'''

dinghanhua

2018-11-11

练习：一段英文文章，统计每个单词的频率，返回出现频率最高的5个单词和次数

'''

import re

art = ' If we want to" run Locust \ / distributed on multiple machines we would also have to specify the master host when starting the slaves (this is not needed when running Locust distributed on a single machine, since the master host defaults to 127.0.0.1):'

'''

怎么判定单词?

1 不是字母的特殊字符作为分隔符分割字符串

2 遍历字符串，取每个word

3 正则匹配

怎么统计个数？

将wordlist的word和word的个数放入dict，排序

'''

word_dict = {} #用于统计 word：个数

word_list = [] #用于存放所有单词

#  找出所有不是字母的字符替换成统一的字符，split()分割之后便是单词

pattern = r'[^a-z]+'
art_new = re.sub(pattern,'-',art.lower()) #所有的非字母替换成-

word_list = art_new.split('-') #转成小写分隔单词

wordlist = list(filter(lambda x : x != '',word_list)) #去掉空串

print('所有的单词列表：',wordlist)

#正则表达式分隔

pattern = r'[^a-z]+' #非字母
word_list = re.split(pattern,art.lower()) #还要去除空串
print(word_list)

# 遍历字符串，获取每个word追加到wordlist （不好）

word =''

word_list2 = []

for letter in art.lower():

    if letter.isalpha(): #如果是字母，追加到word

        word += letter

    else:

        if word != '':

            word_list2.append(word) #不是字母，word不为空的话追加wordlist

            word = '' # word置空

print(word_list2)

# 正则表达式匹配单词

pattern = r'[a-z]+'

word_list3 = re.findall(pattern,art.lower())

print(word_list3)

#统计

for word in set(word_list):

    word_dict[word] = word_list.count(word) #key=单词，value=单词在list里的count

#取最多的前五个

print(sorted(word_dict.items(),key = lambda x:x[1],reverse=True)[0:5]) #dict根据value倒序，取前5个

word_dict = {}.fromkeys(word_list) #先用list生成dict的keys

for word in word_dict.keys():

    word_dict[word] = word_list.count(word)