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Hovmoller Diagram: longitude (x axis), time (y axis), and z(temperature values)

조회 수: 35 (최근 30일)
Hi,
I wuold like to extract temperature from a 3D matrix temp 609x881x372 (lonxlatxtime) and do an Hovmoller plot where I have time on the y-axis and longitude on the x-axis (see figure).
I created a logical matrix of 0 and 1 to identify where my coordinates meet my condition:
coord = (aa_lon>=-7 & aa_lon<=9 & aa_lat>=79 & aa_lat<=79.01);
coord is 609x881 logical but my matrix (temp) from which I want to extract the data is 609x881x372 double.
Hope my question is clear.
Please help me :)

채택된 답변

Mathieu NOE
Mathieu NOE 2023년 9월 13일
이동: Mathieu NOE 2023년 9월 13일
hello again
I modified my code so it should work now on your data
as I understand we have here only 12 months of data , so my display assumes we have here only the first 12 months (1 year) of the entire time vector (which starts in Jan 1991) - you may need to correct this
also I changed a bit the tolerance on the latitude : aa_lat>=79 & aa_lat<=79+0.1
so we pick more data points of the temp matrix ; otherwise your diagram will have a very coarse display
result so far :
code
%% load data 609x881x12 (lonxlatxtime)
load lat.mat
load lon.mat
load time.mat
load temp.mat
%% convert time serial number into Y/M/D format
[y,mo,d,h,mi,s] = datevec(TT);% samplin time = 1 month (data is stored every 15th of the month)
%% spatial filtering
% coord = (aa_lon>=-7 & aa_lon<=9 & aa_lat>=79 & aa_lat<=79.01);
b = (aa_lat>=79 & aa_lat<=79+0.1); % play with tolerance (here 0.1) and see impact on how many points are selected
selected_lon = aa_lon(b);
ind_lon = (selected_lon>=-7 & selected_lon<=9);
final_lon = selected_lon(ind_lon);
%% main code
[m,n,p] = size(temp);
out = [];
for k = 1:p
tmp = temp(:,:,k);
selected_temp = tmp(b); % filtering for lat
selected_temp = selected_temp(ind_lon); % filtering for lon
% concatenation of the selected_temp data
out = [out; selected_temp']; % nb transpose of selected_temp to have data as row
end
% plot
% downsample Y tick spacing to display only years (tick position at january
% / 1st month of the year)
new_y = (1:12:p);
imagesc(final_lon,(1:p),out);
yticks(new_y)
yticklabels(num2str(y(new_y)))
xlabel('Longitude');
ylabel('Time (Years)');
% add horizontal line corresponding to january 15th (first data of the
% year)
hold on
for ci = 1:numel(new_y)
plot(final_lon,new_y(ci)*ones(size(final_lon)),'k--');
end
  댓글 수: 3
Carlotta Dentico
Carlotta Dentico 2023년 9월 20일
Hello! Yes, the code works perfectly :) Thank ou again

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추가 답변 (1개)

Mathieu NOE
Mathieu NOE 2023년 9월 1일
hello
maybe this ?
I created some dummy data to test my code
hope it helps
%% dummy data 609x881x372 (lonxlatxtime)
aa_lon = -45:0.1:60.9-45-0.1;
aa_lat = 0:0.1:88.1-0.1;
p = 10*12;% 10 years of monthly data = 120 time steps
for k = 1:p
data(:,:,k) = rand(609,881) + sin(0.1*k)*ones(609,881);
end
%% spatial filtering
% coord = (aa_lon>=-7 & aa_lon<=9 & aa_lat>=79 & aa_lat<=79.01);
% equivalent to :
a = (aa_lon>=-7 & aa_lon<=9);
b = (aa_lat>=79 & aa_lat<=79.01);
coord = logical(a'*b);
%% main code
[m,n,p] = size(data);
dt = 1; % put here the time increment or use a time vector if it exist
out = [];
% main loop
for k = 1:p
tmp = data(:,:,k);
tmp = tmp(coord); % spatial filtering of the data
time(k) = k*dt;
% time concatenation of the tmp data
out = [out; tmp']; % nb transpose of tmp to have data as row
end
% plot
x_plot = aa_lon(a);
imagesc(x_plot,time,out);
xlabel('Longitude');
ylabel('Time (months)');
  댓글 수: 4
Carlotta Dentico
Carlotta Dentico 2023년 9월 11일
Here the files :) I had to reduce the temperature file, that now is 609x881x12.
Mathieu NOE
Mathieu NOE 2023년 9월 13일
see my answer in the answer section below !

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