<p>最后,基于@Wiktor Stribiżew和@Thm Lee提出的答案,在测试了几个想法之后,我找到了一堆处理不同复杂程度的解决方案。为了减少依赖性,我想继续使用Python标准库中的<code>re</code>模块,下面是代码:</p>
<pre><code>import re
text = "aa b%b( %cc(dd! (:ee ff) gg) %hh ii) "
# Solution 1: don't process parentheses at all
regexA = re.compile(r'(\S+)')
print(regexA.split(text))
# Solution 2: works for non-nested parentheses
regexB = re.compile(r'(%[^(\s]*\([^)]*\)|\S+)')
print(regexB.split(text))
# Solution 3: works for one level of nested parentheses
regexC = re.compile(r'(%[^(\s]*\((?:[^()]*\([^)]*\))*[^)]*\)|\S+)')
print(regexC.split(text))
# Solution 4: works for arbitrary levels of nested parentheses
n, words = 0, []
for word in regexA.split(text):
if n: words[-1] += word
else: words.append(word)
if n or (word and word[0] == '%'):
n += word.count('(') - word.count(')')
print(words)
</code></pre>
<p>以下是生成的输出:</p>
<pre><code>Solution 1: ['', 'aa', ' ', 'b%b(', ' ', '%cc(dd!', ' ', '(:ee', ' ', 'ff)', ' ', 'gg)', ' ', '%hh', ' ', 'ii)', ' ']
Solution 2: ['', 'aa', ' ', 'b%b(', ' ', '%cc(dd! (:ee ff)', ' ', 'gg)', ' ', '%hh', ' ', 'ii)', ' ']
Solution 3: ['', 'aa', ' ', 'b%b(', ' ', '%cc(dd! (:ee ff) gg)', ' ', '%hh', ' ', 'ii)', ' ']
Solution 4: ['', 'aa', ' ', 'b%b(', ' ', '%cc(dd! (:ee ff) gg)', ' ', '%hh', ' ', 'ii)', ' ']
</code></pre>
<p>如OP中所述,对于我的特定数据,括号中的转义空格只能用于以<code>%</code>开头的单词,其他括号(例如我的示例中的单词<code>b%b(</code>)不被认为是特殊的。如果要转义任何一对括号中的空格,只需删除regex中的<code>%</code>字符。以下是修改后的结果:</p>
<pre><code>Solution 1: ['', 'aa', ' ', 'b%b(', ' ', '%cc(dd!', ' ', '(:ee', ' ', 'ff)', ' ', 'gg)', ' ', '%hh', ' ', 'ii)', ' ']
Solution 2: ['', 'aa', ' ', 'b%b( %cc(dd! (:ee ff)', ' ', 'gg)', ' ', '%hh', ' ', 'ii)', ' ']
Solution 3: ['', 'aa', ' ', 'b%b( %cc(dd! (:ee ff) gg)', ' ', '%hh', ' ', 'ii)', ' ']
Solution 4: ['', 'aa', ' ', 'b%b( %cc(dd! (:ee ff) gg) %hh ii)', ' ']
</code></pre>