Variable passed through function doesn't work

1 Ansicht (letzte 30 Tage)
nines
nines am 26 Jan. 2023
Bearbeitet: Fangjun Jiang am 26 Jan. 2023
Hello all,
I have a function where I am passing three different behaviors (drug, before drug, after drug).
function defining_parsing(subj, path, data_type, behavior)
%% loading data
all_folders = dir(fullfile(path, subj, '/dir/')); % loading all the dir folders
% Define the regular expressions for each behavior and ROI
if behavior == 'drug'
expression_behavior = [data_type, '.*', '(s2|s4|s6|s3)', '.*'];
elseif behavior == 'before drug'
expression_behavior = [data_type, '.*', '(w3|w2|w4|w|WA)', '.*'];
elseif behavior == 'after drug'
expression_behavior = [data_type, '.*', '(EO)', '.*'];
end
end
When I pass the variable outside of the function for the above I get the following error:
Arrays have incompatible sizes for this operation.
Error in defining_parsing (line 20)
if behavior == 'drug'
When I pass the the variable for behavior within the function, I do not get the error:
function defining_parsing(subj, path, data_type, roi, behavior)
%%
behavior = 'drug'
%% loading data
all_folders = dir(fullfile(path, subj, '/dir/')); % loading all the dir folders
% Define the regular expressions for each behavior and ROI
if behavior == 'drug'
expression_behavior = [data_type, '.*', '(s2|s4|)', '.*'];
elseif behavior == 'before drug'
expression_behavior = [data_type, '.*', '(w3|w2|)', '.*'];
elseif behavior == 'after drug'
expression_behavior = [data_type, '.*', '(EO)', '.*'];
end
end
I have tried the following:
1) making sure that both the variable input to the function and the variable input with the function are the same (they are both char)
if ~ischar(behavior)
behavior = num2str(behavior);
end
2) making sure that there are no typos.
Do you have any suggestions? Thanks so much!
  6 Kommentare
nines
nines am 26 Jan. 2023
final thing: how do i accept an answer? the accept button isn't coming up next to any of your names!
Fangjun Jiang
Fangjun Jiang am 26 Jan. 2023
Bearbeitet: Fangjun Jiang am 26 Jan. 2023
I moved the comment by @Dyuman Joshi to an Answer. You can accept it now.

Melden Sie sich an, um zu kommentieren.

Akzeptierte Antwort

Dyuman Joshi
Dyuman Joshi am 26 Jan. 2023
Verschoben: Fangjun Jiang am 26 Jan. 2023
How are you calling the function?
Also, Use strcmp or isequal to compare strings
behavior = 'drug'
behavior = 'drug'
isequal(behavior,'drug')
ans = logical
1
strcmp(behavior,'drug')
ans = logical
1
You can also use switch here instead of if-else
switch behavior
case 'drug'
disp('1')
case 'before drug'
disp('2')
case 'after drug'
disp('3')
end
1
  1 Kommentar
Fangjun Jiang
Fangjun Jiang am 26 Jan. 2023
Bearbeitet: Fangjun Jiang am 26 Jan. 2023
This explains the root cause of the error message in the OP's question. When variable "behavior" takes the value of 'before drug' and it is compared to 'drug' in the If statement.
behavior='before drug';
behavior=='drug'
Arrays have incompatible sizes for this operation.

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Kategorien

Mehr zu Develop Apps Using App Designer finden Sie in Help Center und File Exchange

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by