Variable passed through function doesn't work
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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!
Akzeptierte Antwort
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'
isequal(behavior,'drug')
strcmp(behavior,'drug')
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 Kommentar
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'
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