OP: Правило для полужирного текста отсутствует. при условии, что поврежденный текст закончится safe
:
import re
def removePunctuation(text):
for elem in text:
if elem.startswith('a01'):
print(elem)
else:
elem = elem.rpartition('safe')[2]
print(elem)
test_list = ['xian/gps_201610010000644016240301624032416162641013323634045015307 0ustar bigdata_safebigdata_safea01b8439e1e42ffcd286241b04d9b1b5,f11440a64a0f084fe346a398c62aa9ad,1475277482,108.92466,34.27657',
'a01b8439e1e42ffcd286241b04d9b1b5,f11440a64a0f084fe346a398c62aa9ad,1475277488,108.92527,34.27658',
'a01b8439e1e42ffcd286241b04d9b1b5,f11440a64a0f084fe346a398c62aa9ad,1475277506,108.9276,34.27659',
'a01b8439e1e42ffcd286241b04d9b1b5,f11440a64a0f084fe346a398c62aa9ad,1475277476,108.92399,34.27655',
'a01b8439e1e42ffcd286241b04d9b1b5,f11440a64a0f084fe346a398c62aa9ad,1475277515,108.9291,34.2766']
removePunctuation(test_list)
ВЫХОД:
a01b8439e1e42ffcd286241b04d9b1b5,f11440a64a0f084fe346a398c62aa9ad,1475277482,108.92466,34.27657
a01b8439e1e42ffcd286241b04d9b1b5,f11440a64a0f084fe346a398c62aa9ad,1475277488,108.92527,34.27658
a01b8439e1e42ffcd286241b04d9b1b5,f11440a64a0f084fe346a398c62aa9ad,1475277506,108.9276,34.27659
a01b8439e1e42ffcd286241b04d9b1b5,f11440a64a0f084fe346a398c62aa9ad,1475277476,108.92399,34.27655
a01b8439e1e42ffcd286241b04d9b1b5,f11440a64a0f084fe346a398c62aa9ad,1475277515,108.9291,34.2766