统计文件夹下面的文本文件中频率最高的单词

通过该练习，需要熟悉lambda在字典排序中的使用，可参考这篇内容http://www.cnblogs.com/kaituorensheng/archive/2012/08/07/2627386.html

# coding:utf-8
import re
import os


def get_myfiles(path):

    filepath = os.listdir(path)
    files = []
    for fp in filepath:
        # 将path路径下面的文件进行拼接
        fppath = path + '/' + fp
        fppattern = fppath.split('.')
        if os.path.isfile(fppath) and fppattern[2] == 'txt':
            files.append(fppath)
        elif os.path.isdir(fppath):
            files += get_myfiles(fppath)
    return files


def get_word(files):

    worddict = {}
    for filename in files:
        f = open(filename, 'rb')
        s = f.read()
        words = re.findall(r'[a-z0-9A-z]+', s)
        for word in words:
            worddict[word] = worddict[word] + 1 if word in worddict else 1
        f.close()
    wordsort = sorted(worddict.items(), key=lambda e: e[1], reverse=True)
    return wordsort

if __name__ == '__main__':
    files = get_myfiles('.')
    wordsort = get_word(files)
    num = 1
    # 因为wordsort已经倒序排列，因而只需要按照前面的几个值进行比对
    for i in range(len(wordsort)-1):
        if wordsort[i][1] == wordsort[i+1][1]:
            num += 1
        else:
            break
    # 通过用变量num记录最大值的个数，后面只需要根据num的值按顺序读取字典
    for i in range(num):
        print wordsort[i]

 下面通过构造类方法来实现该功能

import re
import io


class Counter:

    def __init__(self, path):
        """
        :param path:文件路径
        """
        self.mapping = {}
        with io.open(path, 'r', encoding='utf-8') as f:
            content = f.read()
            words = [s.lower() for s in re.findall(r'\w+', content)]
            for word in words:
                self.mapping[word] = self.mapping.get(word, 0) + 1

    def most_num(self, n):
        assert n > 0, "n should be large than 0"
        return sorted(self.mapping.items(), key=lambda s: s[1], reverse=True)[:n]

if __name__ == '__main__':
    most_num_5 = Counter('yes.txt').most_num(6)
    for item in most_num_5:
        print item