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15. Strings

15. Strings. Operations Subscripting Concatenation Search Numeric-String Conversions Built-Ins: int2str,num2str, str2double. Previous Dealings. N = input(‘ Enter Degree: ’) title(‘ The Sine Function ’) disp( sprintf(‘N = %2d’,N) ).

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15. Strings

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  1. 15. Strings Operations Subscripting Concatenation Search Numeric-String Conversions Built-Ins: int2str,num2str, str2double

  2. Previous Dealings N = input(‘Enter Degree: ’) title(‘The Sine Function’) disp( sprintf(‘N = %2d’,N) )

  3. A a 7 * > @ x ! A String is an Array of Characters ‘Aa7*>@ x!’ This string has length 9.

  4. Why are Stirngs Important? • Numerical Data often encoded as strings 2. Genomic calculation/search

  5. Numerical Data is Often Encoded in Strings For example, a file containing Ithaca weather data begins with the string W07629N4226 Longitude:76o 29’ West Latitude:42o 26’ North

  6. What We Would Like to Do W07629N4226 Get hold of the substring ‘07629’ Convert it to floating format so that it can be involved in numerical calculations.

  7. Format Issues 9 as an IEEE floating point number: 9 as a character: 0100000blablahblah01001111000100010010 Different Representation 01000otherblabla

  8. Genomic Computations Looking for patterns in a DNA sequence: ‘ATTCTGACCTCGATC’ ACCT

  9. Genomic Computations Quantifying Differences: ATTCTGACCTCGATC ATTGCTGACCTCGAT Remove?

  10. Working With Strings

  11. Strings Can Be Assignedto Variables S = ‘N = 2’ N = 2; S = sprintf(‘N = %1d’,N) ‘N = 2’ S sprintf produces a formatted string using fprintf rules

  12. Strings Have a Length s = ‘abc’; n = length(s); % n = 3 s = ‘’; % the empty string n = length(s) % n = 0 s = ‘ ‘; % single blank n = length(s) % n = 1

  13. Concatenation This: S = ‘abc’; T = ‘xy’ R = [S T] is the same as this: R = ‘abcxy’

  14. Repeated Concatenation This: s = ‘’; for k=1:5 s = [s ‘z’]; end is the same as this: z = ‘zzzzz’

  15. Replacing and AppendingCharacters s = ‘abc’; s(2) = ‘x’ % s = ‘axc’ t = ‘abc’ t(4) = ‘d’ % t = ‘abcd’ v = ‘’ v(5) = ‘x’ % v = ‘ x’

  16. Extracting Substrings s = ‘abcdef’; x = s(3) % x = ‘c’ x = s(2:4) % x = ‘bcd’ x = s(length(s)) % x = ‘f’

  17. Colon Notation s( : ) Starting Location Ending Location

  18. Replacing Substrings s = ‘abcde’; s(2:4) = ‘xyz’ % s = ‘axyze’ s = ‘abcde’ s(2:4) = ‘wxyz’ % Error

  19. Question Time s = ‘abcde’; for k=1:3 s = [ s(4:5) s(1:3)]; end What is the final value of s ? A abcde B. bcdea C. eabcd D. deabc

  20. Problem: DNA Strand x is a string made up of the characters ‘A’, ‘C’, ‘T’, and ‘G’. Construct a string Y obtained from x by replacinig each A by T, each T by A, each C by G, and each G by C x: ACGTTGCAGTTCCATATG y: TGCAACGTCAAGGTATAC

  21. function y = Strand(x) % x is a string consisting of % the characters A, C, T, and G. % y is a string obtained by % replacing A by T, T by A, % C by G and G by C.

  22. Comparing Strings Built-in function strcmp strcmp(s1,s2) is true if the strings s1 and s2 are identical.

  23. How y is Built Up x: ACGTTGCAGTTCCATATG y: TGCAACGTCAAGGTATAC Start: y: ‘’ After 1 pass:y: T After 2 passes: y: TG After 3 passes: y: TGC

  24. for k=1:length(x) if strcmp(x(k),'A') y = [y 'T']; elseif strcmp(x(k),'T') y = [y 'A']; elseif strcmp(x(k),'C') y = [y 'G']; else y = [y 'C']; end end

  25. A DNA Search Problem Suppose S and T are strings, e.g., S: ‘ACCT’ T: ‘ATGACCTGA’ We’d like to know if S is a substring of T and if so, where is the first occurrance?

  26. function k = FindCopy(S,T) % S and T are strings. % If S is not a substring of T, % then k=0. % Otherwise, k is the smallest % integer so that S is identical % to T(k:k+length(S)-1).

  27. A DNA Search Problem S: ‘ACCT’ T: ‘ATGACCTGA’ strcmp(S,T(1:4))  False

  28. A DNA Search Problem S: ‘ACCT’ T: ‘ATGACCTGA’ strcmp(S,T(2:5))  False

  29. A DNA Search Problem S: ‘ACCT’ T: ‘ATGACCTGA’ strcmp(S,T(3:6))  False

  30. A DNA Search Problem S: ‘ACCT’ T: ‘ATGACCTGA’ strcmp(S,T(4:7)))  True

  31. Pseudocode First = 1; Last = length(S); while S is not identical to T(First:Last) First = First + 1; Last = Last + 1; end

  32. Subscript Error S: ‘ACCT’ T: ‘ATGACTGA’ strcmp(S,T(6:9)) There’s a problem if S is not a substring of T.

  33. Pseudocode First = 1; Last = length(s); while Last<=length(T) && ... ~strcmp(S,T(First:Last)) First = First + 1; Last = Last + 1; end

  34. Post-Loop Processing Loop ends when this is false: Last<=length(T) && ... ~strcmp(S,T(First:Last))

  35. Post-Loop Processing if Last>length(T) % No Match found k=0; else % There was a match k=First; end The loop ends for one of two reasons.

  36. Numeric/StringConversion

  37. String-to-Numeric Conversion An example… Convention: W07629N4226 Longitude: 76o 29’ West Latitude: 42o 26’ North

  38. String-to-Numeric Conversion S = ‘W07629N4226’ s1 = s(2:4); x1 = str2double(s1); s2 = s(5:6); x2 = str2double(s2); Longitude = x1 + x2/60 There are 60 minutes in a degree.

  39. Numeric-to-String Conversion x = 1234; s = int2str(x); % s = ‘1234’ x = pi; s = num2str(x,’%5.3f’); % s =‘3.142’

  40. Problem Given a date in the format ‘mm/dd’ specify the next day in the same format

  41. y = Tomorrow(x) x y 02/28 03/01 07/13 07/14 12/31 01/01

  42. Get the Day and Month month = str2double(x(1:2)); day = str2double(x(4:5)); Thus, if x = ’02/28’ then month is assigned the numerical value of 2 and day is assigned the numerical value of 28.

  43. L = [31 28 31 30 31 30 31 31 30 31 30 31]; if day+1<=L(month) % Tomorrow is in the same month newDay = day+1; newMonth = month;

  44. L = [31 28 31 30 31 30 31 31 30 31 30 31]; else % Tomorrow is in the next month newDay = 1; if month <12 newMonth = month+1; else newMonth = 1; end

  45. The New Day String Compute newDay (numerical) and convert… d = int2str(newDay); if length(d)==1 d = ['0' d]; end

  46. The New Month String Compute newMonth (numerical) and convert… m = int2str(newMonth); if length(m)==1; m = ['0' m]; end

  47. The Final Concatenation y = [m '/' d];

  48. Some other useful string functions str= ‘Cs 1112’; length(str) % 7 isletter(str) % [1 1 0 0 0 0 0] isspace(str) % [0 0 1 0 0 0 0] lower(str) % ‘cs 1112’ upper(str) % ‘CS 1112’ ischar(str) % Is str a char array? True (1) strcmp(str(1:2),‘cs’) % Compare strings str(1:2) & ‘cs’. False (0) strcmp(str(1:3),‘CS’) % False (0)

  49. ascii code Character : : : : 65 ‘A’ 66 ‘B’ 67 ‘C’ : : 90 ‘Z’ : : ascii code Character : : : : 48 ‘0’ 49 ‘1’ 50 ‘2’ : : 57 ‘9’ : : ASCII characters(American Standard Code for Information Interchange)

  50. Character vs ASCII code str= ‘Age 19’ %a 1-d array of characters code= double(str) %convert chars to ascii values str1= char(code) %convert ascii values to chars

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