2013-03-18 20 views
-2
P0C5Z8: MLGVINRMAKKFNYKLPSMVALTLVGSAVTAHQVQAAETTQDQTTNKNVLDSNKVKATTEQAKAEVKNPTQNISGTQVYQDPAIVQPKTANNKTGNAQVS 
Q53630: MSNNFKDDFEKNRQSIDTNSHQDHTEDVEKDQSELEHQDTIENTEQQFPPRNAQRRKRRRDLATNHNKQVHNESQTSEDNVQNEAGTIDDRQVESSHSTE 
P0C6P1: MKKKLLVLTMSTLFATQLINSNHANASVTESVDKKFVVPESGINKIIPTYNEFKKAPKVNVSNLTDNKNFVASEDKLKKISDPSAASKIVDKNFVVPESK 
O69174: MPIITDVYAREVLDSRGNPTVEVEVLTESGAFGRALVPSGASTGEHEAVELRDGDKSRYLGKGVTKAVENVNEIIAPEIIEGEFSVLDQVSIDKMMIALD 
P0A031: MLEFEQGFNHLATLKVIGVGGGGNNAVNRMIDHGMNNVEFIAINTDGQALNLSKAESKIQIGEKLTRGLGAGANPEIGKKAAEESREQIEDAIQGADMVF 
P0C1S5: MTKHYLNSKYQSEQRSSAMKKITMGTASIILGSLVYIGADSQQVNAATEATNATNNQSTQVSQATSQPINFQVQKDGSSEKSHMDDYMQHPGKVIKQNNK 
P0C1S0: MSDQHNLKEQLCFSLYNAQRQVNRYYSNKVFKKYNLTYPQFLVLTILWDESPVNVKKVVTELALDTGTVSPLLKRMEQVDLIKRERSEVDQREVFIHLTD 

我有順序的這份名單列表....但我想知道,如果這個序列(TGTCV)在這整個名單 我怎麼會找很明顯它與代碼?如何找到特定的順序在一個巨大的序列

+1

使用'string.find'功能,它會返回索引,如果它發現,否則返回-1 – avasal 2013-03-18 08:00:58

+0

你明白什麼意思? – pradyunsg 2013-03-18 08:23:18

回答

3
lol = """P0C5Z8: MLGVINRMAKKFNYKLPSMVALTLVGSAVTAHQVQAAETTQDQTTNKNVLDSNKVKATTEQAKAEVKNPTQNISGTQVYQDPAIVQPKTANNKTGNAQVS 
Q53630: MSNNFKDDFEKNRQSIDTNSHQDHTEDVEKDQSELEHQDTIENTEQQFPPRNAQRRKRRRDLATNHNKQVHNESQTSEDNVQNEAGTIDDRQVESSHSTE 
P0C6P1: MKKKLLVLTMSTLFATQLINSNHANASVTESVDKKFVVPESGINKIIPTYNEFKKAPKVNVSNLTDNKNFVASEDKLKKISDPSAASKIVDKNFVVPESK 
O69174: MPIITDVYAREVLDSRGNPTVEVEVLTESGAFGRALVPSGASTGEHEAVELRDGDKSRYLGKGVTKAVENVNEIIAPEIIEGEFSVLDQVSIDKMMIALD 
P0A031: MLEFEQGFNHLATLKVIGVGGGGNNAVNRMIDHGMNNVEFIAINTDGQALNLSKAESKIQIGEKLTRGLGAGANPEIGKKAAEESREQIEDAIQGADMVF 
P0C1S5: MTKHYLNSKYQSEQRSSAMKKITMGTASIILGSLVYIGADSQQVNAATEATNATNNQSTQVSQATSQPINFQVQKDGSSEKSHMDDYMQHPGKVIKQNNK 
P0C1S0: MSDQHNLKEQLCFSLYNAQRQVNRYYSNKVFKKYNLTYPQFLVLTILWDESPVNVKKVVTELALDTGTVSPLLKRMEQVDLIKRERSEVDQREVFIHLTD""" 

print "TGTCV" in lol 
+0

太酷了!從來不知道這是可能的。現在我有更多的理由去愛Python – 2013-03-18 08:13:20

+0

強大的單線...喜歡它 – Anuragh27crony 2013-03-18 08:44:30

1

通過簡單地使用string.count

>>> string.count('TGTCV') 
0 

,或者嘗試使用正則表達式

>>> regex = re.compile("TGTCV") 
>>> r = regex.search(string) 
# Run findall 
>>> print len(regex.findall(string)) # count of 'TGTCV' 
0