# word frequency in a text
# tested with Python24 vegaseat 25aug2005
# Chinese wisdom ...
str1 = """Man who run in front of car, get tired.
Man who run behind car, get exhausted."""
print "Original string:"
print str1
print
# create a list of words separated at whitespaces
wordList1 = str1.split(None)
# strip any punctuation marks and build modified word list
# start with an empty list
wordList2 = []
for word1 in wordList1:
# last character of each word
lastchar = word1[-1:]
# use a list of punctuation marks
if lastchar in [",", ".", "!", "?", ";"]:
word2 = word1.rstrip(lastchar)
else:
word2 = word1
# build a wordList of lower case modified words
wordList2.append(word2.lower())
print "Word list created from modified string:"
print wordList2
print
# create a wordfrequency dictionary
# start with an empty dictionary
freqD2 = {}
for word2 in wordList2:
freqD2[word2] = freqD2.get(word2, 0) + 1
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Python 统计一个英文字符串中每个字符出现的次数 三种方法
方法一
定义字符串转化为list函数strchls
def strchls (str):
    list = []  # 集合
    i = 0
    for w in str:
        # print(w)
        if w != ' ':
            list.append(w)
    return list

定义list 转化为字典序列，并把字符作为key(不重复）
def countw(list):
    count_word = {}
    for w in list:
        # j = 1
        if w not in count_word.keys():
            count_word[w] = 1  # 编号 w:j
        else:
            count_word[w] += 1  # 编号 w:j
    return count_word

函数调用
str = 'how are you'
ls = strchls (str)
print(countw(ls))

方法二
#去空格，转化为list，然后再转化为字典即可
str =  'how are you'
list = []
list2 = []
dict={}
i= 0
for w in str:
    if w!=' ':
        list.append(w)
# print(list)
for w in list:
    c = list.count(w)
    dict[w] = c
print(dict)

方法三
直接去空格，转化为字典，利用字典的key值唯一
str =  'how are you'
dict = {}
for w1 in str:
    if w1 != ' ':
        for w1 in list:
            c = list.count(w1)
            dict[w1] = c
print(dict)

英语字符和中文字符的区别在于：大小写字符和字符个数（中文中是一个词语）

 

统计英语字符：

def getText():

    txt=open(‘halmet.txt’,’r’).read()

    txt=txt.lower()   #文中所有英语小写

    for ch in ‘!@#$%^&*()<>?”:{}|’:

        txt = txt.replace(ch,’’)

    return txt

 

halmetTxt=getText()

words=halmetTxt.split()

counts={}

for word in words:

    counts[word]=couts.get(word,0)+1

items=list(counts.items())

item.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))

 

 

jieba库：中文分词

jiebe.lcut(s)  精确模式

jieba.lcut(s,cut_all=True) 全模式

jieba.lcut_for_search(s)  搜索引擎模式

 

中文字符（使用jieba库）：

import jieba

f=open(‘红楼梦.txt’.’r’)

txt=f.read()

f.close()

words=jieba.lcut(txt)

counts]{}

for word in words:

     if len(word)==1:

         continue

     else:

         counts[word]=counts.get(word,0)+1

items=list(counts.items())

item.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))

 

 

如查找人物名字的出现频次将会出现许多无关的答案。所以可以用excludes进一步完整代码。

添加的部分我在下面都加了括号。

 

import jiebe

(exclude={‘’,’’,’’})

f=open(‘红楼梦.txt’.’r’)

txt=f.read()

f.close()

words=jieba.lcut(txt)

counts]{}

for word in words:

     if len(word)==1:

         continue

     else:

         counts[word]=counts.get(word,0)+1

(for word in excludes:

      del(counts[word])   )

items=list(counts.items())

item.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))

 

 

进一步可用wordcloud库进一步进行展示。

C语言-输入一行英文字符串统计单词个数，单词之间以空格隔开



判断是否为单词的依据，通过一个flag标识，flag为0表示字符为空格，接下来出现不是空格，会将flag置为1

只有当flag为1且当前字符不为空格，才会算作一个单独单词，保证连续多个字符为一个单词。

#include<stdio.h>
int main() {
    //定义一个string字符数组来接收输入的字符串
    //定义一个c字符用来比较当前字符是否是空格
    char string[100],c;
    //count用来统计单词个数，
    // 使用flag来标识如何为一个新的单词，为一个新单词的标准是前面的一个字符为空格，且初始值为0
    int i, count = 0, flag = 0;
    gets(string);//输入一个字符串给字符数组string
    for (i = 0; (c = string[i]) != '\0'; i++) {
        if (c == ' ')//如果是空格字符，使word置0
        {
            flag = 0;
        } else if (flag == 0)//如果不是空格字符，且flag原来的值为0，即前面是空格，使flag置1
        {
            flag = 1;count++;
        }
    }
    printf("count=%d\n", count);
    return 0;
}