列出一级“拆包”

Question

I have a function which returns Sympy points:

result = [Point3D(500, 500, 10), Point3D(-500, 500, 10), Point3D(-500, -500, 10), Point3D(500, -500, 10)]

But I need to get the values of the point, not the point instance, so I do this:

a, b, c, d = [], [], [], []
for i in range(3):
    a.append(result[0][i])
for i in range(3):
    b.append(result[1][i])
for i in range(3):
    c.append(result[2][i])
for i in range(3):
    d.append(result[3][i])

# And then pack it:
result = [tuple(a), tuple(b), tuple(c), tuple(d)]

I know this is a horrible way of getting the desired input, which is:

result = [(500, 500, 10), (-500, 500, 10), (-500, -500, 10), (500, -500, 10)]

How can I do it in a better way?

Answer 1

Use a list comprehension:

result = [tuple([p[0], p[1], p[2]]) for p in result]

I just checked the documentation of Point3D , and it seems you can have a more elegant solution:

result = [tuple([p.x, p.y, p.z]) for p in result]

Or as @OscarBenjamin has indicated in the comments, which probably is the simplest solution, you can use:

result = [p.args for p in result]

Answer 2

Small note: A new method has been added (by me) to sympy's point class which returns all the coordinates of the point:

result = [p.coordinates for p in result]