为什么pyplot.contour()要求Z为2D数组? [英] Why does pyplot.contour() require Z to be a 2D array?

问题描述

matplotlib.pyplot.contour()函数采用3个输入数组X，Y和Z.
数组X和Y指定点的x和y坐标，而Z指定在点处求值的感兴趣函数的相应值.

The matplotlib.pyplot.contour() function takes 3 input arrays X, Y and Z.
The arrays X and Y specify the x- and y-coordinates of points, while Z specifies the corresponding value of the function of interest evaluated at the points.

我了解到np.meshgrid()使得生成contour()的参数数组变得容易:

I understand that np.meshgrid() makes it easy to produce arrays which serve as arguments to contour():

X = np.arange(0,5,0.01)
Y = np.arange(0,3,0.01)

X_grid, Y_grid = np.meshgrid(X,Y)
Z_grid = X_grid**2 + Y_grid**2

plt.contour(X_grid, Y_grid, Z_grid)  # Works fine

这很好.方便地，这也很好:

This works fine. And conveniently, this works fine too:

plt.contour(X, Y, Z_grid)  # Works fine too

但是，为什么Z输入 是二维数组?

However, why is the Z input required to be a 2D-array?

为什么即使指定了所有正确对齐的相同数据，也不允许以下内容?

Why is something like the following disallowed, even though it specifies all the same data aligned appropriately?

plt.contour(X_grid.ravel(), Y_grid.ravel(), Z_grid.ravel())  # Disallowed

还有，当仅指定 Z(没有对应的X和Y)时的语义是什么?

Also, what are the semantics when only Z is specified (without the corresponding X and Y)?

推荐答案

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Looking at the documentation of contour one finds that there are a couple of ways to call this function, e.g. contour(Z) or contour(X,Y,Z). So you'll find that it does not require any X or Y values to be present at all.

但是，为了绘制轮廓，函数必须知道下面的网格. Matplotlib的contour基于矩形网格.但是即使这样，允许contour(z)且z是一维数组，也将使我们无法知道应如何绘制该字段.在contour(Z)的情况下，其中Z是2D数组，其形状明确地设置了绘图的网格.

However in order to plot a contour, the underlying grid must be known to the function. Matplotlib's contour is based on a rectangular grid. But even so, allowing contour(z), with z being a 1D array, would make it impossible to know how the field should be plotted. In the case of contour(Z) where Z is a 2D array, its shape unambiguously sets the grid for the plot.

一旦知道了网格，是否可选的X和Y数组是否被展平就无关紧要；实际上，文档告诉我们的是:

Once that grid is known, it is rather unimportant whether optional X and Y arrays are flattened or not; which is actually what the documentation tells us:

X和Y必须都为与Z形状相同的2D，或者它们都必须为1D，以使len(X)是Z中的列数，而len(Y)是行数在Z中.

X and Y must both be 2-D with the same shape as Z, or they must both be 1-D such that len(X) is the number of columns in Z and len(Y) is the number of rows in Z.

很明显，有点像 plt.contour(X_grid.ravel(), Y_grid.ravel(), Z_grid.ravel())无法生成等高线图，因为有关网格形状的所有信息都将丢失，并且等高线函数无法知道如何解释数据.例如.如果len(Z_grid.ravel()) == 12，则基础网格的形状可以是(1,12), (2,6), (3,4), (4,3), (6,2), (12,1)中的任何一个.

It is also pretty obvious that someting like plt.contour(X_grid.ravel(), Y_grid.ravel(), Z_grid.ravel()) cannot produce a contour plot, because all the information about the grid shape is lost and there is no way the contour function could know how to interprete the data. E.g. if len(Z_grid.ravel()) == 12, the underlying grid's shape could be any of (1,12), (2,6), (3,4), (4,3), (6,2), (12,1).

一种可能的解决方法当然是允许一维数组并引入参数shape，例如plt.contour(x,y,z, shape=(6,2)).但是事实并非如此，因此您必须忍受Z必须是2D的事实.

A possible way out could of course be to allow for 1D arrays and introduce an argument shape, like plt.contour(x,y,z, shape=(6,2)). This however is not the case, so you have to live with the fact that Z needs to be 2D.

但是，如果您正在寻找一种获得具有平坦(散乱的)数组的计数图的方法，则可以使用

However, if you are looking for a way to obtain a countour plot with flattened (ravelled) arrays, this is possible using plt.tricontour().

plt.tricontour(X_grid.ravel(), Y_grid.ravel(), Z_grid.ravel()) 

此处，将使用Delaunay三角剖分在内部生成一个三角形网格.因此，即使完全随机化的点也将产生不错的结果，如下图所示，将其与contour赋予的相同随机点进行比较.

Here a triangular grid will be produced internally using a Delaunay Triangualation. Therefore even completely randomized points will produce a nice result, as can be seen in the following picture, where this is compared to the same random points given to contour.

(这里是生成此图片的代码)

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