python(14)---统计词频数

　　说到统计词频数，python比cuda要好用多了，首先是python有相关的库供我们使用，

还有python调用相关的函数，省去我们在cuda中需要写很多的函数。汉语的词频统计

和英语有些不同，汉语需要用到jiba库。

　　英文词频统计的思路：1.首先我们将文中的特殊字符，替换为空格；

　　　　　　　　　　　　2.进行词语分割；

　　　　　　　　　　　　3.遍历文中的词语，计算出现的次数。

　　汉语词频统计思路：整体的思路和英文是一样的，但是汉语的词语分割会出现一些

我们不需要的目标词语，所以我们需要将这些词语排除掉。

代码如下：

#CalHamletV1.py
def getText():
    txt = open("hamlet.txt", "r").read()
    txt = txt.lower()
    for ch in '!"#$%&()*+,-./:;<=>?@[\]^_‘{|}~':
        txt = txt.replace(ch, " ")   #将文本中特殊字符替换为空格
    return txt
 
hamletTxt = getText()
words  = hamletTxt.split()
counts = {}
for word in words:           
    counts[word] = counts.get(word,0) + 1
items = list(counts.items())
items.sort(key=lambda x:x[1], reverse=True) 
for i in range(10):
    word, count = items[i]
    print ("{0:<10}{1:>5}".format(word, count))

#CalThreeKingdomsV2.py
import jieba
excludes = {"将军","却说","荆州","二人","不可","不能","如此","商议","军士","如何","军马","主公","左右",
            "天下","次日"}
txt = open("三国演义.txt", "r", encoding='utf-8').read()
words  = jieba.lcut(txt)
counts = {}
for word in words:
    if len(word) == 1:
        continue
    elif word == "诸葛亮" or word == "孔明曰":
        rword = "孔明"
    elif word == "关公" or word == "云长":
        rword = "关羽"
    elif word == "玄德" or word == "玄德曰":
        rword = "刘备"
    elif word == "孟德" or word == "丞相":
        rword = "曹操"
    else:
        rword = word
    counts[rword] = counts.get(rword,0) + 1
for word in excludes:
    del counts[word]
items = list(counts.items())
items.sort(key=lambda x:x[1], reverse=True) 
for i in range(10):
    word, count = items[i]
    print ("{0:<10}{1:>5}".format(word, count))

运行结果：