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- from __future__ import print_function
- import string
- import re
- import unittest
-
- # You want to build a word cloud, an infographic where the size of a
- # word corresponds to how often it appears in the body of text.
-
- # To do this, you'll need data. Write code that takes a long string and
- # builds its word cloud data in a dictionary, where the keys are
- # words and the values are the number of times the words occurred.
-
- # Think about capitalized words. For example, look at these sentences:
-
- # 'After beating the eggs, Dana read the next step:'
- # 'Add milk and eggs, then add flour and sugar.'
-
- # What do we want to do with "After", "Dana", and "add"? In this
- # example, your final dictionary should include one "Add" or "add" with
- # a value of 22. Make reasonable (not necessarily perfect) decisions
- # about cases like "After" and "Dana".
-
- # Assume the input will only contain words and standard punctuation.
-
-
- # ignore all punctuation except sentence enders . ! and ?
-
- # lowercase any word that starts a sentence IFF it is also in the
- # corpus of words as a lowercase word.
-
- # How?
-
- # split corpus into sentences (strings ending with a . ? or !)
- # strip sentences into words
- # push all words into a case sensitive, word frequency counting, dict
- # scan the dict
- # if a cap word is in the dict as cap and downcase then downcase
- # return dict
-
- # we do make two passes through the input stream, but if this were a
- # real problem, I'd use a lexing and parsing library to implement a
- # real world's problem's requirements.
-
-
- def word_cloud(input):
- """map string of words into a dict of word frequencies"""
-
- sentence_enders = r"\.|!|\?"
- sentences = re.split(sentence_enders, input)
-
- freq = {}
- for sentence in sentences:
- words = re.split(r"[^a-zA-Z0-9-]+", sentence)
- for word in words:
- count = freq.get(word, 0)
- freq[word] = count + 1
-
- def is_cap(word):
- ch = word[0:1]
- return ch in string.uppercase
-
- for word, count in freq.items():
- if is_cap(word) and word.lower() in freq:
- count = freq[word]
- freq[word.lower()] += count
- del freq[word]
-
- return freq
-
-
- class TestWordCloud(unittest.TestCase):
-
- def test_examples(self):
- """test the given example"""
- test = 'After beating the eggs, Dana read the next step:' + \
- 'Add milk and eggs, then add flour and sugar-free diet coke.'
- soln = {
- 'After': 1,
- 'Dana': 1,
- 'add': 2,
- 'and': 2,
- 'beating': 1,
- 'coke': 1,
- 'diet': 1,
- 'eggs': 2,
- 'flour': 1,
- 'milk': 1,
- 'next': 1,
- 'read': 1,
- 'step': 1,
- 'sugar-free': 1,
- 'the': 2,
- 'then': 1,
- }
-
- cloud = word_cloud(test)
- self.assertDictEqual(soln, cloud)
-
- def test_more_examples(self):
- "test some additional examples"
-
- tests = [
- ["We came, we saw, we conquered...then we ate Bill's "
- "(Mille-Feuille) cake."
- "The bill came to five dollars.",
- {
- 'Mille-Feuille': 1,
- 'The': 1,
- 'ate': 1,
- 'bill': 2,
- 'cake': 1,
- 'came': 2,
- 'conquered': 1,
- 'dollars': 1,
- 'five': 1,
- 's': 1,
- 'saw': 1,
- 'then': 1,
- 'to': 1,
- 'we': 4
- }
- ]
- ]
-
- for test, soln in tests:
- cloud = word_cloud(test)
- self.assertDictEqual(soln, cloud)
-
-
- if __name__ == "__main__":
- unittest.main()
- suite = unittest.TestLoader().loadTestsFromTestCase(TestWordCloud)
- unittest.TextTestRunner(verbosity=2).run(suite)
我很好地理解你的逻辑，你首先是在字典中添加所有单词。然后，您正在做大量的工作来检查字典，并将以大写字母开出的单词移动到小写字母中的同一个单词，并删除以大写字母开出的单词。
我的建议是：首先降低所有键的低位。然后字典不需要后处理。 - freq = {}
- for sentence in sentences:
- words = re.split(r"[^a-zA-Z0-9-]+", sentence)
- for word in words:
- count = freq.get(word.lower(), 0)
- freq[word.lower()] = count + 1
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发表于 2021-5-21 04:08:40
