Python程序:外文词频词典
Python程序:外文词频词典
提问于 2014-12-10 17:43:58
我正在尝试构建一个简单的程序，它接受一个文本文件，构建一个以单词作为键的dict()，以及每个单词出现的次数(单词频率)的值。
我已经了解到，collections.Counter函数可以很容易地做到这一点(包括其他方法)。我的问题是，我希望字典按频率排序，这样我就能打印出第N个最频繁的单词。最后，我还需要为字典提供一种方法，以便稍后将不同类型的值(单词定义的字符串)关联起来。
基本上，我需要输出以下内容的东西：
Number of words: 5
[mostfrequentword: frequency, definition]
[2ndmostfrequentword: frequency, definition]
etc. 到目前为止，这是我所拥有的，但它只计算单词的频率，我不知道如何按频率排序字典，然后打印第N个最频繁的单词：
wordlist ={}
def cleanedup(string):
alphabet = 'abcdefghijklmnopqrstuvwxyz'
cleantext = ''
for character in string.lower():
if character in alphabet:
cleantext += character
else:
cleantext += ' '
return cleantext
def text_crunch(textfile):
for line in textfile:
for word in cleanedup(line).split():
if word in wordlist:
wordlist[word] += 1
else:
wordlist[word] = 1
with open ('DQ.txt') as doc:
text_crunch(doc)
print(wordlist['todos'])回答 1
Stack Overflow用户
回答已采纳
发布于 2014-12-10 18:01:07
一个更简单的代码版本，它可以很好地实现您想要的功能:)
import string
import collections
def cleanedup(fh):
for line in fh:
word = ''
for character in line:
if character in string.ascii_letters:
word += character
elif word:
yield word
word = ''
with open ('DQ.txt') as doc:
wordlist = collections.Counter(cleanedup(doc))
print wordlist.most_commond(5)带有正则表达式的替代解决方案：
import re
import collections
def cleandup(fh):
for line in fh:
for word in re.findall('[a-z]+', line.lower()):
yield word
with open ('DQ.txt') as doc:
wordlist = collections.Counter(cleanedup(doc))
print wordlist.most_commond(5)或者：
import re
import collections
def cleandup(fh):
for line in fh:
for word in re.split('[^a-z]+', line.lower()):
yield word
with open ('DQ.txt') as doc:
wordlist = collections.Counter(cleanedup(doc))
print wordlist.most_commond(5)页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/27407485
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