topic Re: Count the number of Ts in a sequence in Discussions

Count the number of Ts in a sequence

padhikari — Fri, 30 Nov 2018 15:46:17 GMT

Hello,

I am trying to find a formula that will count highest number times a letter is repeated consecuteivley in a sequence. I have attached an example where I am trying to write formula for poly Ts column, and it will generate count of Ts in a sequence consectively.

Thank you,

Pratish

Re: Count the number of Ts in a sequence

gzmorgan0 — Thu, 29 Nov 2018 09:51:08 GMT

A regular expression expert might have a nice pattern to scan and find all matches, but that is beyond my REGEX skills.

I have provided two solutions that can be done using column formulas. Both might need some explanation.

The first uses nested character functions, the second uses the ShortestEditScript() function. By the way, you did not specify if you are counting T sequences prior to (N1), both use the entire string. The example table is attached and explanations are below

Assume s2 = "(N1:25252525)AACCAA(N1)GACGTTAACAGTTCTTTG";

Character functions: Words(), Sort List(), Reverse(), list[n], Length()

Length(Reverse(Sort List(words(s2,"ACGN():0123456789")))[1]);

Here is the log output for the respective functions

//:*/
words(s2,"ACGN():0123456789")
/*:
{"TT", "TT", "TTT"}
//:*/
Reverse(Sort List(words(s2,"ACGN():0123456789")))
/*:
{"TTT", "TT", "TT"}
//:*/
Reverse(Sort List(words(s2,"ACGN():0123456789")))[1]
/*:
"TTT"
//:*/
Length(Reverse(Sort List(words(s2,"ACGN():0123456789")))[1]);
/*:
3

ShortestEditScript() is an interesting function. Script 9_Extra_ShortestEditDistance.jsl written for JSL Companion, Applications of the JMP Scripting Language Second Edition document 4 different methods for using this powerful and useful function. For this example, I am using Sequnces() and requesting matrix output. It would take too much space to document this completely, in this forum, so I will just show the results and add a few comments.

 msed = Shortest Edit Script( Strings(s2, Repeat("T",length(s2)),matrix(1)) );
 maximum(msed[loc(msed[0,1]==0),4]);

The two strings being compared are s2 and a string of all T's created by function Repeat("T", length(s2) ).

 msed = Shortest Edit Script( Strings(s2, Repeat("T",length(s2)),matrix(1)) );

/*:
[-1 1 . 27,
  0 28 1 2,
 -1 30 . 5,
  0 35 3 2,
 -1 37 . 1,
  0 38 5 3,
 -1 41 . 1,
  1 . 8 34]
  
/* The matrix output  n x 4 where n = nrow(msed)
Column1:  -1 | 1| 0  -1-->remove, 1-->insert, 0-->common
Column2:  position in the 1st string .-->missing / not found
Column3:  position in the 2nd string .-->missing / not found
Column4:  length
*/

So now it is a matter of finding the locations in the 1st column of the matrix with 0's (matches/common/T's) this can be done with the loc() function. Then the length of the matching sequence is in the 4th column, so just find the maximum. Note msed[0,1] represents the 1st column of the matrix msed.

loc(msed[0,1]==0)
/*:
[2, 4, 6] 
//:*/
msed[loc(msed[0,1]==0),4]
/*:
[2, 2, 3]
//:*/
 maximum(msed[loc(msed[0,1]==0),4]);
/*:
3

It will be interesting to see other solutions.

Re: Count the number of Ts in a sequence

pmroz — Thu, 29 Nov 2018 13:46:55 GMT

Here's a simple brute force approach; not sure of the performance relative to @gzmorgan0's methods.

s2 = "(N1:25252525)AACCAA(N1)GACGTTAACAGTTCTTTG";
len = length(s2);
tstring = repeat("T", len);
maxlen  = 0;
for (i = len, i >= 1, i--,
	if (contains(s2, tstring),
		maxlen = i;
		break();
		,
		tstring = substr(tstring, 2);
	);
);
show(maxlen);

Re: Count the number of Ts in a sequence

gzmorgan0 — Thu, 29 Nov 2018 18:27:54 GMT

Table formula using ShortestEditScript() used s2 in a portion of the formula that should have been Sequence. Table with corrected function is attached.

Re: Count the number of Ts in a sequence

ian_jmp — Fri, 30 Nov 2018 18:33:17 GMT

In the spirit of 'other solutions' here's another brute force one:

NamesDefaultToHere(1);

// Given a string and a single character, finds the longest sequence of that character
// and returns the length and starting position of that sequence. If the sequence
// occurs more than once, only the first is identified
findLongestRepeatedCharacter =
Function({str, char}, {Default Local},
	n = Length(str);
	count = 0;
	currentCount = 1;
	// Traverse the string except for the last character
	for (i = 1, i <= n-1, i++,
		thisChar = Substr(str, i, 1);
		nextChar = Substr(str, i+1, 1);
		// If the current character and the next are both 'char' ...
		if((thisChar == char & nextChar == char),
			// ... increment 'currentCount'
			currentCount++,
			// ... else if they're not ...
			if(currentCount > count,
				// ... record 'currentCount' if it's bigger than we've seen so far
				count = currentCount;
				);
			// ... and reset 'currentCount'
			currentCount = 1;
			);
		);
	// Build the sequence we've found
	seq = Repeat(char, count);
	// Find where it occurs
	pos = Munger(str, 1, seq);
	// Return the results
	if (pos == 0,
		EvalList({0, pos}),
		EvalList({count, pos})
		);
	);

// Try it out
str = "(N1:25252525)AACCAA(N1)GACGTTAACAGTTCTTTG";
Print(findLongestRepeatedCharacter(str, "T"));
Print(findLongestRepeatedCharacter(str, "A"));
Print(findLongestRepeatedCharacter(str, "X"));