think it over,
If I'm wrong I am right,
Don't need to look no further,
This ain't lust,
I know this is love but,
If I tell the world,
I'll never say enough,
Cause it was not said to you,
And that's exactly what I need to do,
If I'm in love with you,
cat song1.txt
Come Mr. DJ song pon de replay
Come Mr. DJ won't you turn the music up
All the gal pon the dance floor wantin' some more what
Come Mr. DJ won't you turn the music up
cat song2.txt
And they say
She's in the class A team
Stuck in her daydream
Been this way since eighteen
But lately her face seems
Slowly sinking, wasting
Crumbling like pastries
cat song3.txt
Ooh whoa, ooh whoa, ooh whoa You know you love me, you know you care
Just shout whenever and I'll be there
You are my love, you are my heart
And we will never, ever, ever be apart
Are we an item? Girl quit playin'
We're just friends, what are you sayin'
Said there's another, look right in my eyes
My first love, broke my heart for the first time
And I was like baby, baby, baby oh
Like baby, baby, baby no
Like baby, baby, baby oh
I thought you'd always be mine (Mine)
Baby, baby, baby oh
Like baby, baby, baby no
Like baby, baby, baby ooh
I thought you'd always be mine
Oh for you, I would have done whatever
And I just can't believe we ain't together
And I wanna play it cool
But I'm losin' you
I'll buy you anything
I'll buy you any ring
And I'm in pieces, baby fix me
And just shake me, til you wake me from this bad dream
I'm goin' down, down, down, down
And I can't believe my first love won't be around
cat song4.txt
The birds they sang At the break of day
Start again
I heard them say
Don't dwell on what
Has passed away
Or what is yet to be.
Ah the wars they will
Be fought again
The holy dove
She will be caught again
Bought and sold
And bought again
The dove is never free.
Ring the bells that still can ring
Forget your perfect offering
There is a crack in everything
That's how the light gets in.Each is from one of the artists in the dataset but they are not from a song in the dataset.
As a first step in this analysis, write a Python script total_words.py which counts the total number of words in its stdin.
For the purposes of this program (and the following programs) we will define a word to be a maximal, non-empty, contiguous, sequence of alphabetic characters ([a-zA-Z]).
Any characters other than [a-zA-Z] separate words.
So for example the phrase "The soul's desire" contains 4 words: ("The", "soul", "s", "desire")
./total_words.py < lyrics/Justin_Bieber.txt
46589 words
./total_words.py < lyrics/Metallica.txt
38096 words
./total_words.py < lyrics/Rihanna.txt
53157 wordsIf your word counts are a little too high, you might be counting empty strings.
A word is defined for these exercises to be a maximal, non-empty, contiguous, sequence of alphabetic characters (
[a-zA-Z]).You can assume your input is only ASCII.
Your answer must be Python only. You can not use other languages such as Shell, Perl or C.
You may not run external programs.
When you think your program is working, you can use autotest to run some simple automated tests:
2041 autotest total_wordsWhen you are finished working on this exercise, you must submit your work by running give:
give cs2041 lab08_total_words total_words.pybefore Tuesday 09 April 12:00 (midday) (2024-04-09 12:00:00) to obtain the marks for this lab exercise.
Sample solution for total_words.py
#!/usr/bin/env python3
"""
count words in stdin
written by andrew@unsw.edu.au for COMP(2041|9044)
"""
import re, sys
word_count = 0
for line in sys.stdin:
line_words = re.findall(r"[a-zA-Z]+", line)
for word in line_words:
word_count += 1
print(word_count, "words")
Alternative solution for total_words.py
#!/usr/bin/env python3
"""
count words in stdin
written by andrew@unsw.edu.au for COMP(2041|9044)
"""
import re, sys
word_count = 0
for line in sys.stdin:
line_words = re.findall(r"[a-zA-Z]+", line)
line_word_count = len(line_words)
word_count += line_word_count
print(word_count, "words")
Alternative solution for total_words.py
#!/usr/bin/env python3
"""
count words in stdin
written by andrew@unsw.edu.au for COMP(2041|9044)
"""
import re
import sys
all_input = sys.stdin.read()
words = re.findall(r"[a-zA-Z]+", all_input)
word_count = len(words)
print(word_count, "words")
Alternative solution for total_words.py
#!/usr/bin/env python3
"""
count words in stdin
written by d.brotherston@unsw.edu.au for COMP(2041|9044)
"""
from sys import stdin
from re import split
def main():
input = stdin.read()
words = split(r'[^a-zA-Z]', input)
words = list(filter(None, words)) # remove empty strings
print(f"{len(words)} words")
if __name__ == "__main__":
main()
Alternative solution for total_words.py
#!/usr/bin/env python3
"""
count words in stdin
written by d.brotherston@unsw.edu.au for COMP(2041|9044)
"""
from sys import stdin
from re import split
print(f"{len(list(filter(None, split(r'[^a-zA-Z]', stdin.read()))))} words")
Exercise: How many times does a word occur in standard input
Write a Python script count_word.py that counts the number of times a specified word is found in its stdin
The word you should count will be specified as a command line argument.
