Insight Through Computing 15. Strings Operations Subscripting Concatenation Search Numeric-String...

Insight Through Computing

15. Strings

OperationsSubscriptingConcatenationSearchNumeric-String Conversions

Built-Ins: int2str,num2str, str2double


Previous Dealings

N = input(‘Enter Degree: ’)

title(‘The Sine Function’)

disp( sprintf(‘N = %2d’,N) )


A String is an Array of Characters

‘Aa7*>@ x!’

A a 7 * > @ x !

This string has length 9.


Why are Stirngs Important?

1. Numerical Data often encoded as strings

2. Genomic calculation/search


Numerical Data is Often Encoded in Strings

For example, a file containingIthaca weather data begins with the string

W07629N4226

Longitude: 76o 29’ WestLatitude: 42o 26’ North


What We Would Like to Do

W07629N4226

Get hold of the substring ‘07629’

Convert it to floating format so thatit can be involved in numerical

calculations.


Format Issues

9 as an IEEE floating point number:

9 as a character:

0100000blablahblah01001111000100010010

01000otherblablaDifferent Representation


Genomic Computations

Looking for patterns in a DNA sequence:

‘ATTCTGACCTCGATC’ACCT


Genomic Computations

Quantifying Differences:

ATTCTGACCTCGATCATTGCTGACCTCGAT

Remove?


Working With Strings


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


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


Concatenation

This: S = ‘abc’;

T = ‘xy’

R = [S T]

is the same as this: R = ‘abcxy’


Repeated Concatenation

This: s = ‘’;

for k=1:5

s = [s ‘z’];

end

is the same as this:

z = ‘zzzzz’


Replacing and AppendingCharacters

s = ‘abc’;s(2) = ‘x’ % s = ‘axc’

t = ‘abc’t(4) = ‘d’ % t = ‘abcd’

v = ‘’v(5) = ‘x’ % v = ‘ x’


Extracting Substrings

s = ‘abcdef’;

x = s(3) % x = ‘c’

x = s(2:4) % x = ‘bcd’

x = s(length(s)) % x = ‘f’


Colon Notation

s( : )

Starting Location

Ending Location


Replacing Substrings

s = ‘abcde’;

s(2:4) = ‘xyz’ % s = ‘axyze’

s = ‘abcde’

s(2:4) = ‘wxyz’ % Error


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


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: ACGTTGCAGTTCCATATGy: TGCAACGTCAAGGTATAC


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.


Comparing Strings

Built-in function strcmp

strcmp(s1,s2) is true if the strings s1 and s2 are identical.


How y is Built Up

x: ACGTTGCAGTTCCATATGy: TGCAACGTCAAGGTATAC

Start: y: ‘’ After 1 pass: y: TAfter 2 passes: y: TGAfter 3 passes: y: TGC


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


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?


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).



S: ‘ACCT’

T: ‘ATGACCTGA’

strcmp(S,T(1:4)) False



S: ‘ACCT’

T: ‘ATGACCTGA’




S: ‘ACCT’

T: ‘ATGACCTGA’




S: ‘ACCT’

T: ‘ATGACCTGA’

strcmp(S,T(4:7))) True


Pseudocode

First = 1; Last = length(S);

while S is not identical to T(First:Last) First = First + 1;

Last = Last + 1;

end


Subscript Error

S: ‘ACCT’

T: ‘ATGACTGA’

strcmp(S,T(6:9))

There’s a problem if S is not a substring of T.


Pseudocode

First = 1; Last = length(s);

while Last<=length(T) && ... ~strcmp(S,T(First:Last))

First = First + 1;

Last = Last + 1;

end


Post-Loop Processing

Loop ends when this is false:

Last<=length(T) && ...

~strcmp(S,T(First:Last))


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.


Numeric/StringConversion


String-to-Numeric Conversion

An example…

Convention: W07629N4226

Longitude: 76o 29’ West Latitude: 42o 26’ North


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.


Numeric-to-String Conversion

x = 1234;

s = int2str(x); % s = ‘1234’

x = pi;

s = num2str(x,’%5.3f’); % s =‘3.142’


Problem

Given a date in the format ‘mm/dd’

specify the next day in the same format


y = Tomorrow(x)

x y

02/28 03/01

07/13 07/14

12/31 01/01


Get the Day and Month

month = str2double(x(1:2));

day = str2double(x(4:5));

Thus, if x = ’02/28’ then month is assignedthe numerical value of 2 and day is assigned the numerical value of 28.


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;


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


The New Day String

Compute newDay (numerical) and convert…

d = int2str(newDay);if length(d)==1 d = ['0' d];end


The New Month String

Compute newMonth (numerical) and convert…

m = int2str(newMonth);

if length(m)==1;

m = ['0' m];

end


The Final Concatenation

y = [m '/' d];


Some other useful string functionsstr= ‘Cs 1112’;

length(str) % 7isletter(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)


ASCII characters(American Standard Code for Information Interchange)

ascii code Character: :: :65 ‘A’66 ‘B’67 ‘C’: :90 ‘Z’: :

ascii code Character

: :: :48 ‘0’49 ‘1’50 ‘2’: :57 ‘9’: :


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


Arithmetic and relational ops on characters

• ‘c’-‘a’ gives 2• ‘6’-‘5’ gives 1• letter1=‘e’; letter2=‘f’; • letter1-letter2 gives -1

• ‘c’>’a’ gives true• letter1==letter2 gives false

• ‘A’ + 2 gives 67• char(‘A’+2) gives ‘C’


Example: toUpperWrite a function toUpper(cha) to convert character cha to upper case if cha is a lower case letter. Return the converted letter. If cha is not a lower case letter, simply return the character cha.

Hint: Think about the distance between a letter and the base letter ‘a’ (or ‘A’). E.g.,

a b c d e f g h …

A B C D E F G H …

Of course, do not use Matlab function upper!

distance = ‘g’-‘a’ = 6 = ‘G’-‘A’


function up = toUpper(cha)% up is the upper case of character cha.% If cha is not a letter then up is just cha.

up= cha;

cha is lower case if it is between ‘a’ and ‘z’



up= cha;

if ( cha >= 'a' && cha <= 'z' )

% Find distance of cha from ‘a’

end



up= cha;

if ( cha >= 'a' && cha <= 'z' )

% Find distance of cha from ‘a’ offset= cha - 'a';

% Go same distance from ‘A’ end



up= cha;

if ( cha >= 'a' && cha <= 'z' )

% Find distance of cha from ‘a’ offset= cha - 'a';

% Go same distance from ‘A’ up= char('A' + offset);end

Date post:	22-Dec-2015
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Insight Through Computing 15. Strings Operations Subscripting Concatenation Search Numeric-String...

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