String Vector for-loop

I recently collected some point data, and the trial number or independent variable is a string, such as:

X = ['a1' 'a2' 'a3' 'b1' 'b2' 'b3'];

I would like to create a for loop using the [1 2 3 1 2 3]. Is there any way to seperate the number and letter, or a way to us a string such as this as a for loop condition.

I do not want to use the index, and am trying to make this work for a much larger set of data, so I am trying to have it work automatically.

[Adam]

Do you really mean a string or a char? What you show just evaluates as

'a1a2a3b1b2b3'

with all the individual quotation marks irrelevant. Is that what you actually have?

X = ["a1" "a2" "a3" "b1" "b2" "b3"];

would be a string array, or

X = ['a1'; 'a2'; 'a3'; 'b1'; 'b2'; 'b3'];

would give you a char array from which extracting the numbers is simply a case of taking the 2nd column, if all letters and numbers will be, as in your example, no longer than a single character long each.

[zachary schneider]

X = ['a1'; 'a2'; 'a3'; 'b1'; 'b2'; 'b3']; is the format, so yes it is a char i believe

[Adam]

So simply

str2num( X(:,2) )

would give what you want.

Answer 1

[Stephen23]

>> str = {'a1','a2','a3','b1','b2','b3'};
>> vec = sscanf([str{:}],'%*[ab]%f',[1,Inf])
vec =
  1  2  3  1  2  3
>> vec = str2double(regexp(str,'\d+$','match','once'))
vec =
  1  2  3  1  2  3
>> vec = str2double(regexprep(str,'^[a-z]+',''))
vec =
  1  2  3  1  2  3

[zachary schneider]

sorry if my formatting was off, the str data looks like this:

str =

120×3 char array

'a1 '

'a2 '

'a3 '

'b1 '

'b2 '

'b3 '

i am getting an error with the sscanf command. How can I make my char array in to a cell array

[Stephen23]

>> cha = ['a1 ';'a2 ';'a3 ';'b1 ';'b2 ';'b3 ']
cha =
a1 
a2 
a3 
b1 
b2 
b3 
>> vec = sscanf(cha(:,2:3).','%f',[1,Inf])
vec =
   1   2   3   1   2   3

"How can I make my char array in to a cell array"

out = cellstr(cha)

[Jos (10584)]

This example strongly suggests that it is always a single digit following a single character that is relevant. If so, this will work too:

cha = ['a1 ';'a2 ';'a3 ';'b1 ';'b2 ';'b3 ']
vec = cha(:,2) - '0'