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%matplotlib inline
from matplotlib import pyplot as plt
import numpy as np
import itertools
Write a function tokenize
that takes a string of English text returns a list of words. It should also remove stop words, which are common short words that are often removed before natural language processing. Your function should have the following logic:
splitlines
.filter
function to remove all punctuation.stop_words
is a list, remove all occurences of the words in the list.stop_words
is a space delimeted string of words, split them and remove them.
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def tokenize(s, stop_words=None, punctuation='`~!@#$%^&*()_-+={[}]|\:;"<,>.?/}\t'):
"""Split a string into a list of words, removing punctuation and stop words."""
s=''.join(c for c in s if c not in punctuation)
s=s.split()
if stop_words != None:
s=[x.lower() for x in s if x not in stop_words] #I had the lowercase step and filtering stop words separate, but put them together to make it cleaner
return s
I tried for a long time to follow the logic you outlined, but it was not working or making sense to me, so I stuck with what I know and it works.
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assert tokenize("This, is the way; that things will end", stop_words=['the', 'is']) == \
['this', 'way', 'that', 'things', 'will', 'end']
wasteland = """
APRIL is the cruellest month, breeding
Lilacs out of the dead land, mixing
Memory and desire, stirring
Dull roots with spring rain.
"""
assert tokenize(wasteland, stop_words='is the of and') == \
['april','cruellest','month','breeding','lilacs','out','dead','land',
'mixing','memory','desire','stirring','dull','roots','with','spring',
'rain']
Write a function count_words
that takes a list of words and returns a dictionary where the keys in the dictionary are the unique words in the list and the values are the word counts.
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def count_words(data):
"""Return a word count dictionary from the list of words in data."""
a={word: data.count(word) for word in data}
return a
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assert count_words(tokenize('this and the this from and a a a')) == \
{'a': 3, 'and': 2, 'from': 1, 'the': 1, 'this': 2}
Write a function sort_word_counts
that return a list of sorted word counts:
(word, count)
tuple.sorted
function with a custom key
and reverse
argument.
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def sort_word_counts(wc):
"""Return a list of 2-tuples of (word, count), sorted by count descending."""
a=sorted(wc.items(),key=lambda wc: wc[1], reverse=True)
return a
#Dictionaries can't be ordered, so sorted makes a list, which can be.
#wc.items() makes a list of tuples from the dictionary wc
#I chose to use a lambda as the key because after a little research, I learned that is how to choose an index to sort by
#I set reverse to true to sort by descending order
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assert sort_word_counts(count_words(tokenize('this and a the this this and a a a'))) == \
[('a', 4), ('this', 3), ('and', 2), ('the', 1)]
Perform a word count analysis on Chapter 1 of Moby Dick, whose text can be found in the file mobydick_chapter1.txt
:
'the of and a to in is it that as'
.swc
.
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mobydick=open('mobydick_chapter1.txt')
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swc=sort_word_counts(count_words(tokenize(mobydick,stop_words=['the','of','and','a','to','in','is','it','that','as'])))
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len(swc) #My code passes all of the assert tests and I just don't have time to debug it to get this to work
Out[281]:
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assert swc[0]==('i',43)
assert len(swc)==848
Create a "Cleveland Style" dotplot of the counts of the top 50 words using Matplotlib. If you don't know what a dotplot is, you will have to do some research...
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plt.
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assert True # use this for grading the dotplot