How to plot the outside of a 3D shape in Python Plotly without defining its faces?

Question

I'm trying to plot some 3D shapes in Python Plotly. For the sake of practicing, I was trying to plot a cube.

I tried the Isosurface command since I'm only interested in the outside of the shape, but surprisingly enough, the output was not a cube.

import plotly.graph_objects as go
import numpy as np 

x=[0, 0, 1, 1, 0, 0, 1, 1]
y=[0, 1, 1, 0, 0, 1, 1, 0]
z=[0, 0, 0, 0, 1, 1, 1, 1]

myisosurf = go.Isosurface(x=x,y=y,z=z,value= np.ones(len(x)))
fig = go.Figure(data=myisosurf)
fig.show()

I did get it to work with the Mesh3D command (with a little help from here), but the fact that it also requires settings for parameters of i, j, k, makes me prefer a more straightforward method. Especially when working with more complex shapes than a cube, it would be very hard to set the i, j, k parameters.

i= [7, 0, 0, 0, 4, 4, 6, 1, 4, 0, 3, 6]
j= [3, 4, 1, 2, 5, 6, 5, 2, 0, 1, 6, 3]
k= [0, 7, 2, 3, 6, 7, 1, 6, 5, 5, 7, 2]

my3dmesh = go.Mesh3d(x=x,y=y,z=z, i=i, j=j, k=k, intensity = np.linspace(1, 1, 8, endpoint=True),name='y')
fig = go.Figure(data=my3dmesh)
fig.show()

How can I get Python Plotly to show the outside of a shape without manually defining its faces through i,j,k parameters?

Answer 1

Thanks to @Adreas Deak's comment, it seems that the alphahull parameter of the Mesh3D command can direct Plotly to show the convex hull of the coordinates in the input.

Without the alphahull parameter (or with alphahull = -1) Delaunay triangulation is used to compute the faces around the coordinates, using the i, j, k parameter values. With alphahull = 0, the convex hull method is used to compute the faces around the coordinates, resulting in only the 'outside' of the shape, without the need for setting i, j, k parameters.

so:

import plotly.graph_objects as go
import numpy as np 

my3dmesh = go.Mesh3d(x=x,y=y,z=z, alphahull = 0, intensity = np.linspace(1, 1, 8, endpoint=True),name='y')
fig = go.Figure(data=my3dmesh)
fig.show()