Question

我一直在研究HTTLCS，但在完成问题时遇到了一些困难。

解决问题不是问题，但我将结果作为字符串而不是元组数据类型返回。

这是我的代码：

def wordCount(paragraph):
    splited = paragraph.split()
    wordnum = len(splited)
    eWord = []
    for aWord in splited:
        if "e" in aWord:
            eWord.append(aWord)
    eWordnum = len(eWord)
    percent = round(eWordnum / wordnum * 100,2)
    return "Your text contains", wordnum, "words, of which" , eWordnum , "(" , percent , "%)" , "contains an 'e'." 



print(wordCount(p))

Python输出('Your text contains', 108, 'words, of which', 50, '(', 46.3, '%)', "contains an 'e'.")这是一个元组，而不是字符串。

我知道我可以将print放在函数的末尾，并在没有print（）语句的情况下调用函数，但是如何使用return语句解决这个问题？

Answer 1

这是因为你在return语句中使用逗号，Python将其解释为元组。请尝试使用format()：

def wordCount(paragraph):
    splited = paragraph.split()
    wordnum = len(splited)
    eWord = []
    for aWord in splited:
        if "e" in aWord:
            eWord.append(aWord)
    eWordnum = len(eWord)
    percent = round(eWordnum / wordnum * 100,2)
    return "Your text contains {0} words, of which {1} ({2}%) contains an 'e'".format(wordnum, eWordnum, percent)

>>> wordCount("doodle bugs")

"Your text contains 2 words, of which 1 (0.0%) contains an 'e'"

Answer 2

return "Your text contains %s words, of which %s (%s%%) contains an 'e'." % (wordnum, eWordnum, percent)

Answer 3

return "Your text contains " + str(wordnum) + 
       " words, of which " + str(eWordnum) + 
       " (" + str(percent) +  "%)" + " contains an 'e'."

或

return "Your text contains %s words, of which %s (%s%%) contains an 'e'." 
        % (wordnum, eWordnum, percent)

在第一种情况下，您执行字符串连接，您必须将wordnum，eWordnum和其他数字变量转换为str （通过执行str(variableName)）允许连接（并且没有运行时错误）

在第二种情况下，你做一个字符串替换，这意味着你给了某种“占位符”%s（这意味着字符串），你用下面的元组参数替换它们%符号

如果你通过,单独返回一个元素，你将返回一个元组（如你所见）

Answer 4

return "Your text contains {0} words, of which {1} ({2}%) contains an 'e'.".format(wordnum,eWordnum,percent)

Answer 5

for循环可能有效，但您必须格式化字符串以向其添加空格。

for item in tuplename: print item,

确保将逗号保留在项目后面，因为它会将其打印在同一行上。

Answer 6

def wordCount(paragraph):
    splited = paragraph.split()
    wordnum = len(splited)
    eWord = []
    for aWord in splited:
        if "e" in aWord:
            eWord.append(aWord)
    eWordnum = len(eWord)
    percent = round(eWordnum / wordnum * 100,2)
    dummy =  "Your text contains {0} words, of which {1} {2} contains an 'e'.".format(wordnum,eWordnum, percent) 
    return dummy


print(wordCount(p))

Answer 7

试试这个：

return "Your text contains %(wordnum)s words, of which %(ewordnum)s (%(percent)s %%), contains an 'e'."%locals()

使用%(variable_name)s作为string formatting通常更容易维护。

Answer 8

这个怎么样

return "Your text contains " + wordnum + " words, of which " + eWordnum + " (" + percent + "%) " + " contains an 'e'."

用“+”替换逗号，这应该有用。

Python：将一个元组从return语句转换为字符串

8 个答案: