从python中给定的字符串列表中查找公共超弦
我的输入中有我的字符串从python中给定的字符串列表中查找公共超弦,python,Python,我的输入中有我的字符串 'ATTAGACCTG', 'CCTGCCGGAA', 'AGACCTGCCG', 'GCCGGAATAC' 在输出中,我想要公共的最短超弦 ATTAGACCTGCCGGAATAC 我已经使用lambda表达式完成了它,但我希望它不使用lambda表达式 from itertools import * print min((reduce(lambda s,w:(w+s[max(i*(s[:i]==w[-i:])for i in range(99)):],s)[w in
'ATTAGACCTG', 'CCTGCCGGAA', 'AGACCTGCCG', 'GCCGGAATAC'
在输出中,我想要公共的最短超弦
ATTAGACCTGCCGGAATAC
我已经使用lambda表达式完成了它,但我希望它不使用lambda表达式
from itertools import *
print min((reduce(lambda s,w:(w+s[max(i*(s[:i]==w[-i:])for i in range(99)):],s)[w in s],p)
for p in permutations(input())),key=len)
我在没有使用lambda表达式的情况下尝试了它,但得到了错误的输出
from itertools import permutations
def solve(*strings):
"""
Given a list of strings, return the shortest string that contains them all.
"""
return min((simplify(p) for p in permutations(strings)), key=len)
def prefixes(s):
"""
Return a list of all the prefixes of the given string (including itself), in ascending order (from shortest to longest).
"""
return [s[:i+1] for i in range(len(s))]
return [(i,s[:i+1]) for i in range(len(s))][::-1]
def simplify(strings):
"""
Given a list of strings, concatenate them wile removing overlaps between
successive elements.
"""
ret = ''
for s in strings:
if s in ret:
break
for i, prefix in reversed(list(enumerate(prefixes(s)))):
if ret.endswith(prefix):
ret += s[i+1:]
break
else:
ret += s
return ret
print solve('ATTAGACCTG', 'CCTGCCGGAA', 'AGACCTGCCG', 'GCCGGAATAC')
我的输出错误:
ATTAGACCTGCCGGAA
看来
if s in ret:
break
应该是
if s in ret:
continue
我想那会解决的
另外,第二个return语句是多余的-它不是模拟了反转(list(enumerate(前缀))
最后,我想我更喜欢您最初的map reduce解决方案 简单看一下,你就知道前缀中只有一个返回语句会起作用,对吧?你已经读过了吗?看起来你也在做DNA测序;)