Your program should ignore the case of words.
./count_word.py death < lyrics/Metallica.txt
death occurred 69 times
./count_word.py death < lyrics/Justin_Bieber.txt
death occurred 0 times
./count_word.py love < lyrics/Ed_Sheeran.txt
love occurred 218 times
./count_word.py love < lyrics/Rage_Against_The_Machine.txt
love occurred 4 timesStart with your code from the previous activity.
A word is defined for these exercises to be a maximal, non-empty, contiguous, sequence of alphabetic characters (
[a-zA-Z]).You can assume your input is only ASCII.
Your answer must be Python only. You can not use other languages such as Shell, Perl or C.
You may not run external programs.
When you think your program is working, you can use autotest to run some simple automated tests:
2041 autotest count_wordWhen you are finished working on this exercise, you must submit your work by running give:
give cs2041 lab08_count_word count_word.pybefore Tuesday 09 April 12:00 (midday) (2024-04-09 12:00:00) to obtain the marks for this lab exercise.
Sample solution for count_word.py
#!/usr/bin/env python3
"""
read stdin counting occurrences of word given as command-line argument
written by andrewt@unsw.edu.au for COMP(2041|9044)
"""
import re, sys
if len(sys.argv) != 2:
print(f"Usage: {argv[0]} <word>")
sys.exit(1)
specified_word = sys.argv[1].lower()
count = 0
for line in sys.stdin:
line = line.lower()
words = re.findall(r'[a-z]+', line)
for word in words:
if word == specified_word:
count += 1
print(specified_word, "occurred", count, "times")
Alternative solution for count_word.py
#!/usr/bin/env python3
"""
read stdin counting occurrences of word given as command-line argument
written by andrewt@unsw.edu.au for COMP(2041|9044)
"""
import re
import sys
specified_word = sys.argv[1].lower()
specified_word_lowercase = specified_word.lower()
all_input = sys.stdin.read()
all_input_lowercase = all_input.lower()
words = re.findall(r"[a-z]+", all_input_lowercase)
count = words.count(specified_word)
print(specified_word, "occurred", count, "times")
Alternative solution for count_word.py
#!/usr/bin/env python3
"""
read stdin counting occurrences of word given as command-line argument
written by d.brotherston@unsw.edu.au for COMP(2041|9044)
"""
from sys import argv, stdin
from re import split
from collections import Counter
def toWords(input):
return list(filter(None, split(r'[^a-zA-Z]', input)))
def main():
if len(argv) < 2:
print(f"Usage: {argv[0]} <word>")
word = argv[1].lower()
words = toWords(stdin.read())
words = list(map(str.lower, words)) # ignore case
counter = Counter(words)
print(f"{word} occurred {counter[word]} times")
if __name__ == "__main__":
main()
Exercise: Do you use that word often?
Write a Python script frequency.py thar prints the frequency with which each artist uses a word specified as an argument.
So if Justin Bieber uses the word "love" 493 times in the 46583 words of his songs, then its frequency is 493/46583 = 0.0105832599875491.
