Variable passed through function doesn't work

1 次查看（过去 30 天）
nines
nines 2023-1-26
编辑： Fangjun Jiang 2023-1-26
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 个评论
显示 4更早的评论
nines
nines 2023-1-26
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 2023-1-26
编辑：Fangjun Jiang 2023-1-26
I moved the comment by @Dyuman Joshi to an Answer. You can accept it now.

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Dyuman Joshi
Dyuman Joshi 2023-1-26
移动：Fangjun Jiang 2023-1-26
Ran in:
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 个评论
Fangjun Jiang
Fangjun Jiang 2023-1-26
编辑：Fangjun Jiang 2023-1-26
Ran in:
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.

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Semiconductors and Converters 的更多信息

标签

Community Treasure Hunt

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

Start Hunting!

Translated by