python - 语料库中每个文本的平均句子长度(python3 和 nltk)

标签 python nlp nltk average iterable

我正在分析 NLTK 包中的就职地址语料库,作为 Python 编程类(class)简介的一部分。我想找出语料库中每个文本的平均句子长度(以便以后可以比较它们),但我似乎被困在这里。

我创建了这个函数:

def averageSentence(text):
    sents = inaugural.sents(fileids=['fileid_here.txt']  
    avg = sum(len(word) for word in sents) / len(sents)  
    print(avg)

(如果我是正确的)应该给我单个文本的平均句子长度。现在,我知道我需要一个for循环。难道我不应该能够使用我刚刚定义的这个函数创建一个相对简单且直接的 for 循环吗?这非常令人沮丧。

编辑:这是我已经走了多远:

for fileid in inaugural.fileids():
    avg_sents = averageSentence(fileid)
    print = sum(avg_sents) / avg_sents

最佳答案

尝试:

>>> from __future__ import division
>>> from nltk.corpus import inaugural
>>> total_lens = 0
>>> for i, sent in enumerate(inaugural.sents()):
...     total_lens += len(sent)
... 
>>> total_lens 
145735
>>> i
4867
>>> avg_sent_len = total_lens / i
>>> avg_sent_len
29.943497020752
>>> avg_sent_len = total_lens / (i+1)
>>> avg_sent_len
29.9373459326212

请注意,当分母足够大时 +1 并不那么重要。

<小时/>

所有文本的微平均句子长度

以下代码是一行代码,但不鼓励这样做,因为您可能已经实现了生成器两次:

>>> sum(len(sent) for sent in inaugural.sents()) / len(inaugural.sents())
29.9373459326212
<小时/>

所有文本的马可平均句子长度:

>>> sum(sum(len(sent) for sent in inaugural.sents(fileids=[fileid])) / len(inaugural.sents(fileids=[fileid])) for fileid in inaugural.fileids()) / len(inaugural.fileids())
32.84054349411484
<小时/>

每个文本的平均句子长度:

>>> from __future__ import division
>>> from nltk.corpus import inaugural
>>> inaugural.fileids()
[u'1789-Washington.txt', u'1793-Washington.txt', u'1797-Adams.txt', u'1801-Jefferson.txt', u'1805-Jefferson.txt', u'1809-Madison.txt', u'1813-Madison.txt', u'1817-Monroe.txt', u'1821-Monroe.txt', u'1825-Adams.txt', u'1829-Jackson.txt', u'1833-Jackson.txt', u'1837-VanBuren.txt', u'1841-Harrison.txt', u'1845-Polk.txt', u'1849-Taylor.txt', u'1853-Pierce.txt', u'1857-Buchanan.txt', u'1861-Lincoln.txt', u'1865-Lincoln.txt', u'1869-Grant.txt', u'1873-Grant.txt', u'1877-Hayes.txt', u'1881-Garfield.txt', u'1885-Cleveland.txt', u'1889-Harrison.txt', u'1893-Cleveland.txt', u'1897-McKinley.txt', u'1901-McKinley.txt', u'1905-Roosevelt.txt', u'1909-Taft.txt', u'1913-Wilson.txt', u'1917-Wilson.txt', u'1921-Harding.txt', u'1925-Coolidge.txt', u'1929-Hoover.txt', u'1933-Roosevelt.txt', u'1937-Roosevelt.txt', u'1941-Roosevelt.txt', u'1945-Roosevelt.txt', u'1949-Truman.txt', u'1953-Eisenhower.txt', u'1957-Eisenhower.txt', u'1961-Kennedy.txt', u'1965-Johnson.txt', u'1969-Nixon.txt', u'1973-Nixon.txt', u'1977-Carter.txt', u'1981-Reagan.txt', u'1985-Reagan.txt', u'1989-Bush.txt', u'1993-Clinton.txt', u'1997-Clinton.txt', u'2001-Bush.txt', u'2005-Bush.txt', u'2009-Obama.txt']
>>> for fileid in inaugural.fileids():
...     avg = sum(len(sent) for sent in inaugural.sents(fileids=[fileid])) / len(inaugural.sents(fileids=[fileid]))
...     print fileid, avg
... 
1789-Washington.txt 64.0833333333
1793-Washington.txt 36.75
1797-Adams.txt 69.8648648649
1801-Jefferson.txt 46.0714285714
1805-Jefferson.txt 52.9777777778
1809-Madison.txt 60.2380952381
1813-Madison.txt 39.5151515152
1817-Monroe.txt 30.2704918033
1821-Monroe.txt 38.0542635659
1825-Adams.txt 42.5675675676
1829-Jackson.txt 48.32
1833-Jackson.txt 42.2333333333
1837-VanBuren.txt 43.9052631579
1841-Harrison.txt 43.6428571429
1845-Polk.txt 33.9607843137
1849-Taylor.txt 53.7272727273
1853-Pierce.txt 35.1634615385
1857-Buchanan.txt 34.808988764
1861-Lincoln.txt 29.0217391304
1865-Lincoln.txt 29.0740740741
1869-Grant.txt 30.2195121951
1873-Grant.txt 33.5909090909
1877-Hayes.txt 46.1694915254
1881-Garfield.txt 28.9196428571
1885-Cleveland.txt 41.5454545455
1889-Harrison.txt 30.2547770701
1893-Cleveland.txt 37.1206896552
1897-McKinley.txt 33.6230769231
1901-McKinley.txt 24.5
1905-Roosevelt.txt 33.0606060606
1909-Taft.txt 36.7672955975
1913-Wilson.txt 28.0147058824
1917-Wilson.txt 27.6
1921-Harding.txt 25.2080536913
1925-Coolidge.txt 22.5482233503
1929-Hoover.txt 24.6202531646
1933-Roosevelt.txt 24.2705882353
1937-Roosevelt.txt 21.03125
1941-Roosevelt.txt 22.5882352941
1945-Roosevelt.txt 24.5
1949-Truman.txt 21.7931034483
1953-Eisenhower.txt 22.5609756098
1957-Eisenhower.txt 20.8369565217
1961-Kennedy.txt 29.7307692308
1965-Johnson.txt 18.2446808511
1969-Nixon.txt 22.8773584906
1973-Nixon.txt 29.3913043478
1977-Carter.txt 26.0377358491
1981-Reagan.txt 22.0551181102
1985-Reagan.txt 23.380952381
1989-Bush.txt 18.7103448276
1993-Clinton.txt 22.9012345679
1997-Clinton.txt 21.9821428571
2001-Bush.txt 18.8144329897
2005-Bush.txt 25.0105263158
2009-Obama.txt 24.3392857143
<小时/>

所有文本的平均宏观平均字长:

>>> sum([sum(len(sent) for sent in inaugural.sents(fileids=[fileid])) for fileid in inaugural.fileids()]) / len(inaugural.fileids())
2602.410714285714

关于python - 语料库中每个文本的平均句子长度(python3 和 nltk),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/35900029/

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