Indexing string in loop

Hi everyone!

I have a problem with a loop: I am trying to write a script that runs a few tasks for several files so I tried to index each name in the list, but when I print out

fullfile(datapath,sprintf('%s',author),'A',sprintf('%s',mask))

I see that a = 1 doesn't mean the whole name, but only the first letter. I tried different versions of this but I don't manage to index to the whole name. Any help would be very appreciated!

datapath = 'D:\DATA\test\';
author_list = ['Peterson', 'Jacobs'];
mask_list = ['mask_peterson.nii,1','mask_jacobs.nii,1'];

for a = 1:2
    for m = 1:2
        author = author_list(a)
        mask = mask_list(m)
        
        matlabbatch{1}.spm.util.imcalc.input = {fullfile(datapath,sprintf('%s',author),'A',sprintf('%s',mask))};
        matlabbatch{1}.spm.util.imcalc.output = sprintf('1%s_thr',author);
        matlabbatch{1}.spm.util.imcalc.outdir = {fullfile(datapath,sprintf('%s',author),'A')};
        %...
        %...
        %...
    end
end

[Stephen23]

Square brackets are a concatenation operator, they concatenate arrays together. So your code:

['Peterson', 'Jacobs']

concatenates two character vectors into one vector, and so is exactly equivalent to writing this:

'PetersonJacobs'

I doubt that is very useful for you. Most likely you should be storing those character vectors in a cell array:

https://www.mathworks.com/help/matlab/cell-arrays.html

[RP]

Thank you for the answer and the link, you are right, this works perfectly!

Answer 1

[Walter Roberson]

datapath = 'D:\DATA\test\';
author_list = {'Peterson', 'Jacobs'};
mask_list = {'mask_peterson.nii,1','mask_jacobs.nii,1'};

for a = 1:2
    for m = 1:2
        author = author_list{a}
        mask = mask_list{m}
        
        matlabbatch{1}.spm.util.imcalc.input = {fullfile(datapath,sprintf('%s',author),'A',sprintf('%s',mask))};
        matlabbatch{1}.spm.util.imcalc.output = sprintf('1%s_thr',author);
        matlabbatch{1}.spm.util.imcalc.outdir = {fullfile(datapath,sprintf('%s',author),'A')};
        %...
        %...
        %...
    end
end

[RP]

Thank you so much, this works perfectly as well! Seems like there are two options (this one with cell arrays and the one suggested below by Cris LaPierre), I suppose both are similar in terms of speed so they can be used interchangeably? Obviously speed is not a concern in my short code but it would be good to know for the future

[Stephen23]

@Rose Potter: good question. In general there is not a huge subjective difference in speed, so you should pick whichever one is best for your data processing. A few distinctions:

• If you are manipulating characters, character arrays/vectors gives direct access to the character code.
• The string class is a container class, one string element contains a character vector of any length.
• The string class has a number of overloaded operators and convenience syntaxes designed to make it easier to work with.
• The string class was introduced in R2016b, so if you need to run your code on earlier versions then use a cell array & character vectors.

Read more: https://www.mathworks.com/help/matlab/characters-and-strings.html

[Walter Roberson]

N = 10000;
C = cell(1,N);

for K = 1 : N
   C{K} = char(randi([32 127], 1, randi([5 200])));
end

fprintf('conversion speed\n');

conversion speed

tic
S = string(C);
toc

Elapsed time is 0.017607 seconds.

fprintf('basic indexing speed, cells\n');

basic indexing speed, cells

tic
for K = 1 : N
  C{K};
end
toc

Elapsed time is 0.011222 seconds.

fprintf('last character indexing speed, cells\n');

last character indexing speed, cells

tic
for K = 1 : N
  C{K}(end);
end
toc

Elapsed time is 0.011354 seconds.

fprintf('basic indexing speed, strings\n')

basic indexing speed, strings

tic
for K = 1 : N
  S(K);
end
toc

Elapsed time is 0.019268 seconds.

fprintf('last character indexing speed, strings using {}\n')

last character indexing speed, strings using {}

tic
for K = 1 : N
   S{K}(end);
end
toc

Elapsed time is 0.022054 seconds.

fprintf('last character indexing speed, strings extractAfter\n');

last character indexing speed, strings extractAfter

tic
for K = 1 : N
  extractAfter(S(K), strlength(S(K)));
end
toc

Elapsed time is 0.034476 seconds.

fprintf('last character, cell regexp\n');

last character, cell regexp

tic
regexp(C, '.$', 'once', 'match');
toc

Elapsed time is 0.030094 seconds.

fprintf('last character, string regexp\n');

last character, string regexp

tic
regexp(S, '.$', 'once', 'match');
toc

Elapsed time is 0.032894 seconds.

fprintf('last character, string regexppattern')

last character, string regexppattern

tic
extractAfter(S, regexpPattern('.$'));
toc

Elapsed time is 0.032256 seconds.

So cell is anywhere from just barely faster, to a couple of times as fast (eg using the official extractAfter strlength() instead of using lower level {}(end) )

Answer 2

[Cris LaPierre]

One way is to use strings instead of char arrays.

datapath = "D:/DATA/test";
author_list = ["Peterson", "Jacobs"];
mask_list = ["mask_peterson.nii,1","mask_jacobs.nii,1"];

for a = 1:length(author_list)
    for m = 1:length(mask_list)
        input = fullfile(datapath,author_list(1),'A',mask_list(m))
    end
end

input = "D:/DATA/test/Peterson/A/mask_peterson.nii,1"

input = "D:/DATA/test/Peterson/A/mask_jacobs.nii,1"

input = "D:/DATA/test/Peterson/A/mask_peterson.nii,1"

input = "D:/DATA/test/Peterson/A/mask_jacobs.nii,1"

[RP]

Thank you so much, this works! Just one stupid question - it works only with e.g. "Peterson", but when I use 'Peterson' I get each character spelled out instead of the whole string. I couldn't find anything about this online, is it always like this that I need to use " when I mean the whole string and ' for characters? Or is it more complex?

[Walter Roberson]

A = ['PQR'; 'STU']

A = 2×3 char array
    'PQR'
    'STU'

class(A)

ans = 'char'

size(A)

ans = 1×2

2 3

A(:,1)

ans = 2×1 char array
    'P'
    'S'

B = ["PQR"; "STU"]

B = 2×1 string array
    "PQR"
    "STU"

class(B)

ans = 'string'

size(B)

ans = 1×2

2 1

B(:,1)

ans = 2×1 string array
    "PQR"
    "STU"

Apostrophe creates character vectors, which are vectors of class char(). Indexing by a scalar gets you one character. A(:,1) is asking for the first column of A, which is a 2 x 1 array of char. Character vectors can be used as-if they are numeric in a number of different contexts: for example,

A(1,1)+0

ans = 80

Here the single numeric character code stored at A(1,1), corresponding to 'P', is automatically converted to decimal, and then 0 is added to the result.. so 'P' is character #80.

double-quote on the other hand, creates string() objects. Using () indexing on string objects gets you entire strings. B(:,1) gets you the first column of string objects. There are number of operations defined for string objects that are different than for character arrays. Such as

B(1,1)+0

ans = "PQR0"

what has happened here is that for string objects, the + operator is defined as concatenation. And also, + between a string object and a numeric value is defined as formatting the numeric value as its representable string object -- so numeric value 0 to "0" the representation. Then the "PQR" + "0" is appending, giving "PQR0"

There are uses for both methods of operating with characters.

[Cris LaPierre]

"One way is to use strings instead of char arrays."

• string array
• char array