./frequency.py love
165/ 16359 = 0.010086191 Adele
189/ 34080 = 0.005545775 David Bowie
218/ 18207 = 0.011973417 Ed Sheeran
493/ 46589 = 0.010581897 Justin Bieber
217/ 27016 = 0.008032277 Keith Urban
212/ 26192 = 0.008094075 Leonard Cohen
57/ 38096 = 0.001496220 Metallica
4/ 18985 = 0.000210693 Rage Against The Machine
494/ 53157 = 0.009293226 Rihanna
89/ 26188 = 0.003398503 Taylor Swift
./frequency.py death
1/ 16359 = 0.000061128 Adele
9/ 34080 = 0.000264085 David Bowie
3/ 18207 = 0.000164772 Ed Sheeran
0/ 46589 = 0.000000000 Justin Bieber
1/ 27016 = 0.000037015 Keith Urban
16/ 26192 = 0.000610874 Leonard Cohen
69/ 38096 = 0.001811214 Metallica
23/ 18985 = 0.001211483 Rage Against The Machine
0/ 53157 = 0.000000000 Rihanna
0/ 26188 = 0.000000000 Taylor SwiftMake sure your Python script produces exactly the output above.
Start with your code from the previous activity.
A print like this will produce the correct output format:
program
print(f"{var1:4}/{var2:6} = {var3:.9f} {var4}")Use a dict of dicts indexed by artist then word to store the word counts.
Use the glob module to find all the files that match a glob string.
This loop executes once for each .txt file in the directory lyrics.
program
for file in glob.glob("lyrics/*.txt"):
print(file);A word is defined for these exercises to be a maximal, non-empty, contiguous, sequence of alphabetic characters (
[a-zA-Z]).You can assume your input is only ASCII.
Your answer must be Python only. You can not use other languages such as Shell, Perl or C.
You may not run external programs.
When you think your program is working, you can use autotest to run some simple automated tests:
2041 autotest frequencyWhen you are finished working on this exercise, you must submit your work by running give:
give cs2041 lab08_frequency frequency.pybefore Tuesday 09 April 12:00 (midday) (2024-04-09 12:00:00) to obtain the marks for this lab exercise.
Sample solution for frequency.py
#!/usr/bin/env python3
"""
print frequencies of specified words in lyrics files
implemented using dicts & regex
see identify_artist.py for a better version of this
code decomposed into functions so it readable & maintainable.
written by andrewt@unsw.edu.au for COMP(2041|9044)
"""
import glob
import re
import sys
frequency = {}
for pathname in glob.glob("lyrics/*"):
artist = re.sub(r".*/", "", pathname)
artist = re.sub(r".txt$", "", artist)
artist = re.sub(r"_", " ", artist)
frequency[artist] = {}
with open(pathname, encoding="utf-8") as f:
for line in f:
line = line.lower()
for word in re.findall(r"[a-z]+", line):
if word not in frequency[artist]:
frequency[artist][word] = 0
frequency[artist][word] += 1
for word in sys.argv[1:]:
word = word.lower()
for artist in sorted(frequency):
if word in frequency[artist]:
f = frequency[artist][word]
else:
f = 0
n = sum(frequency[artist].values())
print(f"{f:4}/{n:6} = {f/n:.9f} {artist}")
Alternative solution for frequency.py
#!/usr/bin/env python3
"""
print frequencies of specified words in lyrics files
Implemented using using counters & os.path functions
written by andrewt@unsw.edu.au for COMP(2041|9044)
"""
import collections, glob, os, re, sys
frequency = {}
for pathname in glob.glob("lyrics/*.txt"):
filename = os.path.basename(pathname)
filename_without_extension = os.path.splitext(filename)[0]
artist = filename_without_extension.replace("_", " ")
with open(pathname, encoding="utf-8") as f:
lyrics = f.read().lower()
words = re.findall(r"[a-z]+", lyrics)
frequency[artist] = collections.Counter(words)
for word in sys.argv[1:]:
for artist in sorted(frequency):
f = frequency[artist][word.lower()]
n = sum(frequency[artist].values())
print(f"{f:4}/{n:6} = {f/n:.9f} {artist}")
Alternative solution for frequency.py
#!/usr/bin/env python3
"""
print frequencies of specified words in lyrics files
written by d.brotherston@unsw.edu.au for COMP(2041|9044)
"""
from sys import argv
from re import split
from collections import Counter
from glob import glob
from pathlib import Path
def toWords(input):
return list(filter(None, split(r'[^a-zA-Z]', input)))
def main():
if len(argv) < 2:
print(f"Usage: {argv[0]} <word>")
word = argv[1].lower()
lyric_counts = {}
for filename in sorted(glob('lyrics/*.txt')):
with open(filename) as f:
words = toWords(f.read())
words = list(map(str.lower, words))
lyric_counts[Path(filename).stem] = Counter(words)
for artist, counts in lyric_counts.items():
total_words = sum(counts.values())
frequency = counts[word] / total_words
print(f"{counts[word]:4}/{total_words:6} = {frequency:.9f} {artist.replace('_', ' ')}")
if __name__ == "__main__":
main()
Exercise: When numbers get very small, logarithms are your friend
Now suppose we have the song line "truth is beauty".
