python 实验题目：统计词频给出一段文本，例如： “who have an apple apple is free free is money you know“，请统计单词出现的个数

一种思路：两个单词之间夹着一个空格，所以空格数比单词数少一
但是又两种特殊情况：

 #对于空格很多不正常的情况：
    string1 = "who have an apple                apple is free free is money you know"
    # 对于文本中出现正常的标点符号的情况：
    string2 = "who have an apple ， apple ，，，，，， is free free is money you know"

下面是解决方法：设置一个flag来判断

代码如下：

if __name__ == '__main__':
    count = 0#设置一个计数器，计数单词个数
    count1 = 0
    count2 = 0
    flag = 0#设置一个flag用来标记空格是否值出现一次
    string = "who have an apple apple is free free is money you know"
    #对于空格很多不正常的情况：
    string1 = "who have an apple                apple is free free is money you know"
    # 对于文本中出现正常的标点符号的情况：
    string2 = "who have an apple ， apple ，，，，，， is free free is money you know"
    for s in string:
        if(("a" <= s and s <= "z") or ("A" <= s and s <= "Z")):
            flag = 0
        else:
            if(flag == 0):
                count += 1
                flag = 1
    for s in string1:
        if(("a" <= s and s <= "z") or ("A" <= s and s <= "Z")):
            flag = 0
        else:
            if(flag == 0):
                count1 += 1
                flag = 1
    for s in string2:
        if(("a" <= s and s <= "z") or ("A" <= s and s <= "Z")):
            flag = 0
        else:
            if(flag == 0):
                count2 += 1
                flag = 1
    print("正常文本情况下string的单词个数为：%s"%(count + 1))
    print("空格异常文本情况下string1的单词个数为：%s"%(count1 + 1))
    print("含有标点符号的情况下string2的单词个数为：%s"%(count2 + 1))

运行结果如下：不要搬我的运行截图，代码复制完直接就可以运行了，改改数据

点个吧