How can I plot contour lines on the generated map?
    4 visualizaciones (últimos 30 días)
  
       Mostrar comentarios más antiguos
    
    Behrooz Daneshian
 el 20 de Feb. de 2023
  
    
    
    
    
    Comentada: Walter Roberson
      
      
 el 20 de Feb. de 2023
            Hi all,
Using the first section of code below, I could generate map of Alaska state. I need to draw contour lines (second section of the code) on the generated map.  Can anyone help me with this regard? Please use .shp file in the zipped folder,
%%%%% generating Alaska map
S=shaperead('Location of the .shp file\tl_2018_02_anrc.shp','UseGeoCoords',true);
geoshow(S,"DisplayType","multipoint")
xlim([-179.148909,-130]);
ylim([51.214183,71.365162]);
%%%%%%%% generating contour lines based on the latitude and longitude of
%%%%%%%% weather stations within the Alaska sate
load('POFDE.mat');
data = cell2mat(POFDE);
[lat,lon] = meshgrid(unique(data(:,1)),unique(data(:,2)));
for i = 1:(size(data, 2)-2)
    figure
    I = scatteredInterpolant(data(:,[1 2]), data(:,i+2));
    contourm(lat,lon,min(1,max(0,I(lat,lon))),'ShowText','on')
    colorbar
    title(['The probability of frost depth exceedance of \Omega_{\delta} = ' num2str(i) ' feet'])
    exportgraphics(gca, sprintf('FrostPlot_%d_feet.png', i))
end
1 comentario
  Walter Roberson
      
      
 el 20 de Feb. de 2023
				You do not need meshgrid() for contour() . contour() will accept vectors of coordinates.
meshgrid() or ndgrid() are useful in cases where the Z coordinate is being calculated, or the Z value is being interpolated, but these days many of the graphing functions accept marginal vectors.
Respuesta aceptada
  Voss
      
      
 el 20 de Feb. de 2023
        In your code, you make the map then create a new figure for each contour. Those new figures aren't going to have the map; the map was in the first figure. To have a map in each figure, call geoshow after figure.
unzip('tl_2018_02_anrc.zip')
%%%%% generating Alaska map
% S=shaperead('Location of the .shp file\tl_2018_02_anrc.shp','UseGeoCoords',true);
S=shaperead('tl_2018_02_anrc.shp','UseGeoCoords',true);
%%%%%%%% generating contour lines based on the latitude and longitude of
%%%%%%%% weather stations within the Alaska sate
load('POFDE.mat');
data = cell2mat(POFDE);
[lat,lon] = meshgrid(unique(data(:,1)),unique(data(:,2)));
for i = 1:(size(data, 2)-2)
    figure
    geoshow(S,"DisplayType","multipoint")
    xlim([-179.148909,-130]);
    ylim([51.214183,71.365162]);
    I = scatteredInterpolant(data(:,[1 2]), data(:,i+2));
    contourm(lat,lon,min(1,max(0,I(lat,lon))),'ShowText','on')
    colorbar
    title(['The probability of frost depth exceedance of \Omega_{\delta} = ' num2str(i) ' feet'])
    exportgraphics(gca, sprintf('FrostPlot_%d_feet.png', i))
end
4 comentarios
  Walter Roberson
      
      
 el 20 de Feb. de 2023
				At approximately x = -171 y = 57 (lower left quardrant), there appears to be a single weather station. If you restrict interpolation to be from data inside the shape, then that weather station would be surrounded by a barrier of all NaN, and when you do 2D interpolation with NaN around you, the result is NaN unless you happen to query exactly at the location of the weather station. For example if you were to ask for the reading 3 cm to the east of the weather station, the interpolated result would have to be NaN because the adjacent readings would be from outside the shape, and NaN+something is NaN. 
Más respuestas (0)
Ver también
Categorías
				Más información sobre Weather and Atmospheric Science en Help Center y File Exchange.
			
	Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!







