def insidelastparentheses(str):
  return str.rpartition('(')[2].partition(')')[0]

lst = ['NM_001001182.3 Mus musculus bromodomain adjacent to zinc finger domain, 2B (Baz2b), mRNA',
       'xxx(12) yyy(second)',
       'yyy(12) zzz(second) aaa(last)',
       'yyy(12) zzz(second) aaa(3) bb(4)',
       'yyy(12) zzz(second) aaa(3) bb(4) cc(5) haha',
      ]

for str in lst:
  print(insidelastparentheses(str))

執行結果:

Baz2b
second
last
4
5

附上比較笨的做法：

s = 'NM_001001182.3 Mus musculus bromodomain adjacent to zinc finger domain, 2B (Baz2b), mRNA'
length = len(s)
start = 0
end = 0

for index in range(0, length):
    if s[length - 1 - index] == ')':
        end = length - 1 - index
    if s[length - 1 - index] == '(':
        start = length - 1 - index

result = s[start + 1:end]
print(result)

使用 Regular Expression 解法:

import re

x = 'NM_00100(1182.3) Mus musculus (bromodomain) adjacent to zinc finger domain, 2B (Baz2b), mRNA'

p1 = re.compile(r'\(\w*\)')
list1 = re.findall(p1, x)

#last one
if len(list1) > 0:
    print(list1[-1][1:-1])

import re
s = 'NM_001001182.3 Mus musculus bromodomain adjacent to zinc finger domain, 2B (Baz2b), mRNA'

p1 = re.findall('\(([^)]+)', s)
print(p1[-1])