你能用concordance做正则表达式吗？

Question

在NLTK introduction book中，它们向我们展示了如何通过协调来获得给定单词的上下文。但我想要更复杂一点的东西。我能把文字围绕某个图案吗？像这样：

text.concordances(", [A-Za-z]+ , ")~所有单词都用空格和逗号括起来

Answer 1

简而言之，nltk无法从当前状态的regex创建一致性。从nltk的ConcordanceIndex类（或其子类）创建一致性的困难在于，该类接受标记列表作为参数（并围绕这些标记构建），而不是全文字符串。在

我想我的建议是创建您自己的类，它接受字符串作为参数而不是令牌。下面是一个松散地基于nltk的ConcordanceIndex类的类，可以作为起点：

import re


class RegExConcordanceIndex(object):
    "Class to mimic nltk's ConcordanceIndex.print_concordance."

    def __init__(self, text):
        self._text = text

    def print_concordance(self, regex, width=80, lines=25, demarcation=''):
        """
        Prints n <= @lines contexts for @regex with a context <= @width".
        Make @lines 0 to display all matches.
        Designate @demarcation to enclose matches in demarcating characters.
        """ 
        concordance = []
        matches = re.finditer(regex, self._text, flags=re.M)
        if matches:
            for match in matches:
                start, end = match.start(), match.end()
                match_width = end - start
                remaining = (width - match_width) // 2
                if start - remaining > 0:
                    context_start = self._text[start - remaining:start]
                    #  cut the string short if it contains a newline character
                    context_start = context_start.split('\n')[-1]
                else:
                    context_start = self._text[0:start + 1].split('\n')[-1]
                context_end = self._text[end:end + remaining].split('\n')[0]
                concordance.append(context_start + demarcation + self._text
                                   [start:end] + demarcation + context_end)
                if lines and len(concordance) >= lines:
                    break
            print("Displaying %s matches:" % (len(concordance)))
            print '\n'.join(concordance)
        else:
            print "No matches"

现在可以这样测试类：

^{pr2}$