Given that David Bowie uses:
the word "truth" with frequency 0.000146714
the word "is" with frequency 0.005897887
the word "beauty" with frequency 0.000264085
we can estimate the probability of Bowie writing the phrase "truth is beauty" as:
program
0.000146714 * 0.005897887 * 0.000264085 = 2.2851343535638401e-10We could similarly estimate probabilities for each of the other 9 artists
and then determine which of the 10 artists is most likely to sing "truth is beauty"
(it's Leonard Cohen).
A sidenote: we are actually making a large simplifying assumption in calculating this probability.
It is often called the bag of words model.
Multiplying probabilities like this quickly leads to very small numbers and may result in arithmetic underflow of our floating point representation.
A common solution to this underflow is instead to work with the log of the numbers.
So instead we will calculate the log of the probability of the phrase. You do this by adding the log of the probabilities of each word.
For example, you calculate the log-probability of Bowie singing the phrase "Truth is beauty." like this:
program
log(0.000146714) + log(0.005897887) + log(0.000264085) = -22.19942610926425Log-probabilities can be used directly to determine the most likely artist, as the artist with the highest log-probability will also have the highest probability.
Another problem is that we might be given a word that an artist has not used in the dataset we have.
You should avoid this when estimating probabilities by adding 1 to the count of occurrences of each word.
So for example we'd estimate the probability of Ed Sheeran using the word fear as (0+1)/18205 and the probability of Metallica using the word fear as (39+1)/38096.
This is a simple version of Additive smoothing.
Write a Python script log_probability.py which given a phrase (sequence of words) as arguments, prints the estimated log of the probability that each artist would use this phrase.
./log_probability.py truth is beauty
-23.11614 Adele
-21.90679 David Bowie
-23.10075 Ed Sheeran
-21.70202 Justin Bieber
-23.45248 Keith Urban
-18.58417 Leonard Cohen
-21.08903 Metallica
-21.98171 Rage Against The Machine
-22.51582 Rihanna
-24.40992 Taylor Swift
./log_probability.py death and taxes
-22.64301 Adele
-22.42756 David Bowie
-21.66227 Ed Sheeran
-25.56650 Justin Bieber
-23.20281 Keith Urban
-20.97467 Leonard Cohen
-20.90589 Metallica
-20.26248 Rage Against The Machine
-25.84396 Rihanna
-23.90310 Taylor SwiftMake sure your output matches the above exactly
Start with your code from the previous activity.
A print like this will produce the correct output format:
program
print(f"{var1:10.5f} {var2}")Use the natural logarithm (base e) -
math.logreturns this by default, if you don't specify a base.A word is defined for these exercises to be a maximal, non-empty, contiguous, sequence of alphabetic characters (
[a-zA-Z]).You can assume your input is only ASCII.
Your answer must be Python only. You can not use other languages such as Shell, Perl or C.
You may not run external programs.
When you think your program is working, you can use autotest to run some simple automated tests:
2041 autotest log_probabilityWhen you are finished working on this exercise, you must submit your work by running give:
give cs2041 lab08_log_probability log_probability.pybefore Tuesday 09 April 12:00 (midday) (2024-04-09 12:00:00) to obtain the marks for this lab exercise.
