清空筛选条件
清空筛选条件

Interpolating 3D Gridded data at specific cordinates

2 次查看（过去 30 天）
Mark
Mark 2022-11-22
评论： Mark 2022-11-22
I am working with 3D gridded data with a dimension of (Longitude,Latitude,Height) and size (73x37x41). I want to plot temperature vs longitude at a slice of latitude = 0 degree and Height at 100 km and also temperature vs latitude at a slice of longitude = 0 degrees and height = 100 km.The original data does not provide data at these specific point i.e ( latitude = 0 and height = 100 km and longitude= 0 degree and height = 100 km). I know I have to interpolate my data to plot at these specific point but I am having problem in interpolating.
My data is something like this
[Long,Lat,Height] = ndgrid(0:5:360,-90:5:90,0:15:600);
Temp = rand(73,37,41)*1000;

请先登录，再回答此问题。

采纳的回答

Matt J
Matt J 2022-11-22
编辑：Matt J 2022-11-22
Ran in:
[Long,Lat,Height] = deal(0:5:360,-90:5:90,0:15:600); %Fake data
Temp = rand(73,37,41)*1000;
LUT=griddedInterpolant({Long,Lat,Height}, Temp); %interpolation object
TempvLong=LUT({0,Long,100}); %slices
TempvLat=LUT({Lat,0,100});
figure(1)
plot(Long(:),TempvLong(:));
xlabel 'Longitude'; ylabel 'Temperature'
figure(2);
plot(Lat(:),TempvLat(:));
xlabel 'Latitude'; ylabel 'Temperature'
  6 个评论
显示 4更早的评论
Matt J
Matt J 2022-11-22
Both 100 and 500 are well outside the range of Alt values given, and are in different units, so LUT is just doing meaningless extrapolation.
>> mnmx(Alt)
Maximum entry = 664122.1875
Minimum entry = 96420.9922
NaNs Present = NO

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Author Block Masks 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by