Question

如何从带有数字的矩阵生成等轴测图？

我需要想法

示例：

矩阵：

[[3,2]，

[1,1]]

此 Isometric http://image.prntscr.com/image/83c2d1c43fd94b7bb1ba101a8db17ef3.png

每个数字是高度，3代表3个立方体高度第一个文件，2个代表2个立方体高度第一个文件第二个元素

由于

Answer 1

这是一个非常好的问题。我认为matplotlib并不直接提供任何这样的功能，但我们当然可以通过它的六个表面模拟一个立方体。获取这些表面后，我们可以使用here提供的一段代码。然后，我们需要在矩阵定义的位置绘制立方体。为了使绘图看起来是等距的，我们使用一种解决方法，在mpl3d轴的立方边界框的角上绘制不可见点。最后，我们需要使轴不可见。

import matplotlib as mpl
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
import matplotlib.pyplot as plt

def cuboid_data(center, size=(1,1,1)):
    # code taken from
    # https://stackoverflow.com/questions/30715083/python-plotting-a-wireframe-3d-cuboid?noredirect=1&lq=1
    # suppose axis direction: x: to left; y: to inside; z: to upper
    # get the (left, outside, bottom) point
    o = [a - b / 2 for a, b in zip(center, size)]
    # get the length, width, and height
    l, w, h = size
    x = [[o[0], o[0] + l, o[0] + l, o[0], o[0]],  # x coordinate of points in bottom surface
         [o[0], o[0] + l, o[0] + l, o[0], o[0]],  # x coordinate of points in upper surface
         [o[0], o[0] + l, o[0] + l, o[0], o[0]],  # x coordinate of points in outside surface
         [o[0], o[0] + l, o[0] + l, o[0], o[0]]]  # x coordinate of points in inside surface
    y = [[o[1], o[1], o[1] + w, o[1] + w, o[1]],  # y coordinate of points in bottom surface
         [o[1], o[1], o[1] + w, o[1] + w, o[1]],  # y coordinate of points in upper surface
         [o[1], o[1], o[1], o[1], o[1]],          # y coordinate of points in outside surface
         [o[1] + w, o[1] + w, o[1] + w, o[1] + w, o[1] + w]]    # y coordinate of points in inside surface
    z = [[o[2], o[2], o[2], o[2], o[2]],                        # z coordinate of points in bottom surface
         [o[2] + h, o[2] + h, o[2] + h, o[2] + h, o[2] + h],    # z coordinate of points in upper surface
         [o[2], o[2], o[2] + h, o[2] + h, o[2]],                # z coordinate of points in outside surface
         [o[2], o[2], o[2] + h, o[2] + h, o[2]]]                # z coordinate of points in inside surface
    return x, y, z

def plotCubeAt(pos=(0,0), N=0, ax=None):
    # Plotting N cube elements at position pos
    if ax !=None:
        if N > 0:
            for n in range(N):
                X, Y, Z = cuboid_data( (pos[0],pos[1],n) )
                ax.plot_surface(X, Y, Z, color='b', rstride=1, cstride=1, alpha=1)

def plotIsoMatrix(ax, matrix):
    # plot a Matrix 
    # where matrix[i,j] cubes are added at position (i,j) 
    for i  in range(matrix.shape[0]):
            for j in range(matrix.shape[1]):
                plotCubeAt(pos=(i,j), N=matrix[i,j], ax=ax)

    l = max(matrix.shape[0], matrix.shape[1], matrix.max())
    bb = np.array([(0,0,0), (0,l,0), (l,0,0), (l,l,0),(0,0,l), (0,l,l), (l,0,l), (l,l,l)])
    ax.plot(bb[:,0], bb[:,1], bb[:,2], "w", alpha=0.0)            



if __name__ == '__main__':
    fig = plt.figure()
    ax = fig.gca(projection='3d')
    ax.set_aspect('equal')
    matrix = np.array([[3,2],[1,1]])
    plotIsoMatrix(ax, matrix)
    ax.set_axis_off()
    plt.show()

基于矩阵绘制等距图

1 个答案: