在Matlab中绘制2D网格 [英] Drawing 2D grid in matlab

问题描述

我试图使用 x> = -1 和 y< == 1 的matlab获取2D网格，步长为0.1 但是我正在获得没有适当步长的3D网格.有什么想法吗?

I am trying to get a 2D grid using matlab with x >= -1 and y <= 1 with step size of 0.1 But I'm getting 3D grid with no proper step sizes. Any ideas?

我的代码:

[x, y] = meshgrid(-1:0.1:5, 0:0.1:1);
surf(x,y)

推荐答案

您是否只想绘制一堆2D点?您使用 plot .以您的示例为例，您将获取x,y点，并简单地为每个点放置点标记.在执行此操作之前，请先将它们转换为一维数组:

Do you just want to plot a bunch of 2D points? You use plot. Using your example, you would take your x,y points and simply put dot markers for each point. Convert them into 1D arrays first before you do this:

[X,Y] = meshgrid(-1:0.1:5, 0:0.1:1);
X = X(:);
Y = Y(:);
plot(X,Y,'b.');
xlabel('X'); % // Label the X and Y axes
ylabel('Y');

这就是我得到的:

如果要将此网格旋转一个角度，则可以使用旋转矩阵，并将其与每对(x,y)坐标相乘.如果您想从线性代数中回想起，要逆时针旋转点，可以执行以下矩阵乘法:

If you want to rotate this grid by an angle, you would use a rotation matrix and multiply this with each pair of (x,y) co-ordinates. If you recall from linear algebra, to rotate a point counter-clockwise, you would perform the following matrix multiplication:

[x'] = [cos(theta) -sin(theta)][x]
[y']   [sin(theta)  cos(theta)][y]

x,y是原始坐标，而x',y'是旋转角度theta之后的输出坐标.如果要旋转-30度(顺时针旋转30度)，只需指定theta = -30 degrees.请记住，cos和sin的角度为弧度，因此实际上是-pi/6弧度.您需要做的是将每个点放入2D矩阵中.然后，您将使用旋转矩阵并将其应用于每个点.这样，您将向量化解决方案，而不是使用for循环.因此，您可以这样做:

x,y are the original co-ordinates while x',y' are the output co-ordinates after rotation of an angle theta. If you want to rotate -30 degrees (which is 30 degrees clockwise), you would just specify theta = -30 degrees. Bear in mind that cos and sin take in their angles as radians, so this is actually -pi/6 in radians. What you need to do is place each of your points into a 2D matrix. You would then use the rotation matrix and apply it to each point. This way, you're vectorizing the solution instead of... say... using a for loop. Therefore, you would do this:

theta = -pi/6; % // Define rotation angle
rot = [cos(theta) -sin(theta); sin(theta) cos(theta)]; %// Define rotation matrix
rotate_val = rot*[X Y].'; %// Rotate each of the points
X_rotate = rotate_val(1,:); %// Separate each rotated dimension
Y_rotate = rotate_val(2,:);
plot(X_rotate, Y_rotate, 'b.'); %// Show the plot
xlabel('X');
ylabel('Y');

这就是我得到的:

如果要执行其他转换，例如缩放每个轴，则只需将X或Y坐标乘以适当的比例即可:

If you wanted to perform other transformations, like scaling each axis, you would just multiply either the X or Y co-ordinates by an appropriate scale:

X_scaled = scale_x*X;
Y_scaled = scale_y*Y;

X_scaled和Y_scaled是坐标的缩放版本，而scale_x和scale_y是所需每个维度的缩放比例.如果要平移坐标，则可以将每个尺寸相加或相减一些数字:

X_scaled and Y_scaled are the scaled versions of your co-ordinates, with scale_x and scale_y are the scales in each dimension you want. If you wanted to translate the co-ordinates, you would add or subtract each of the dimensions by some number:

X_translate = X + X_shift; %// Or -
Y_translate = Y + Y_shift; %// Or -

X_translate和Y_translate是转换后的坐标，而X_shift和Y_shift是每个尺寸上想要的偏移量.请注意，您可以根据自己的需要执行+或-.

X_translate and Y_translate are the translated co-ordinates, while X_shift and Y_shift are the amount of shifts you want per dimension. Note that you either do + or -, depending on what you want.

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