我正在解决这个编码问题“计算每个Z行中每个Y单词的元音超过X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个或更多元音的单词。但限制是我只需要考虑备用的Zth行以及这些Zth行中的备用Yth字。例如,假设我需要计算每3行中有2个或更多元音的每3个单词。所以这里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,根据Z=3
的标准,即每3行计算一次,因此要考虑的行是line number 3, 6, 9
。在这些行中,我们还需要计算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
。根据这个标准,只有在第6行中,匹配的单词有2个或更多元音,并且有单词"leash"
和"Collins"
,所以输出是WordCount=2和LineCount=1。你知道吗
这是我第一次用Python编写任何东西,因此我编写了以下基本代码:
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));
因为这是我的第一个Python代码,所以我想检查如何在性能和行优化方面优化这段代码。有什么技巧可以缩短多重while循环和for循环吗?你知道吗
感谢您的帮助!你知道吗
使用转换表可以更有效地执行此操作。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()方法对元音进行计数。剩下的只是遍历行和词的列表,您可以通过列表理解和范围索引来完成:
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