Question

让我超越我的确切代码：这是短模块

class SentenceSplitter:

def __init__(self, filename=None):
    self._raw_text = self.raw_text(filename)
    self._sentences = self.to_sentences()


def raw_text(self, filename):
    text = ''
    with open(filename, 'r') as file:
        for line in file.readlines():
            line = line.strip()
            text += ''.join(line.replace(line, line+' '))
    file.close()
    text = text.strip() # Deal with the last whitespace
    return text

def to_sentences(self):
    """ Sentence boundaries occur at '.', '!', '?' except that,
    there are some not-sentence boundaries that
    may occur before/after the period.
    """
    raw_text = self._raw_text
    sentences = []
    sentence = ''
    boundary = None

    for char in raw_text:
        sentence += ''.join(char)
        if char == '!' or char == '?':
            sentences.append(sentence)
            sentence = ''

        """ The sign -> refers to 'followed by' """
        elif char == '.':
            i = raw_text.index(char) # slicing previous/following characters
            boundary = True

        if boundary:
            sentences.append(sentence)
            sentence = ''

    return sentences

主要：

import textchange

ss = textchange.SentenceSplitter(filename='text.txt')
print(ss._sentences)

第一个if语句后的docstring

""" The sign -> refers to 'followed by' """

我评论了它，程序运行，否则没有。 elif语句中有更多代码，但在确定它仍然抛出错误后将其删除。这是追溯：

Traceback (most recent call last):
File "D:\Programs\Python 3.3.2\Tutorials\46 Simple Python Exercises.py", line 26, in        
<module>
import textchange
File "D:\Programs\Python 3.3.2\Tutorials\textchange.py", line 51
elif char == '.':
   ^
SyntaxError: invalid syntax

Answer 1

文档字符串只是在函数的 start 中找到的字符串文字。他们仍然必须遵守缩进规则。

您的字符串未正确缩进elif块;通过从if块中删除之前，您完全结束了if - elif - else块，并且允许<{1}} < / em>跟随。

使用常规正常评论，以elif开头;包含仅注释的行不受缩进规则的限制：

#

或缩进字符串（它完全由Python作为代码执行，但是没有分配，因此再次丢弃）：

if char == '!' or char == '?': sentences.append(sentence) sentence = '' # The sign -> refers to 'followed by' elif char == '.': i = raw_text.index(char) # slicing previous/following characters boundary = True

docstring阻止elif语句

1 个答案: