我正在学习haskell。我正在从文本文件中读取一个字符串,并且需要将此字符串变成char列表。我需要在Char列表中转换此字符串
输入文件是这样的:
Individuo A; TACGATCAAAGCT
Individuo B; AATCGCAT
Individuo C; TAAATCCGATCAAAGAGAGGACTTA
我需要转换这个字符串
S1 = "AAACCGGTTAAACCCGGGG" in S1 =
["A","A","A","C","C","G","G","T","T","A","A","A","C","C","C","G","G","G","G"]
or S1 =
['A','A','A','C','C','G','G','T','T','A','A','A','C','C','C','G','G','G','G']
但它们是由 “;” 分隔
我该怎么办?
我该怎么办?
得到两份名单后,我送他们到这个代码:
lcsList :: Eq a => [a] -> [a] -> [a]
lcsList [] _ = []
lcsList _ [] = []
lcsList (x:xs) (y:ys) = if x == y
then x : lcsList xs ys
else
let lcs1 = lcsList (x:xs) ys
lcs2 = lcsList xs (y:ys)
in if (length lcs1) > (length lcs2)
then lcs1
else lcs2
'String'只是'[Char]'的一个类型别名。我不明白你的问题。 – Pubby
我的问题,我从一个文件中读取这个字符串,并且需要通过它作为这个代码的一个例子 lcsList :: Eq a => [a] - > [a] - > [a] lcsList [] _ = [ ] lcsList _ [] = [] lcsList(X:XS)(Y:YS)=如果x ==ý 则x:XS YS lcsList 别的 让LCS1 lcsList =(X:XS)YS LCS2 lcsList = xs(y:ys) in if(length lcs1)>(length lcs2) 然后LCS1 其他LCS2 –