Hexagonal Self-Organizing map in Python
Asked Answered
M

2

8

I am looking for hexagonal self-organizing map on Python.

hexagonal tiling

  1. ready module. If one exists.
  2. way to plot hexagonal cell
  3. algorithms to work with hexagonal cells as array or smth else

About: A self-organizing map (SOM) or self-organizing feature map (SOFM) is a type of artificial neural network that is trained using unsupervised learning to produce a low-dimensional (typically two-dimensional)

Moultrie answered 25/2, 2010 at 14:27 Comment(2)
@S.Lott: A self organizing map is an AI algorithm. See en.wikipedia.org/wiki/Self-organizing_mapRearmost
What are you looking for? Is this an AI question on self-organizing algorithms, or a graphics question on drawing hexagons or a data representation question on how to represent a hexagonal tiling of a surface?Denticle
W
7

I don't have an answer for point 1, but some hints for point 2 and 3. In your context, you're not modelling a physical 2D space but a conceptual space with tiles that have 6 neighbors. This can be modelled with square tiles arranged in columns with the odd colums shifted vertically by half the size of a square. I'll try an ASCII diagram:

 ___     ___     ___     
|   |___|   |___|   |___
|___|   |___|   |___|   |
|   |___|   |___|   |___|
|___|   |___|   |___|   |
|   |___|   |___|   |___|
|___|   |___|   |___|   |
    |___|   |___|   |___|

You can see easily that each square has 6 neighbors (except the ones on the edges of course). This gets easily modeled as a 2D array of squares, and the rules to compute the coordinates of the square at at position (i, j), i being the row and j the column are quite simple:

if j is even:

(i+1, j), (i-1, j), (i, j-1), (i, j+1), (i-1, j-1), (i+1, j-1)

if j is odd:

(i+1, j), (i-1, j), (i, j-1), (i, j+1), (i+1, j-1), (i+1, j+1)

(the 4 first terms are identical)

Worrywart answered 26/2, 2010 at 8:3 Comment(0)
S
8

I know this discussion is 4 years old, however I haven't find a satisfactory answer over the web.

If you have something as a array mapping the input to the neuron and a 2-d array related to the location for each neuron.

For example consider something like this:

hits = array([1, 24, 14, 16,  6, 11,  8, 23, 15, 16, 15,  9, 20,  1,  3, 29,  4,
              32, 22,  7, 26, 26, 35, 23,  7,  6, 11,  9, 18, 17, 22, 19, 34,  1,
              36,  3, 31, 10, 22, 11, 21, 18, 29,  3,  6, 32, 15, 30, 27],
             dtype=int32)
centers = array([[ 1.5       ,  0.8660254 ],
                 [ 2.5       ,  0.8660254 ],
                 [ 3.5       ,  0.8660254 ],
                 [ 4.5       ,  0.8660254 ],
                 [ 5.5       ,  0.8660254 ],
                 [ 6.5       ,  0.8660254 ],
                 [ 1.        ,  1.73205081],
                 [ 2.        ,  1.73205081],
                 [ 3.        ,  1.73205081],
                 [ 4.        ,  1.73205081],
                 [ 5.        ,  1.73205081],
                 [ 6.        ,  1.73205081],
                 [ 1.5       ,  2.59807621],
                 [ 2.5       ,  2.59807621],
                 [ 3.5       ,  2.59807621],
                 [ 4.5       ,  2.59807621],
                 [ 5.5       ,  2.59807621],
                 [ 6.5       ,  2.59807621],
                 [ 1.        ,  3.46410162],
                 [ 2.        ,  3.46410162],
                 [ 3.        ,  3.46410162],
                 [ 4.        ,  3.46410162],
                 [ 5.        ,  3.46410162],
                 [ 6.        ,  3.46410162],
                 [ 1.5       ,  4.33012702],
                 [ 2.5       ,  4.33012702],
                 [ 3.5       ,  4.33012702],
                 [ 4.5       ,  4.33012702],
                 [ 5.5       ,  4.33012702],
                 [ 6.5       ,  4.33012702],
                 [ 1.        ,  5.19615242],
                 [ 2.        ,  5.19615242],
                 [ 3.        ,  5.19615242],
                 [ 4.        ,  5.19615242],
                 [ 5.        ,  5.19615242],
                 [ 6.        ,  5.19615242]])

So I'do this using a the following method:

from matplotlib import collections, transforms
from matplotlib.colors import colorConverter
from matplotlib import cm
import matplotlib.pyplot as plt
import numpy as np

def plot_map(hits, n_centers, w=10):
    """
    Plot Map
    """

    fig = plt.figure(figsize=(w, .7 * w))
    ax = fig.add_subplot(111)
    hits_count = np.histogram(hits, bins=n_centers.shape[0])[0]
    # Discover difference between centers
    collection = RegularPolyCollection(
        numsides=6, # a hexagon 
        rotation=0, sizes=( (6.6*w)**2 ,),
        edgecolors = (0, 0, 0, 1),
        array= hits_count,
        cmap = cm.winter,
        offsets = n_centers,
        transOffset = ax.transData,
    )
    ax.axis('off')
    ax.add_collection(collection, autolim=True)
    ax.autoscale_view()
    fig.colorbar(collection)
    return ax

_ = plot_map(som_classif, matrix)

Finally I got this output:

enter image description here

EDIT

An updated version of this code on https://mcmap.net/q/1325290/-matplotlib-regularpolygon-collection-location-on-the-canvas

Situla answered 15/5, 2014 at 21:35 Comment(0)
W
7

I don't have an answer for point 1, but some hints for point 2 and 3. In your context, you're not modelling a physical 2D space but a conceptual space with tiles that have 6 neighbors. This can be modelled with square tiles arranged in columns with the odd colums shifted vertically by half the size of a square. I'll try an ASCII diagram:

 ___     ___     ___     
|   |___|   |___|   |___
|___|   |___|   |___|   |
|   |___|   |___|   |___|
|___|   |___|   |___|   |
|   |___|   |___|   |___|
|___|   |___|   |___|   |
    |___|   |___|   |___|

You can see easily that each square has 6 neighbors (except the ones on the edges of course). This gets easily modeled as a 2D array of squares, and the rules to compute the coordinates of the square at at position (i, j), i being the row and j the column are quite simple:

if j is even:

(i+1, j), (i-1, j), (i, j-1), (i, j+1), (i-1, j-1), (i+1, j-1)

if j is odd:

(i+1, j), (i-1, j), (i, j-1), (i, j+1), (i+1, j-1), (i+1, j+1)

(the 4 first terms are identical)

Worrywart answered 26/2, 2010 at 8:3 Comment(0)

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