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Brace indexing is not supported for variables of this type

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Behzad Navidi
Behzad Navidi on 23 Jan 2020 at 8:53
Commented: Alex Mcaulley on 24 Jan 2020 at 8:05
Hey all, When I try to use this code:
output = cell(size(new_precipitation));
for i = 1:length(new_precipitation)
mn2t_mat = new_precipitation{i}; % extract matrix of ith element
yd = max(max(mn2t_mat)); % first find maximum for each hour
y = reshape(yd,24,1,size(yd,3)/24); % reshape it by day
output{i} = max(y); % find maximum of each day
end
This error appeared:
Brace indexing is not supported for variables of this type.
I was search for some answer but can't find how to fix it. Do you have any idea how to overcome this error?
I attached a new_precipitation.mat file in the google drive which has 10 MB in this link:
Thank you in advance
Caption:
I have a 3D matrix named new_precipitation (latitude x longitude x time). In essence time dimension is in the hourly time steps for a year (365-day x 24 hours = 8760). Whole data are for temperature and I want to find the maximum temperature for each day instead of hourly.
24 values for each day that represents temperature so I need to find a maximum of 24 values in order to have the maximum temperature of each day.
So I need to find maximum value for every 24 by 24-time steps and consider it for days.
I'm sorry I forgot to change the name so new_precipitation is a wrong variable name of new_temperature.

  2 Comments

Behzad Navidi
Behzad Navidi on 23 Jan 2020 at 9:15
class(new_precipitation)
ans =
'double'
Thank you

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Accepted Answer

ME
ME on 23 Jan 2020 at 9:13
I think Walter was onto the right thing in his comment. new_precipitation is of type 'double' and so doesn't support brace (i.e. { } ) indexing. For that variable you'll want to change to ordinary round brackets (i.e. ( ) ). In other words, you'll need to change the third line to something like:
mn2t_mat = new_precipitation(i); % extract matrix of ith element

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Alex Mcaulley
Alex Mcaulley on 23 Jan 2020 at 10:20
I think you want something like this:
output = zeros(size(new_precipitation)./[1,1,24]);
for i = 1:24:size(new_precipitation,3)
mn2t_mat = new_precipitation(:,:,i:i+23); % extract matrix corresponding to 24 hours of one day
yd = max(mn2t_mat,[],3); %calculate the maximum for each latitude, longitude
output(:,:,ceil(i/24)) = yd; % save the values of day i
end
Behzad Navidi
Behzad Navidi on 23 Jan 2020 at 14:34
Dear Alex Mcaulley
Thank you for your answer. I run this code and everything is ok. Sincerely I want to ask you if you please could send me a description of this code? because I think about it a couple of hours and as I'm not a professional I cant recognize what you do.
Is this code capture maximum values for every 24 by 24 rows and save it as a value for the days?
Thank you so much.
Alex Mcaulley
Alex Mcaulley on 24 Jan 2020 at 8:05
The best way to understand the code is to put breakpoints and go though it line per line seeing what happens in each iteration (reading the documentation in case you don't understand any function)
output = zeros(size(new_precipitation)./[1,1,24]); %Preallocation of the output matrix of size [20,16,365]
for i = 1:24:size(new_precipitation,3) %Loop to extract the values corresponding to 1 day
mn2t_mat = new_precipitation(:,:,i:i+23); %In mn2t_mat you have the values corresponding to 24 hours of 1 day
yd = max(mn2t_mat,[],3); %Here, we calculate the maximum for each latitude, longitude in this day
output(:,:,ceil(i/24)) = yd; % save the values of that day
end

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