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changed default to minimum for word_frequency

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This commit is contained in:
Joshua Chin 2015-07-07 15:03:26 -04:00
parent 0b25caaf24
commit 9aa773aa2b
2 changed files with 5 additions and 5 deletions
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2
tests/test.py
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@ -46,7 +46,7 @@ def test_twitter():
def test_defaults():
eq_(word_frequency('esquivalience', 'en'), 0)
eq_(word_frequency('esquivalience', 'en', default=1e-6), 1e-6)
eq_(word_frequency('esquivalience', 'en', minimum=1e-6), 1e-6)
def test_most_common_words():

8
wordfreq/__init__.py
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@ -243,7 +243,7 @@ def half_harmonic_mean(a, b):
@lru_cache(maxsize=CACHE_SIZE)
def word_frequency(word, lang, wordlist='combined', default=0.):
def word_frequency(word, lang, wordlist='combined', minimum=0.):
"""
Get the frequency of `word` in the language with code `lang`, from the
specified `wordlist`. The default wordlist is 'combined', built from
@ -261,7 +261,7 @@ def word_frequency(word, lang, wordlist='combined', default=0.):
Words that we believe occur at least once per million tokens, based on
the average of these lists, will appear in the word frequency list.
If you look up a word that's not in the list, you'll get the `default`
If you look up a word that's not in the list, you'll get the `minimum`
value, which itself defaults to 0.
If a word decomposes into multiple tokens, we'll return a smoothed estimate
@ -273,12 +273,12 @@ def word_frequency(word, lang, wordlist='combined', default=0.):
tokens = tokenize(word, lang)
if len(tokens) == 0:
return default
return minimum
for token in tokens:
if token not in freqs:
# If any word is missing, just return the default value
return default
return minimum
value = freqs[token]
if combined_value is None:
combined_value = value