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

Insight Through Computing

15. Strings

OperationsSubscriptingConcatenationSearchNumeric-String Conversions

Built-Ins: int2str,num2str, str2double

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Previous Dealings

N = input(‘Enter Degree: ’)

title(‘The Sine Function’)

disp( sprintf(‘N = %2d’,N) )

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A String is an Array of Characters

‘Aa7*>@ x!’

A a 7 * > @ x !

This string has length 9.

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Why are Stirngs Important?

1. Numerical Data often encoded as strings

2. Genomic calculation/search

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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

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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.

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Format Issues

9 as an IEEE floating point number:

9 as a character:

0100000blablahblah01001111000100010010

01000otherblablaDifferent Representation

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Genomic Computations

Looking for patterns in a DNA sequence:

‘ATTCTGACCTCGATC’ACCT

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Genomic Computations

Quantifying Differences:

ATTCTGACCTCGATCATTGCTGACCTCGAT

Remove?

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Working With Strings

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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

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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

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Concatenation

This: S = ‘abc’;

T = ‘xy’

R = [S T]

is the same as this: R = ‘abcxy’

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Repeated Concatenation

This: s = ‘’;

for k=1:5

s = [s ‘z’];

end

is the same as this:

z = ‘zzzzz’

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Replacing and AppendingCharacters

s = ‘abc’;s(2) = ‘x’ % s = ‘axc’

t = ‘abc’t(4) = ‘d’ % t = ‘abcd’

v = ‘’v(5) = ‘x’ % v = ‘ x’

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Extracting Substrings

s = ‘abcdef’;

x = s(3) % x = ‘c’

x = s(2:4) % x = ‘bcd’

x = s(length(s)) % x = ‘f’

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Colon Notation

s( : )

Starting Location

Ending Location

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Replacing Substrings

s = ‘abcde’;

s(2:4) = ‘xyz’ % s = ‘axyze’

s = ‘abcde’

s(2:4) = ‘wxyz’ % Error

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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

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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

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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.

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Comparing Strings

Built-in function strcmp

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

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How y is Built Up

x: ACGTTGCAGTTCCATATGy: TGCAACGTCAAGGTATAC

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

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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

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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?

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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).

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A DNA Search Problem

S: ‘ACCT’

T: ‘ATGACCTGA’

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

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A DNA Search Problem

S: ‘ACCT’

T: ‘ATGACCTGA’

strcmp(S,T(2:5)) False

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A DNA Search Problem

S: ‘ACCT’

T: ‘ATGACCTGA’

strcmp(S,T(3:6)) False

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A DNA Search Problem

S: ‘ACCT’

T: ‘ATGACCTGA’

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

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Pseudocode

First = 1; Last = length(S);

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

Last = Last + 1;

end

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Subscript Error

S: ‘ACCT’

T: ‘ATGACTGA’

strcmp(S,T(6:9))

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

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Pseudocode

First = 1; Last = length(s);

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

First = First + 1;

Last = Last + 1;

end

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Post-Loop Processing

Loop ends when this is false:

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

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

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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.

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Numeric/StringConversion

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String-to-Numeric Conversion

An example…

Convention: W07629N4226

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

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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.

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Numeric-to-String Conversion

x = 1234;

s = int2str(x); % s = ‘1234’

x = pi;

s = num2str(x,’%5.3f’); % s =‘3.142’

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Problem

Given a date in the format ‘mm/dd’

specify the next day in the same format

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y = Tomorrow(x)

x y

02/28 03/01

07/13 07/14

12/31 01/01

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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.

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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;

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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

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The New Day String

Compute newDay (numerical) and convert…

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

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The New Month String

Compute newMonth (numerical) and convert…

m = int2str(newMonth);

if length(m)==1;

m = ['0' m];

end

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The Final Concatenation

y = [m '/' d];

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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)

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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’: :

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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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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’

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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’

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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’

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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;

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

% Find distance of cha from ‘a’

end

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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;

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

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

% Go same distance from ‘A’ end

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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;

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

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

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