<p>仅通过<code>re</code>模块。你知道吗</p>
<pre><code>>>> s = 'n76a+q80a+l83a+i153a+l203f+r207a+s211a+s215w+f216a+e283l'
>>> m = re.findall(r'(?=\b([^+]+\+[^+]+))', s) # This regex would helps to do a overlapping match. See the demo (https://regex101.com/r/jO6zT2/13)
>>> m
['n76a+q80a', 'q80a+l83a', 'l83a+i153a', 'i153a+l203f', 'l203f+r207a', 'r207a+s211a', 's211a+s215w', 's215w+f216a', 'f216a+e283l']
>>> l = []
>>> for i in m:
if abs(int(re.search(r'^\D*(\d+)', i).group(1)) - int(re.search(r'^\D*\d+\D*(\d+)', i).group(1))) < 10:
l.append(i.replace('+', '_'))
else:
l.append(i)
>>> re.sub(r'([a-z0-9]+)\1', r'\1',''.join(l))
'n76a_q80a_l83a+i153a+l203f_r207a_s211a_s215w_f216a+e283l'
</code></pre>
<p>通过定义一个单独的函数。你知道吗</p>
<pre><code>import re
def aa_dist(x):
l = []
m = re.findall(r'(?=\b([^+]+\+[^+]+))', x)
for i in m:
if abs(int(re.search(r'^\D*(\d+)', i).group(1)) - int(re.search(r'^\D*\d+\D*(\d+)', i).group(1))) < 10:
l.append(i.replace('+', '_'))
else:
l.append(i)
return re.sub(r'([a-z0-9]+)\1', r'\1',''.join(l))
string = 'n76a+q80a+l83a+i153a+l203f+r207a+s211a+s215w+f216a+e283l'
print aa_dist(string)
</code></pre>
<p><strong>输出:</strong></p>
<pre><code>n76a_q80a_l83a+i153a+l203f_r207a_s211a_s215w_f216a+e283l
</code></pre>