Python 中的示例函数：计算单词数

Question

我对 Python 有点生疏，只是在寻求帮助来实现一个示例函数来计算单词数（这只是 scons 脚本的示例目标，它不做任何“真实”的事情）：

def countWords(target, source, env):
  if (len(target) == 1 and len(source) == 1):
    fin = open(str(source[0]), 'r')
    # do something with "f.read()"
    fin.close()

    fout = open(str(target[0]), 'w')
    # fout.write(something)
    fout.close()
  return None

你能帮我填写一下吗细节？计算单词数的常用方法是读取每一行，分解为单词，并为该行中的每个单词增加字典中的计数器；然后对于输出，按计数递减对单词进行排序。

编辑：我正在使用Python 2.6（准确地说是Python 2.6.5）

Answer 1

from collections import defaultdict

def countWords(target, source, env):
    words = defaultdict(int)
    if (len(target) == 1 and len(source) == 1):
        with open(str(source[0]), 'r') as fin:
            for line in fin:
                for word in line.split():
                    words[word] += 1

        with open(str(target[0]), 'w') as fout:
            for word in sorted(words, key=words.__getitem__, reverse=True):
                fout.write('%s\n' % word)
    return None

Answer 2

在不知道为什么 env 存在的情况下，我只能执行以下操作：

def countWords(target, source, env):
    wordCount = {}
    if len(target) == 1 and len(source) == 1:
        with fin as open(source[0], 'r'):
            for line in f
                for word in line.split():
                    if word in wordCount.keys():
                        wordCount[word] += 1
                    else:
                        wordCount[word] = 0

        rev = {}
        for v in wordCount.values():
            rev[v] = []
        for w in wordCount.keys():
            rev[wordCOunt[w]].append(w)
        with open(target[0], 'w') as f:
            for v in rev.keys():
                f.write("%d: %s\n" %(v, " ".join(rev[v])))

Answer 3

此处有一个有用的示例。它的工作原理大致如您所描述的那样，并且还对句子进行计数。

Answer 4

虽然效率不是很高，但是很简洁！

with open(fname) as f:
   res = {}
   for word in f.read().split():
       res[word] = res.get(word, 0)+1
with open(dest, 'w') as f:
    f.write("\n".join(sorted(res, key=lambda w: -res[w])))

Answer 5

这是我的版本：

import string
import itertools as it
drop = string.punctuation+string.digits

def countWords(target, source, env=''):
    inputstring=open(source).read()
    words = sorted(word.strip(drop)
                   for word in inputstring.lower().replace('--',' ').split())
    wordlist = sorted([(word, len(list(occurances)))
                      for word, occurances in it.groupby(words, lambda x: x)],
                        key = lambda x: x[1],
                      reverse = True)
    with open(target,'w') as results:
        results.write('\n'.join('%16s : %s' % word for word in wordlist))