计算python中具有多个元音的单词和行数

Question

我正在解决这个编码问题“计算每个 Z 行中每个 Y 个单词的元音超过 X 个的单词和行数”

基本上输入字符串有多行，我需要计算其中包含 X 或更多元音的单词。但限制是我只需要考虑备用第 Z 行以及这些第 Z 行中的备用 Yth 单词。例如假设我需要计算每第三行有 2 个或更多元音的第三个单词。所以这里是X=2、Y=3、Z=3。检查以下输入字符串：

"1.When I first brought my cat home.
 2.It cost a lot to adopt her.
 3.I paid forty dollars for it.
 4.And then I had to buy litter, a litterbox.
 5.Also bought food, and dishes for her to eat out of. 
 6.There's a **leash** law for cats in Fort **Collins**.
 7.If they're not in your yard they have to be on a leash. 
 8.Anyway, my cat is my best friend. 
 9.I'm glad I got her. 
 10.She sleeps under the covers with me when it's cold."

输出：字数：2，行数：1

因此基于Z=3 的标准，即每第三行计数一次，因此要考虑的行是line number 3, 6, 9。同样在这些行中，我们需要计算Y=3 i.e. every 3rd word。所以要考虑的词是"forty, it" from line 3、"leash, cats, Collins" from line 6 和"I" from line 9。鉴于此标准，只有在第 6 行中才能找到具有 2 个或更多元音的匹配词，其中包含词 "leash" 和 "Collins"，因此输出为 WordCount = 2 和 LineCount = 1。

这是我第一次用 Python 写东西，所以写了下面的基本代码：

class StringCount:  #Count the number of words & lines that have more than X vowels for every Y words in every Z line. 

    lines = list();
    totalMatchedLines = 0;
    totalMatchedWords = 0;
    matchedChars = 0;

    def __init__(self, inputString, vowelCount, skipWords, skipLines, wordDelimiter, lineDelimiter):
      self.inputString = inputString;
      self.vowelCount = vowelCount;
      self.skipWords = skipWords;
      self.skipLines = skipLines;
      self.wordDelimiter = wordDelimiter;

    def splitLines(self):
      if self.inputString.strip() == "":
        print ("Please enter a valid string!");
        return False;      
      self.lines = self.inputString.splitlines();

    def splitWords(self):      
      self.matchedWords = 0;
      self.matchedLines = 0;
      self.linesLength = len(self.lines);

      if self.linesLength < self.skipLines:
        print ("Input string should be greater than {0}" .format(self.skipLines));
        return False;

      lineCount = self.skipLines - 1;
      wordCount = self.skipWords - 1;
      lineInUse = "";
      words = list();

      while (lineCount < self.linesLength):
        self.matchedWords = 0;
        self.matchedLines = 0;
        self.words = self.lines[lineCount].split();
        self.wordsLength = len(self.words);
        wordCount = self.skipWords - 1;

        while (wordCount < self.wordsLength):
          self.matchedChars = 0;       
          for i in self.words[wordCount].lower():
            if(i=='a' or i=='e' or i=='i' or i=='o' or i=='u'):
              self.matchedChars += 1;              
          if self.matchedChars >= self.vowelCount:
            self.matchedWords += 1;
          wordCount += self.skipWords;

        if self.matchedWords > 0:
          self.matchedLines += 1;

        self.totalMatchedWords += self.matchedWords;
        self.totalMatchedLines += self.matchedLines;
        lineCount += self.skipLines;

      print ("WordCount = %s" % (self.totalMatchedWords));
      print ("LineCount = %s" % (self.totalMatchedLines));

由于这是我的第一个 Python 代码，我想检查如何在性能和线路优化方面优化此代码。有没有什么技巧可以缩短多个 while 循环和一个 for 循环？

感谢任何帮助！

Answer 1

使用转换表会更有效。如果将每个单词中的所有元音转换为字母“a”，则可以使用 count() 方法计算元音。剩下的只是遍历行和单词列表的问题，您可以使用列表推导和范围索引来完成：

vowels     = str.maketrans("aeiouAEIOU","aaaaaaaaaa")
X,Y,Z      = 2,3,3
counts     = [sum(word.translate(vowels).count("a")>=X for word in line.split(" ")[::Y]) for line in lines[Z-1::Z]]
lineCount  = sum(count>0 for count in counts)
wordCount  = sum(counts)

print(lineCount,wordCount) # 1 2