擅长:python、mysql、java
<p>您可以使用:</p>
<pre><code>l = [1, 2, "3=", "fd", "dfdf", "keyword", "ssd", "sdsd", ";", "dds"]
s = "keyword"
def take(last, iterable):
l = []
for x in iterable:
l.append(x)
if last in x:
break
return l
# get all elements on the right of s
right = take(';', l[l.index(s) + 1:])
# get all elements on the left of s using a reversed sublist
left = take('=', l[l.index(s)::-1])
# reverse the left list back and join it to the right list
subl = left[::-1] + right
print(subl)
['3=', 'fd', 'dfdf', 'keyword', 'ssd', 'sdsd', ';']
</code></pre>