Sample solution for log_probability.py
#!/usr/bin/env python3
"""
calculate log probability of an artist using a phrase
concise unreadable/unmaintainable Perl-like implementation
see identify_artist for this code decomposed into readable/maintainable functions
written by andrewt@unsw.edu.au for COMP(2041|9044)
"""
import collections, glob, math, os, re, sys
frequency = {}
for pathname in glob.glob("lyrics/*.txt"):
filename = os.path.basename(pathname)
filename_without_extension = os.path.splitext(filename)[0]
artist = filename_without_extension.replace("_", " ")
with open(pathname, encoding="utf-8") as f:
lyrics = f.read().lower()
words = re.findall(r"[a-z]+", lyrics)
frequency[artist] = collections.Counter(words)
for artist in sorted(frequency):
log_probability = 0
for word in sys.argv[1:]:
word_count = frequency[artist][word.lower()]
total_words = sum(frequency[artist].values())
log_probability += math.log((word_count + 1) / total_words)
print(f"{log_probability:10.5f} {artist}")
Alternative solution for log_probability.py
#!/usr/bin/env python3
"""
calculate log probability of an artist using a phrase
written by d.brotherston@unsw.edu.au for COMP(2041|9044)
"""
from sys import argv
from re import split
from collections import Counter
from glob import glob
from pathlib import Path
from math import log
def toWords(input):
return list(filter(None, split(r'[^a-zA-Z]', input)))
def main():
if len(argv) < 2:
print(f"Usage: {argv[0]} <word..>")
lyric_counts = {}
for filename in sorted(glob('lyrics/*.txt')):
with open(filename) as f:
words = toWords(f.read())
words = list(map(str.lower, words))
lyric_counts[Path(filename).stem] = Counter(words)
for artist, counts in lyric_counts.items():
frequency = 0
for word in argv[1:]:
word = word.lower()
total_words = sum(counts.values())
frequency += log((counts[word] + 1) / total_words)
print(f"{frequency:10.5f} {artist.replace('_', ' ')}")
if __name__ == "__main__":
main()
Exercise: Who sang those words?
Write a Python script identify_artist.py that given 1 or more files (each containing part of a song), prints the most likely artist to have sung those words.
For each file given as argument, you should go through all artists and for each calculate the log-probability that the artist sung those words.
You calculate the log-probability that the artist sung the words in the file, by for each word in the file calculating the log-probability of that artist using that word, and summing all the the log-probabilities.
You should print the artist with the highest log-probability.
Your program should produce exactly this output:
./identify_artist.py song?.txt
song0.txt most resembles the work of Adele (log-probability=-352.4)
song1.txt most resembles the work of Rihanna (log-probability=-254.9)
song2.txt most resembles the work of Ed Sheeran (log-probability=-206.6)
song3.txt most resembles the work of Justin Bieber (log-probability=-1089.8)
song4.txt most resembles the work of Leonard Cohen (log-probability=-493.8)If a word is used is used multiplied times in a file, its log-probablity should be added multiple times.
You do not need to use glob. song?.txt in the above example is expanded by the Shell. The filenames are passed as separate argument in sys.argv.
Start with your code from the previous activity.
A word is defined for these exercises to be a maximal, non-empty, contiguous, sequence of alphabetic characters (
[a-zA-Z]).You can assume your input is only ASCII.
Your answer must be Python only. You can not use other languages such as Shell, Perl or C.
You may not run external programs.
When you think your program is working, you can use autotest to run some simple automated tests:
2041 autotest identify_artistWhen you are finished working on this exercise, you must submit your work by running give:
give cs2041 lab08_identify_artist identify_artist.pybefore Tuesday 09 April 12:00 (midday) (2024-04-09 12:00:00) to obtain the marks for this lab exercise.
