find a string pattern in a string array - regex

Question

I need to count the occurrence of specified patterns in the input strands and produces a report for each pattern.
The input string would contain 1 AA AATTCGAA end
the 1 signifies one pattern to search for and AA is the pattern and the next is the part you would search AA in.
My idea is to :
public static void main(String[] args){
Scanner s = new Scanner(System.in);
System.out.println("How many patterns do you want and enter patterns and DNA Sequence(type 'end' to signify end):");
String DNA = s.nextLine();
process(DNA);
}
public static void process(String DNA){
String number = DNA.replaceFirst(".*?(\\d+).*", "$1");
int N = Integer.parseInt(number);
DNA.toUpperCase();
String[] DNAarray;
DNAarray = DNA.split(" ");
for(int i=1; i<N; i++){
int count=0;
for(int j =0; j < DNAarray.length; j++) {
if(DNAarray[i+N].contains(DNAarray[i])){
count= count++;
}
}
System.out.println("Pattern:"+DNAarray[i]+ "Count:"+count);
}

Answer 1

This should do it:
using System;
using System.Text.RegularExpressions;
public class Program
{
public void Main()
{
Console.WriteLine(PatternCount("1 AA AADDRRSSAA"));
}
public int PatternCount(string sDNA) {
Regex reParts = new Regex("(\\d+)\\s(\\w\\w)\\s(\\w+)");
Match m = reParts.Match(sDNA);
if (m.Success)
{
return Regex.Matches(m.Groups[3].Value, m.Groups[2].Value).Count;
}
else
return 0;
}
}
First RE splits the input into count, pattern and data. (Not sure why you want to limit the number of patterns to search for. This code ignores that. Modify after your needs...)
Second RE equals the pattern wanted and "Matches" counts the number of occurrences. Work from here.
Regards
(I feel good today, doing people's work ;))

Answer 2

Really no need to put the number of searches. And, actually this could be done
with a single regex. I can't remember if Dot-Net supports the \G anchor,
but this is really not necessary anyway. I left it in.
Each Match:
Finds a new key.
Captures the keys sub-string matches at the end.
Advances the search position by just the key.
So, sit in a Find loop.
On each match print the 'Key' capture buffer,
then print the capture collection 'Values' count.
Thats all there is to it.
The regex will search for overlapping keys. To change it to exclusive keys,
change the = to : as shown in the comments.
You can also make it a little more specific. For example, change all the \w's to [A-Z], etc...
The regex:
(?:
^ [ \d]*
| \G
)
(?<Key> \w+ ) #_(1)
[ ]+
(?=
(?: \w+ [ ]+ )*
(?= \w )
(?:
(?= # <- Change the = to : to get non-overlapped matches
(?<Values> \1 ) #_(2)
)
| .
)*
$
)
This is a perl test case
# $str = '2 6 AA TT PP AAATTCGAA';
# $count = 0;
#
# while ( $str =~ /(?:^[ \d]*|\G)(\w+)[ ]+(?=(?:\w+[ ]+)*(?=\w)(?:(?=(\1)(?{ $count++ }))|.)*$)/g )
# {
# print "search = '$1'\n";
# print "found = '$count'\n";
# $count = 0;
#
# }
#
# Output >>
#
# search = 'AA'
# found = '3'
# search = 'TT'
# found = '1'
# search = 'PP'
# found = '0'
#
#