15. Strings
Operations
Subscripting
Concatenation
Search
Numeric-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 containing
Ithaca weather data begins with the string
W07629N4226
Longitude:
Latitude:
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76o 29’ West
42o 26’ North
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.
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Format Issues
9 as an IEEE floating point number:
0100000blablahblah01001111000100010010
9 as a character:
01000otherblabla
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Different
Representation
Genomic Computations
Looking for patterns in a DNA sequence:
‘ATTCTGACCTCGATC’
ACCT
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Genomic Computations
Quantifying Differences:
ATTCTGACCTCGATC
ATTGCTGACCTCGAT
Remove?
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Working With Strings
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Strings Can Be Assigned
to Variables
S = ‘N = 2’
‘N = 2’
S
N = 2;
S = sprintf(‘N = %1d’,N)
sprintf produces a formatted string using fprintf rules
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Strings Have a Length
s = ‘abc’;
n = length(s);
% n = 3
s = ‘’;
n = length(s)
% the empty string
% n = 0
s = ‘ ‘;
n = length(s)
% single blank
% 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 Appending
Characters
s = ‘abc’;
s(2) = ‘x’
%
s = ‘axc’
t = ‘abc’
t(4) = ‘d’
%
t = ‘abcd’
v = ‘’
v(5) = ‘x’
%
v = ‘
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x’
Extracting Substrings
s = ‘abcdef’;
x
x
x
= s(3)
= s(2:4)
= s(length(s))
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%
%
%
x = ‘c’
x = ‘bcd’
x = ‘f’
Colon Notation
s(
Starting
Location
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:
)
Ending
Location
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
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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: ACGTTGCAGTTCCATATG
y: 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: ACGTTGCAGTTCCATATG
y: TGCAACGTCAAGGTATAC
Start:
After 1 pass:
After 2 passes:
After 3 passes:
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y:
y:
y:
y:
‘’
T
TG
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
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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))
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
False
A DNA Search Problem
S: ‘ACCT’
T: ‘ATGACCTGA’
strcmp(S,T(2:5))
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
False
A DNA Search Problem
S: ‘ACCT’
T: ‘ATGACCTGA’
strcmp(S,T(3:6))
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
False
A DNA Search Problem
S: ‘ACCT’
T: ‘ATGACCTGA’
strcmp(S,T(4:7)))
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
True
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/String
Conversion
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String-to-Numeric Conversion
An example…
Convention:
W07629N4226
Longitude:
Latitude:
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76o 29’ West
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.
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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
02/28
07/13
12/31
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y
03/01
07/14
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 assigned
the 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 functions
str= ‘Cs 1112’;
length(str)
isletter(str)
isspace(str)
lower(str)
upper(str)
%
%
%
%
%
7
[1 1 0 0 0 0 0]
[0 0 1 0 0 0 0]
‘cs 1112’
‘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
:
:
65
66
67
:
90
:
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Character
:
:
‘A’
‘B’
‘C’
:
‘Z’
:
ascii code
:
:
48
49
50
:
57
:
Character
:
:
‘0’
‘1’
‘2’
:
‘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
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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: toUpper
Write 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 …
distance = ‘g’-‘a’ = 6 = ‘G’-‘A’
A B C D E F G H …
Of course, do not use Matlab function upper!
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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
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