Python: Using Numpy to remove "for" loop

Question

I am currently having 2 following classes:

class plane():
    def __init__(self):
        self.points = {}
        # Using dictionary is necessary, to be able refer a specific point by using its name (saved as dictionary key)
    def all_points(self):
        all_points = np.empty((3, len(self.points)))
        for k, point in enumerate(self.points.values()): # Line A
            all_points[k,:] = point.coordinate           # Line B
        return all_points

import numpy as np
class point():
    def __init__(self, x, y, z):
        self.coordinate = np.array((x, y, z))

My use case:

a = plane()
a.points[0] = point(0, 1, 2)
a.points[1] = point(1, 2, 3)
a.points[2] = point(2, 3, 4)
a.all_points()

Output:

array([[0., 1., 2.],
       [3., 4., 5.],
       [6., 7., 8.]])

Is it possible to replace the for loop with numpy operation? Since Numpy allows following operation:

x = np.array([[0, 1, 2], [2, 3, 4]])
x + 1

Output:

array([[1, 2, 3],
       [3, 4, 5]])

My idea was: a.points.values().coordniate. This obviously doesn't work as "dict value" object don't have "coordinate" methode.

Any suggestion is welcome. Thank you in advance for any feedback.

Since only a point object has a coordinate attribute, you can't avoid iterating in one way or other over all points. That's a Python level operation. The suggestions are just window dressing. — hpaulj
– hpaulj, Commented Feb 25, 2021 at 16:44

Rishabh Kumar · Accepted Answer · 2021-02-25 14:16:54Z

1

The builtin map can help:

np.array([*map(lambda x:x.coordinate,self.points.values())])

Internally map will be using its own loop. This only gets rid of the for loop in the code. It may have some performance improvements.

Execution:

class plane():
    def __init__(self):
        self.points = {}

    def all_points(self):
        return np.array([*map(lambda x:x.coordinate,self.points.values())])

class point():
    def __init__(self, x, y, z):
        self.coordinate = np.array((x, y, z))



a = plane()
a.points[0] = point(0, 1, 2)
a.points[1] = point(1, 2, 3)
a.points[2] = point(2, 3, 4)
print(a.all_points())

Output:

[[0 1 2]
 [1 2 3]
 [2 3 4]]

answered Feb 25, 2021 at 14:16

Rishabh Kumar

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Comments

Nerveless_child · Accepted Answer · 2021-02-25 14:07:51Z

0

I don't think you can do it without a for loop, but if you don't mind using a list comprehension then this should do it.

def all_points(self):
    all_points = np.stack(
        [self.points[k].coordinate for k in sorted(self.points)])
    return all_points

answered Feb 25, 2021 at 14:07

Nerveless_child

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