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How can I plot contour lines on the generated map?

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Behrooz Daneshian
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
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.

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Voss
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
Warning: Function GEOSHOW expected DisplayType 'multipoint' to match Geometry 'polygon'. GEOSHOW is ignoring the DisplayType value.
Warning: Duplicate data points have been detected and removed - corresponding values have been averaged.
Warning: Function GEOSHOW expected DisplayType 'multipoint' to match Geometry 'polygon'. GEOSHOW is ignoring the DisplayType value.
Warning: Duplicate data points have been detected and removed - corresponding values have been averaged.
Warning: Function GEOSHOW expected DisplayType 'multipoint' to match Geometry 'polygon'. GEOSHOW is ignoring the DisplayType value.
Warning: Duplicate data points have been detected and removed - corresponding values have been averaged.
Warning: Function GEOSHOW expected DisplayType 'multipoint' to match Geometry 'polygon'. GEOSHOW is ignoring the DisplayType value.
Warning: Duplicate data points have been detected and removed - corresponding values have been averaged.
Warning: Function GEOSHOW expected DisplayType 'multipoint' to match Geometry 'polygon'. GEOSHOW is ignoring the DisplayType value.
Warning: Duplicate data points have been detected and removed - corresponding values have been averaged.
  4 comentarios
Behrooz Daneshian
Behrooz Daneshian el 20 de Feb. de 2023
Actually, the contours should rely on data from inside.Please look at the locations of weather stations on map on the attached figure. The probabilities(values shown by contour) is estimated at each weather stations. So contour lines must be estimatd based on the values there.
Walter Roberson
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.

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