Sample solution for identify_artist.py
#!/usr/bin/env python3
"""
identify artists most likely to have sung lyrics using the "bag of words" model
written by andrewt@unsw.edu.au for COMP(2041|9044)
"""
import collections
import glob
import math
import os
import re
import sys
def main():
"""for each file containing lyrics given as command-line arguments
print the most likely artist to have sung these lyrics"""
artist_word_frequency = read_lyrics()
for pathname in sys.argv[1:]:
log_probability, artist = identify_artist(pathname, artist_word_frequency)
print(
f"{pathname} most resembles the work of {artist} (log-probability={log_probability:.1f})"
)
def identify_artist(pathname, artist_word_frequency):
"""given a file containing lyrics and a dict of word counts for artists
returns: tuple of log-probability and most-likely artist
"""
words = read_words(pathname)
probability_artists = []
for (artist, word_frequency) in artist_word_frequency.items():
log_prob = log_probability_words(words, word_frequency)
probability_artists.append((log_prob, artist))
return max(probability_artists)
def log_probability_words(words, word_frequency):
"""returns summed log probablity of words"""
n_words = sum(word_frequency.values())
return sum(log_probability_word(word, word_frequency, n_words) for word in words)
def log_probability_word(word, word_frequency, n_words):
"""returns log probablity for a single word"""
word = word.lower()
return math.log((word_frequency[word] + 1) / n_words)
def read_lyrics():
"""read song lyrics from sub-directory lyrics
returns: dict of word counts for each artist
"""
artist_word_frequency = {}
for pathname in glob.glob("lyrics/*.txt"):
words = read_words(pathname)
artist = extract_artist(pathname)
artist_word_frequency[artist] = collections.Counter(words)
return artist_word_frequency
def read_words(pathname):
"""read pathname and return a list of words it contains"""
with open(pathname, encoding="utf-8") as f:
lyrics = f.read().lower()
words = re.findall(r"[a-z]+", lyrics)
return words
def extract_artist(pathname):
"""given a pathname return the corresponding artist
e.g give "lyric/David_Bowie.txt" return "David Bowie"
"""
filename = os.path.basename(pathname)
filename_without_extension = os.path.splitext(filename)[0]
artist = filename_without_extension.replace("_", " ")
return artist
if __name__ == "__main__":
main()
Alternative solution for identify_artist.py
#!/usr/bin/env python3
"""
identify artists most likely to have sung lyrics using the "bag of words" model
concise unreadable/unmaintainable Perl-like implementation (for comparison)
written by andrewt@unsw.edu.au for COMP(2041|9044)
"""
import collections, glob, math, os, re, sys
alp = {}
for file in glob.glob("lyrics/*.txt"):
with open(file, encoding="utf-8") as f:
words = re.findall(r"[a-zA-Z]+", f.read().lower())
artist = os.path.splitext(os.path.basename(file))[0].replace("_", " ")
n_words = len(words)
alp[artist] = collections.defaultdict(lambda x=math.log(1 / n_words): x)
for word, count in collections.Counter(words).items():
alp[artist][word] = math.log((count + 1) / n_words)
for file in sys.argv[1:]:
with open(file, encoding="utf-8") as f:
words = re.findall(r"[a-zA-Z]+", f.read().lower())
p, artist = max((sum(l[w] for w in words), a) for (a, l) in alp.items())
print(f"{file} most resembles the work of {artist} (log-probability={p:.1f})")
Alternative solution for identify_artist.py
#!/usr/bin/env python3
"""
identify artists most likely to have sung lyrics using the "bag of words" model
written by d.brotherston@unsw.edu.au for COMP(2041|9044)
"""
from sys import argv
from re import split
from collections import Counter
from glob import glob
from pathlib import Path
from math import log
def toWords(input):
return list(filter(None, split(r'[^a-zA-Z]', input)))
def main():
if len(argv) < 2:
print(f"Usage: {argv[0]} <word..>")
lyric_counts = {}
for filename in sorted(glob('lyrics/*.txt')):
with open(filename) as f:
words = toWords(f.read())
words = list(map(str.lower, words))
lyric_counts[Path(filename).stem] = Counter(words)
for file in argv[1:]:
lyric_frequency = {}
with open(file) as f:
words = toWords(f.read())
for artist, counts in lyric_counts.items():
frequency = 0
for word in words:
word = word.lower()
total_words = sum(counts.values())
frequency += log((counts[word] + 1) / total_words)
lyric_frequency[artist] = frequency
for artist, frequency in sorted(lyric_frequency.items(), key=lambda x: x[1], reverse=True):
print(f"{file} most resembles the work of {artist.replace('_', ' ')} (log-probability={frequency:.1f})")
break
if __name__ == "__main__":
main()
Submission
When you are finished each exercises make sure you submit your work by running give.
You can run give multiple times. Only your last submission will be marked.
Don't submit any exercises you haven't attempted.
If you are working at home, you may find it more convenient to upload your work via give's web interface.
Remember you have until Week 9 Tuesday 12:00:00 (midday) to submit your work.
You cannot obtain marks by e-mailing your code to tutors or lecturers.
You check the files you have submitted here.
Automarking will be run by the lecturer several days after the submission deadline, using test cases different to those autotest runs for you. (Hint: do your own testing as well as running autotest.)
After automarking is run by the lecturer you can view your results here. The resulting mark will also be available via give's web interface.
Lab Marks
When all components of a lab are automarked you should be able to view the the marks via give's web interface or by running this command on a CSE machine:
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