python实现linux下指定目录下文件中的单词个数统计
本帖最后由 432qw 于 2016-8-15 11:20 编辑# !/usr/bin/env python
# -*- coding:utf-8 -*-
"""
实现指定目录下的文件单词计数统计：
例如：你输入
nginx.conf(将要统计的文件)--->/usr(nginx.conf的搜索目录)--->输入要得到该文件的前top个单词数目
结果: 得到---> (单词名,该文件的中该单词出现的次数)
"""
import os
import sys
import re
def getfiles():
file_name = raw_input("please enter the file_name you want count the file\n")
path = raw_input("please enter the root path : example: /tmp\n")
if not os.path.exists(path):
print "sorry,the path %s is not exists !!" %(path)
sys.exit(1)
else:
cmd = "find " + path + " -name "+ file_name
print cmd
files = os.popen(cmd).readlines()
files1 = for file in files]
return files1
def countword(top_num):
files = getfiles()
for file in files:
with open(file) as f:
# get the file content and strip the space and line break(acording the os.linesep)
file_content = f.read().replace(os.linesep,"").replace("#","").replace("/","").replace(";","").replace("}","").replace("{","")
file_words = file_content.split(" ")
wordcounts = [(word, file_words.count(word)) for word in file_words]
# get uniq wordcounts
wordcounts = list(set(wordcounts))
# get the max wordcounts
# wordcounts = )]
wordcounts = sorted(wordcounts,key=lambda d: d, reverse=True)
# get the right word format
wordcounts = )]
if len(wordcounts[:int(top_num)]) > 0:
print"文件:" + f.name + "的前" + top_num + "个单词是及个数分别是:\n" + str(wordcounts[:int(top_num)])
else:
print"文件:" + f.name + "是个空文件！！\n" + str(wordcounts[:int(top_num)])
if __name__ == "__main__":
top_num = raw_input("请输入要得到统计单词数目的前多少个\n")
countword(top_num)
页:
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