Indexing string in loop

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RP
RP 2021-6-28
评论： Cris LaPierre 2021-6-29
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
  2 个评论
Stephen23
Stephen23 2021-6-28
编辑：Stephen23 2021-6-28
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
RP 2021-6-29
Thank you for the answer and the link, you are right, this works perfectly!

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回答（2 个）

Walter Roberson
Walter Roberson 2021-6-28
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
  3 个评论
显示 1更早的评论
Stephen23
Stephen23 2021-6-29
@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
Walter Roberson 2021-6-29
Ran in:
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) )

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Cris LaPierre
Cris LaPierre 2021-6-28
Ran in:
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"
  3 个评论
显示 1更早的评论
Walter Roberson
Walter Roberson 2021-6-29
Ran in:
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
Cris LaPierre 2021-6-29
"One way is to use strings instead of char arrays."

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