模式发现算法 [英] pattern finding algorithm

问题描述

我正在检查你们是否能够帮我找到模式的算法。我有大约2000个对齐的短序列（长度为9）。我希望能够提取相同位置的所有常见模式并报告出现次数。


例如以下内容：


ACGCATTCA

ACTGGATAC

TCAGCCATC


我想要以下输出（其中一个句号表示任何字符）


（AC .... T ..）2次出现（序列1和2之间的模式）

（.CG..C）2次出现（序列2和3之间的模式）

（.C ...... 。）2次出现（序列1和3之间的模式）


如你所见，我现在计划这样做的方式需要总和（n-1 ... 1）比较。有没有更有效的方法来做更少的比较？


谢谢

Hi,

I''m checking to see if you guys may be able to help me with an algorithm for finding patterns. I have around 2000 short sequences (of length 9) that are aligned. I want to be able to extract all common patterns on the same positions and report the number of occurrences.

For example in the following:

ACGCATTCA
ACTGGATAC
TCAGCCATC

I would like the following output (where a full stop represents any character)

(AC....T..) 2 occurrences (pattern between sequence 1 and 2)
(.C.G...C) 2 occurrences (pattern between sequence 2 and 3)
(.C.......) 2 occurrences (pattern between sequence 1 and 3)

As you can see, the way that I am planning on doing this now requires sum(n-1...1) comparisons. Is there a more efficient way of doing this with less comparisons?

Thanks

推荐答案

我正在检查你们是否能够帮助我找到寻找模式的算法。我有大约2000个对齐的短序列（长度为9）。我希望能够提取相同位置的所有常见模式并报告出现次数。


例如以下内容：


ACGCATTCA

ACTGGATAC

TCAGCCATC


我想要以下输出（其中一个句号表示任何字符）


（AC .... T ..）2次出现（序列1和2之间的模式）

（.CG..C）2次出现（序列2和3之间的模式）

（.C ...... 。）2次出现（序列1和3之间的模式）


如你所见，我现在计划这样做的方式需要总和（n-1 ... 1）比较。有没有更有效的方法来做到这一点与较少的比较？


谢谢

Hi,

I''m checking to see if you guys may be able to help me with an algorithm for finding patterns. I have around 2000 short sequences (of length 9) that are aligned. I want to be able to extract all common patterns on the same positions and report the number of occurrences.

For example in the following:

ACGCATTCA
ACTGGATAC
TCAGCCATC

I would like the following output (where a full stop represents any character)

(AC....T..) 2 occurrences (pattern between sequence 1 and 2)
(.C.G...C) 2 occurrences (pattern between sequence 2 and 3)
(.C.......) 2 occurrences (pattern between sequence 1 and 3)

As you can see, the way that I am planning on doing this now requires sum(n-1...1) comparisons. Is there a more efficient way of doing this with less comparisons?

Thanks

巧合的是，正则表达式使用 ;完全停止正如你所指定的那样（但在正则表达式中，它被称为点）。我可以提供更多详细信息，但此刻我很匆忙：

Coincidentally enough, regular expressions use the "full stop" as you have specified (but in a regex, it''s called a "dot"). I can give more detail, but am in a rush at the moment:

展开 | 选择 | Wrap | 行号

巧合的是，正则表达式使用完全停止。正如你所指定的那样（但在正则表达式中，它被称为点）。我可以提供更多详细信息，但此刻我很匆忙：

Coincidentally enough, regular expressions use the "full stop" as you have specified (but in a regex, it''s called a "dot"). I can give more detail, but am in a rush at the moment:

展开 | 选择 | Wrap | 行号

我想我已经得到了它：

I think that I''ve got it:

展开 | 选择 | Wrap < span class =codeDivider> | 行号

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