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

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 :)

Answer 1

[Mathieu NOE]

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

[Mathieu NOE]

hello again

problem solved now ?

[Carlotta Dentico]

Hello! Yes, the code works perfectly :) Thank ou again

[Mathieu NOE]

My pleasure !

Answer 2

[Mathieu NOE]

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)');

[Carlotta Dentico]

Hi,

thank you very much for the help.. I tried the code with my data but I have weird results..

I only modified the code to have years instead of months. I think the spatial index is not working correctly.

data = pt_spatial_lev20;

[m,n,p] = size(data);

dt = 1; % put here the time increment or use a time vector if it exist

out = [];

a = (aa_lon>=-7 & aa_lon<=9);

b = (aa_lat>=79 & aa_lat<=79.01);

coord = logical(a'*b);

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,TT,out);

xlabel('Longitude');

ylabel('Time (years)');

datetick('y')

[Mathieu NOE]

hello

can you share your data (as mat file) ?

[Carlotta Dentico]

Here the files :) I had to reduce the temperature file, that now is 609x881x12.

[Mathieu NOE]

see my answer in the answer section below !