Is it possible to find similar orders by multi-dimensional scaling in scikit learn?

Question

I have several files containing 3D positions of 10 points (as plotting in corresponding pictures). I would like to use multi-dimensional scaling to find similar orderings and print out different orderings. For example, here ordering from text file 1, 2 and 4 are completely the same, but different from 3.

import numpy as np

from matplotlib import pyplot as plt
from matplotlib.collections import LineCollection

from sklearn import manifold
from sklearn.metrics import euclidean_distances
from sklearn.decomposition import PCA

A1=[[0.000, 0.000, 0.5],
[0.250, 0.000, 0.5],
[0.125, 0.250, 0.5],
[0.375, 0.250, 0.5],
[0.250, 0.500, 0.5],
[0.500, 0.500, 0.5],
[0.125, 0.750, 0.5],
[0.375, 0.750, 0.5],
[0.000, 1.000, 0.5],
[0.250, 1.000, 0.5]]
A2=[[0.500, 0.000, 0.5],
[0.750, 0.000, 0.5],
[0.375, 0.250, 0.5],
[0.625, 0.250, 0.5],
[0.250, 0.500, 0.5],
[0.500, 0.500, 0.5],
[0.375, 0.750, 0.5],
[0.625, 0.750, 0.5],
[0.500, 1.000, 0.5],
[0.750, 1.000, 0.5]]
A3=[[0.500, 0.000, 0.5],
[0.750, 0.000, 0.5],
[0.625, 0.250, 0.5],
[0.875, 0.250, 0.5],
[0.250, 0.500, 0.5],
[0.500, 0.500, 0.5],
[0.375, 0.750, 0.5],
[0.625, 0.750, 0.5],
[0.500, 1.000, 0.5],
[0.750, 1.000, 0.5]]
A4=[[0.250, 0.000, 0.5],
[0.500, 0.000, 0.5],
[0.375, 0.250, 0.5],
[0.625, 0.250, 0.5],
[0.500, 0.500, 0.5],
[0.750, 0.500, 0.5],
[0.375, 0.750, 0.5],
[0.625, 0.750, 0.5],
[0.250, 1.000, 0.5],
[0.500, 1.000, 0.5]]

print(len(A1), len(A2), len(A3), len(A4))
a1=euclidean_distances(A1)
a2=euclidean_distances(A2)
a3=euclidean_distances(A3)
a4=euclidean_distances(A4)
print(a1)

OUTPUT

Number of different orders: 2
A1
A3

What is your definition of a "similar ordering"? Why are all the z-coordinates in your example 0.5? Can we assume the points all lie in a single plane? — Svetlin Mladenov, Feb 17 '21 at 13:42
@SvetlinMladenov For example, orders 1, 2, and 4 have a V shape while order 3 is not. In all of my files, I all assume z=0.5, then I think a single plane is ok. But I think it is better if it is written generally for 3 dimensions. — Binh Thien, Feb 17 '21 at 14:33

Rafael Valero · Accepted Answer · 2021-02-20T11:33:57.393

Set up the data and libraries:

import numpy as np
import pandas as pd

from matplotlib import pyplot as plt
from matplotlib.collections import LineCollection

from sklearn import manifold
from sklearn.metrics import euclidean_distances
from sklearn.decomposition import PCA

A1=[[0.000, 0.000, 0.5],
[0.250, 0.000, 0.5],
[0.125, 0.250, 0.5],
[0.375, 0.250, 0.5],
[0.250, 0.500, 0.5],
[0.500, 0.500, 0.5],
[0.125, 0.750, 0.5],
[0.375, 0.750, 0.5],
[0.000, 1.000, 0.5],
[0.250, 1.000, 0.5]]
A2=[[0.500, 0.000, 0.5],
[0.750, 0.000, 0.5],
[0.375, 0.250, 0.5],
[0.625, 0.250, 0.5],
[0.250, 0.500, 0.5],
[0.500, 0.500, 0.5],
[0.375, 0.750, 0.5],
[0.625, 0.750, 0.5],
[0.500, 1.000, 0.5],
[0.750, 1.000, 0.5]]
A3=[[0.500, 0.000, 0.5],
[0.750, 0.000, 0.5],
[0.625, 0.250, 0.5],
[0.875, 0.250, 0.5],
[0.250, 0.500, 0.5],
[0.500, 0.500, 0.5],
[0.375, 0.750, 0.5],
[0.625, 0.750, 0.5],
[0.500, 1.000, 0.5],
[0.750, 1.000, 0.5]]
A4=[[0.250, 0.000, 0.5],
[0.500, 0.000, 0.5],
[0.375, 0.250, 0.5],
[0.625, 0.250, 0.5],
[0.500, 0.500, 0.5],
[0.750, 0.500, 0.5],
[0.375, 0.750, 0.5],
[0.625, 0.750, 0.5],
[0.250, 1.000, 0.5],
[0.500, 1.000, 0.5]]

Let's place the data in a handy way and calculate distances.

""""""
# Number of different elemnts
segments_dic = {'A1': A1,
    'A2': A2,
    'A3': A3,
    'A4': A4,}

# To clasify the elements
segments_distances = []
for i in segments_dic.keys():
    segments_distances.append(round(euclidean_distances(segments_dic[i]).sum(),3))

Let's now check which are the closer groups of points:

"""Number of different elements / orders
I will round the results to make them comparable"""
different_elements = np.unique(segments_distances)
print("number of different orders: ",np.unique(segments_distances).__len__())
print("different orders: ", different_elements)
np.unique(segments_distances).__len__()

for i in different_elements:
    print("For element distance ",i," corresponding groups are: ")
    for j in segments_dic.keys():
        if i == round(euclidean_distances(segments_dic[j]).sum(),3):
            print(j)

This is how the outputs looks like:

number of different orders:  2
different orders:  [46.952 48.496]
For element distance  46.952  corresponding groups are: 
A1
A2
A4
For element distance  48.496  corresponding groups are: 
A3

Let see if we can validate this results with pics.

The plots in 2D:

"""Plots"""
for i in segments_dic.keys():
    # Rotate the data
    clf = PCA(n_components=2)
    X = clf.fit_transform(segments_dic[i])
    aux = pd.DataFrame(X)
    fig = plt.figure()
    plt.scatter(aux.iloc[:,0],aux.iloc[:,1])
    plt.title('{}'.format(i))
    fig.savefig('{}_representation.svg'.format(i))

Upload the pics: