Hay*_*yan 5 python plot matplotlib
问题是,我想调整宽高比,即沿z轴拉伸,以便使所有堆叠的图像或多或少可见。有一个简单的方法吗?
好像没有 xe2\x80\x9cproper\xe2\x80\x9d 的方法来做到这一点,但我们可以尝试猴子修补我们的方法来解决这个问题。这里\xe2\x80\x99s是我能做的最好的组装:
\n\nfrom mpl_toolkits.mplot3d import proj3d\n\ndef make_get_proj(self, rx, ry, rz):\n \'\'\'\n Return a variation on :func:`~mpl_toolkit.mplot2d.axes3d.Axes3D.getproj` that\n makes the box aspect ratio equal to *rx:ry:rz*, using an axes object *self*.\n \'\'\'\n\n rm = max(rx, ry, rz)\n kx = rm / rx; ky = rm / ry; kz = rm / rz;\n\n # Copied directly from mpl_toolkit/mplot3d/axes3d.py. New or modified lines are\n # marked by ##\n def get_proj():\n relev, razim = np.pi * self.elev/180, np.pi * self.azim/180\n\n xmin, xmax = self.get_xlim3d()\n ymin, ymax = self.get_ylim3d()\n zmin, zmax = self.get_zlim3d()\n\n # transform to uniform world coordinates 0-1.0,0-1.0,0-1.0\n worldM = proj3d.world_transformation(xmin, xmax,\n ymin, ymax,\n zmin, zmax)\n\n # adjust the aspect ratio ##\n aspectM = proj3d.world_transformation(-kx + 1, kx, ##\n -ky + 1, ky, ##\n -kz + 1, kz) ##\n\n # look into the middle of the new coordinates\n R = np.array([0.5, 0.5, 0.5])\n\n xp = R[0] + np.cos(razim) * np.cos(relev) * self.dist\n yp = R[1] + np.sin(razim) * np.cos(relev) * self.dist\n zp = R[2] + np.sin(relev) * self.dist\n E = np.array((xp, yp, zp))\n\n self.eye = E\n self.vvec = R - E\n self.vvec = self.vvec / proj3d.mod(self.vvec)\n\n if abs(relev) > np.pi/2:\n # upside down\n V = np.array((0, 0, -1))\n else:\n V = np.array((0, 0, 1))\n zfront, zback = -self.dist, self.dist\n\n viewM = proj3d.view_transformation(E, R, V)\n perspM = proj3d.persp_transformation(zfront, zback)\n M0 = np.dot(viewM, np.dot(aspectM, worldM)) ##\n M = np.dot(perspM, M0)\n return M\n return get_proj\n\n# and later in the code:\nax.get_proj = make_get_proj(ax, 1, 1, 2)\nax.set_aspect(1.0)\nRun Code Online (Sandbox Code Playgroud)\n\n\n