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让我过去我拥有的确切代码: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

主要的:

import textchange

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

第一个 if 语句之后的文档字符串

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

我将其注释掉并且程序运行,否则没有。elif 语句中有更多代码,但在确保它仍然抛出错误后将其删除。这是回溯:

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
4

1 回答 1

7

文档字符串只是在函数开头找到的字符串文字。他们仍然必须遵循缩进规则。

您的字符串没有正确缩进该elif块;通过从if之前的块中删除,您完全结束了if- elif-else块并且elif不允许跟随

改用常规的普通注释,以#;开头的行 包含注释的行不受缩进规则的约束:

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

或缩进字符串(它仍然完全由 Python 作为代码执行,但未分配并因此再次被丢弃):

if char == '!' or char == '?':
    sentences.append(sentence)
    sentence = ''

elif char == '.':
    """ The sign -> refers to 'followed by' """
    i = raw_text.index(char) # slicing previous/following characters
    boundary = True
于 2013-08-31T07:50:19.077 回答