uniformly distributed coordinate points
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can anyone help me to generate uniformly distributed coordinate points inside a rectangular field.
Thanking u advance
回答（5 个）
Sarah Wait Zaranek
2011-3-10
You can also do the following if you want a uniform grid:
[X,Y] = meshgrid(linspace(1,1000,100),linspace(1,1000,100));
This will give X coordinates and Y coordinates for each grid point inside a 1000 x 1000 rectangle. The segmenting in terms of x and y are defined by the input vectors, in this case, I am choosing 100 points per side.
scatter(X(:),Y(:))
will show you all the points.
Andreas Goser
2011-3-10
0 个投票
This sounds like a one-liner with RAND or RANDI for me. Do you run into issues with this command or did you just not find it?
Matt Tearle
2011-3-10
xmin = -3;
xmax = pi;
npts = 42;
x = xmin + (xmax-xmin)*rand(npts,1);
Repeat for y.
4 个评论
Basanta
2011-3-10
Matt Tearle
2011-3-10
yes. what i showed generates 42 x-coordinates uniformly distributed between -3 and pi. do the same for y, then scatter(x,y) will show the result.
Oleg Komarov
2011-3-10
To get all at once:
min = [ 1 1];
max = [10 3];
npts = 1000;
xy = bsxfun(@plus, min, bsxfun(@times, max-min,rand(npts,2)));
f = scatter(xy(:,1),xy(:,2),'.');
xlim([0,11])
ylim([0,4])
Matt Tearle
2011-3-10
ahahaha. nice :)
sh
2013-6-17
0 个投票
Hi, thanks for your help. can you find the x-coordinate and y-coordinates of these points by coding only? many thanks
Edward Goodison
2020-8-11
编辑：Edward Goodison
2020-8-11
[X,Y,Z] = meshgrid(linspace(Start,stop, number of points),linspace(Start,stop, number of points), linspace(Start,stop, number of points));
p = [X(:), Y(:), Z(:)]
2 个评论
Rik
2020-8-11
Why not use [X(:), Y(:), Z(:)] instead of the loop?
Edward Goodison
2020-8-11
nice, good improvement, will edit.
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2020-8-11
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