对Python非常陌生,希望你们能给我一些帮助。
我有一本关于第一次世界大战的书,想计算一个国家在书中出现的次数。到目前为止我有这个:
>>> from __future__ import division
>>> import nltk, re, pprint
>>> from urllib import urlopen
>>> url = "http://www.gutenberg.org/files/29270/29270.txt"
>>> raw = urlopen(url).read()
>>> type(raw)
<type 'str'>
>>> len(raw)
1067008
>>> raw[:75]
'The Project Gutenberg EBook of The Story of the Great War, Volume II (of\r\nV'
>>>
标记化。 将字符串分解为单词和标点符号。
>>> tokens = nltk.word_tokenize(raw)
>>> type(tokens)
<type 'list'>
>>> len(tokens)
189743
>>> tokens[:10] //vind de eerste 10 tokens
['The', 'Project', 'Gutenberg', 'EBook', 'of', 'The', 'Story', 'of', 'the', 'Great']
>>>
更正书的开头和结尾
>>> raw.find("PART I")
>>> 2629
>>> raw.rfind("End of the Project Gutenberg")
>>> 1047663
>>> raw = raw[2629:1047663]
>>> raw.find("PART I")
>>> 0
不幸的是,我不知道如何将这本书实现到字数统计中。我理想的结果是这样的:
Germany 2000
United Kingdom 1500
USA 1000
Holland 50
Belgium 150
等等
请帮忙!
最佳答案
Python 有一个内置方法来计算字符串中的子字符串。
from urllib import urlopen
url = "http://www.gutenberg.org/files/29270/29270.txt"
raw = urlopen(url).read()
raw = raw[raw.find("PART I"):raw.rfind("End of the Project Gutenberg")]
countries = ['Germany', 'United Kingdom', 'USA', 'Holland', 'Belgium']
for c in countries:
print c, raw.count(c)
产生
Germany 117
United Kingdom 0
USA 0
Holland 10
Belgium 63
编辑:eumiro 是对的,如果你想计算确切的单词,这是行不通的。如果您想搜索确切的单词,请使用此:
import re
from urllib import urlopen
url = "http://www.gutenberg.org/files/29270/29270.txt"
raw = urlopen(url).read()
raw = raw[raw.find("PART I"):raw.rfind("End of the Project Gutenberg")]
for key, value in {c:len(re.findall(c + '[^A-Za-z]', raw)) for c in countries}.items():
print key, value
编辑:如果您想要行号:
from urllib import urlopen
import re
from collections import defaultdict
url = "http://www.gutenberg.org/files/29270/29270.txt"
raw = urlopen(url).readlines()
count = defaultdict(list)
countries = ['Germany', 'United Kingdom', 'USA', 'Holland', 'Belgium']
for c in countries:
for nr, line in enumerate(raw):
if re.search(c + r'[^A-Za-z]', line):
count[c].append(nr + 1) #nr + 1 so the first line is 1 instead of 0
print c, len(count[c]), 'lines:', count[c]
关于python - 计算标记化网址中的单词数,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/10894228/