Question

这正是我想要做的:

• 两条以上线和“沥青”(无脱氧核糖核酸/RNA序列或类似体,只是每条1 000件)

我已经做了一些与配对(两套)的工作,然而,“gaps”在试图把一个以上奶粉结合起来时给我造成了一些问题。

<<>Example(目前测试一次):

ABCDEF
ABGHCEEF
AJKLBCDYEOF

AB--CDEF
ABGHCEEF
=======================
AB--C-EF

A-B--C--E-F
AJKLBCDYEOF
=======================
A----C--E-F

www.un.org/spanish/ecosoc 另一个(更说明性)的例子:

http://nest.drkameleon.com
http://www.google.com
http://www.yahoo.com

http://nest.drkameleon.com
http://-www.--google--.com

=======================
http://----.------le--.com

http://----.------le--.com
http://-www.-----yahoo.com

=======================
http://----.----------.com

www.un.org/spanish/ecosoc 目前做的是:

Sort the strings (longer strings come first in the list)
Align the first pair : A-B and get the result (let s say R1)
Then align the second pair : R1 and C (result in R2)
Then align the third pair : R2 and D
And so on...

因此,你们的思想是什么? 我怎么办? 是否有更好的办法? (当然必须......)

然而,在Perl/Python或大致如此的情况下,任何类型的法典/参考都不会受到欢迎!

Answer 1

I think you may be able to cast this problem as a more general string diff problem instead of a string alignment. Consider how GNU diff is used for finding differences between two files, and use the same algorithms as are used to perform an N-way diff.

我不敢肯定这种做法的时间/表面复杂性是否适合你们的需要,但至少可以这样考虑这个问题。

Answer 2

There is an algorithm based on Levenshtein algorithm to compute the longest common sequence, with optional spaces. Not sure if that helps.

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