meshgrid

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priya 2011-6-8

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/9072-meshgrid

dimensions of X and Y are as given below

X = 1X89000 Y = 1X 89000

[uxmesh, uymesh] = meshgrid(X,Y);

would you please tell me what is wrong in the above statements? Whenever I execute the above statements, MATLAB keeps crashing..

回答（2 个）

Matt Fig 2011-6-8

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/9072-meshgrid#answer_12496

Are you running out of memory? Why not tell us what prompted you to ask the question???

There is nothing syntactically wrong with the statement. However, you better have

8*2*89000^2 bytes (read: LOTS!!) of free memory, assuming X and Y are doubles...

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Matt Fig 2011-6-8

In that case I would think it would be acceptable to use a FOR loop and loop over the calculated velocities one at a time to find where in the list of true velocities the closest number lies, like:

[err(ii), idx(ii)] = min(abs(cv(ii)-tv));

Walter Roberson 2011-6-8

Current versions of interp2() do not require meshgrid(). The documentation says (at least as of 2008b):

XI and YI can be matrices, in which case interp2 returns the values of Z corresponding to the points (XI(i,j),YI(i,j)). Alternatively, you can pass in the row and column vectors xi and yi, respectively. In this case, interp2 interprets these vectors as if you issued the command meshgrid(xi,yi).

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Walter Roberson 2011-6-8

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https://ww2.mathworks.cn/matlabcentral/answers/9072-meshgrid#answer_12497

Are X and Y vectors? If so, then the above should work, but only if you have enough memory: 89000 x 89000 double precision numbers requires between 59 and 60 gigabytes.