Equidistant points between two points?

Question

I am trying to know equidistant points between two points. For example:

p1 = (1,1)
p2 = (5,5)

The answer that I am expecting is:

def getEquidistantPoints(p1, p2, HowManyParts):
    #some code
    return (array with points)

In this example, with p1, and p2:

A = getEquidistantPoints(p1,p2,4)
A = [(1,1),(2,2),(3,3),(4,4),(5,5)]

Always will be a straight line. HowManyParts in this case is the whole distance that is divided
something like numpy.linspace() but in two dimensions.

If this is supposed to be Python you should use Python syntax, not stuff like `Function` and `//` for a comment. — PM 2Ring, Nov 22 '17 at 20:06
What have you tried so far? Also, I think your example is confusing- is the 3rd parameter supposed to be the number of points found between the two starting points, or is it a number of segments formed by each pair of points? — Surreal Dreams, Nov 22 '17 at 20:06

score 22 · Accepted Answer · edited Dec 22 '19 at 21:43

22

Thanks to linearity of the line connecting two points, you can simply use numpy.linspace for each dimension independently:

import numpy

def getEquidistantPoints(p1, p2, parts):
    return zip(numpy.linspace(p1[0], p2[0], parts+1),
               numpy.linspace(p1[1], p2[1], parts+1))

For example:

>>> list(getEquidistantPoints((1,1), (5,5), 4))
>>> [(1.0, 1.0), (2.0, 2.0), (3.0, 3.0), (4.0, 4.0), (5.0, 5.0)]

edited Dec 22 '19 at 21:43

martineau

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answered Nov 22 '17 at 20:08

randomir

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Exactly what I needed for generating higher (7+) dimensional lines quickly and easily. – Harsh Apr 09 '18 at 20:50
4

Just replace the return line with `zip(*[numpy.linspace(p1[i], p2[i], parts+1) for i in range(len(p1))])` – Harsh Apr 09 '18 at 21:15
@Harsh Nice generalization! – randomir Apr 10 '18 at 02:01

Joe Iddon · Answer 2 · 2017-11-22T20:23:48.173

5

A pure Python solution using linear interpolation:

First create a linear interpolation function:

def lerp(v0, v1, i):
    return v0 + i * (v1 - v0)

and then just use this to interpolate between the x and y coordinates:

def getEquidistantPoints(p1, p2, n):
    return [(lerp(p1[0],p2[0],1./n*i), lerp(p1[1],p2[1],1./n*i)) for i in range(n+1)]

and a test with your values:

>>> getEquidistantPoints((1,1), (5,5), 4)
[(1.0, 1.0), (2.0, 2.0), (3.0, 3.0), (4.0, 4.0), (5.0, 5.0)]

edited Nov 22 '17 at 20:23

answered Nov 22 '17 at 20:08

Joe Iddon

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in Python 2, 1/n would evaluate to zero for all n>1, if I'm not mistaken – Grisha Nov 22 '17 at 20:22
@Grisha Good point, I hadn't considered that, I have updated the solution now so that it will work for both these versions – Joe Iddon Nov 22 '17 at 20:24
A simple(r) way to get "true division" division in Python 2 code is to add a `from __future__ import division` at the top of the script — which won't hurt even though not needed in Python 3. – martineau Dec 22 '19 at 21:52

s.ouchene · Answer 3 · 2021-10-01T10:30:34.377

The easiest way is to use an iterable (tuple, list, etc.). to specify the start and the end arguments of numpy.linspace:

p1 = (1,1)
p2 = (5,5)
HowManyParts = 4
A = np.linspace(p1, p2, HowManyParts+1)
print(A)

output:

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

You can also use complex numbers:

p1 = 1+1j
p2 = 5+5j
HowManyParts = 4
A = np.linspace(p1, p2, HowManyParts+1)
A = np.stack((A.real,A.imag), axis=1)
print(A)

Output:

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

Equidistant points between two points?

3 Answers3