如何将六边形 matplotlib 图转换为交互式散景图？

Question

我一直在使用优秀的 minisom 包，并希望以交互方式绘制反映自组织地图训练过程结果的六角形地图。已经有一个代码示例使用 matplotlib 静态地执行此操作，但要以交互方式执行此操作，我想使用 bokeh。这就是我挣扎的地方。

这是生成包中已有内容的简化 matplotlib 示例的代码 page:

from minisom import MiniSom
import pandas as pd
import numpy as np

import matplotlib.pyplot as plt
from matplotlib.patches import RegularPolygon
from matplotlib import cm

from bokeh.plotting import figure
from bokeh.io import save, show, output_file, output_notebook

output_notebook()

data = pd.read_csv('https://archive.ics.uci.edu/ml/machine-learning-databases/00236/seeds_dataset.txt', 
                    names=['area', 'perimeter', 'compactness', 'length_kernel', 'width_kernel',
                   'asymmetry_coefficient', 'length_kernel_groove', 'target'], sep='\t+')
t = data['target'].values
data = data[data.columns[:-1]]
# data normalisation
data = (data - np.mean(data, axis=0)) / np.std(data, axis=0)
data = data.values

# initialisation and training
som = MiniSom(15, 15, data.shape[1], sigma=1.5, learning_rate=.7, activation_distance='euclidean',
              topology='hexagonal', neighborhood_function='gaussian', random_seed=10)

som.train(data, 1000, verbose=True)

# plot hexagonal topology
f = plt.figure(figsize=(10,10))
ax = f.add_subplot(111)

ax.set_aspect('equal')

xx, yy = som.get_euclidean_coordinates()
umatrix = som.distance_map()
weights = som.get_weights()

for i in range(weights.shape[0]):
    for j in range(weights.shape[1]):
        wy = yy[(i, j)]*2/np.sqrt(3)*3/4
        hex = RegularPolygon((xx[(i, j)], wy), numVertices=6, radius=.95/np.sqrt(3),
                      facecolor=cm.Blues(umatrix[i, j]), alpha=.4, edgecolor='gray')
        ax.add_patch(hex)
for x in data:
    w = som.winner(x) 
    # place a marker on the winning position for the sample xx
    wx, wy = som.convert_map_to_euclidean(w) 
    wy = wy * 2 / np.sqrt(3) * 3 / 4
    plt.plot(wx, wy, markerfacecolor='None',
             markeredgecolor='black', markersize=12, markeredgewidth=2)

plt.show()

matplotlib hexagonal topology plot

我已经尝试将代码转换为散景，但生成的十六进制图（对我来说，原始）看起来需要垂直翻转到点上并拉直倾斜。

tile_centres_column = []
tile_centres_row = []
colours = []
for i in range(weights.shape[0]):
    for j in range(weights.shape[1]):
        wy = yy[(i, j)] * 2 / np.sqrt(3) * 3 / 4
        tile_centres_column.append(xx[(i, j)])
        tile_centres_row.append(wy)
        colours.append(cm.Blues(umatrix[i, j]))
        
weight_x = []
weight_y = []
for x in data:
    w = som.winner(x)
    wx, wy = som.convert_map_to_euclidean(xy=w)
    wy = wy * 2 / np.sqrt(3) * 3/4
    weight_x.append(wx)
    weight_y.append(wy)

# plot hexagonal topology
plot = figure(plot_width=800, plot_height=800,
              match_aspect=True) 
plot.hex_tile(q=tile_centres_column, r=tile_centres_row, 
              size=.95 / np.sqrt(3),
              color=colours,
              fill_alpha=.4,
              line_color='black')
plot.dot(x=weight_x, y=weight_y,
         fill_color='black',
         size=12)

show(plot)

bokeh hexagonal topology plot

如何将其转化为散景图？

Answer 1

在向 minisom 包作者寻求帮助后找到了如何做。完整代码可用 here.

from bokeh.colors import RGB
from bokeh.io import curdoc, show, output_notebook
from bokeh.transform import factor_mark, factor_cmap
from bokeh.models import ColumnDataSource, HoverTool
from bokeh.plotting import figure, output_file

hex_centre_col, hex_centre_row = [], []
hex_colour = []
label = []

# define labels
SPECIES = ['Kama', 'Rosa', 'Canadian']

for i in range(weights.shape[0]):
    for j in range(weights.shape[1]):
        wy = yy[(i, j)] * 2 / np.sqrt(3) * 3 / 4
        hex_centre_col.append(xx[(i, j)])
        hex_centre_row.append(wy)
        hex_colour.append(cm.Blues(umatrix[i, j]))

weight_x, weight_y = [], []
for cnt, i in enumerate(data):
    w = som.winner(i)
    wx, wy = som.convert_map_to_euclidean(xy=w)
    wy = wy * 2 / np.sqrt(3) * 3 / 4
    weight_x.append(wx)
    weight_y.append(wy)
    label.append(SPECIES[t[cnt]-1])
    
# convert matplotlib colour palette to bokeh colour palette
hex_plt = [(255 * np.array(i)).astype(int) for i in hex_colour]
hex_bokeh = [RGB(*tuple(rgb)).to_hex() for rgb in hex_plt]

output_file("resulting_images/som_seed_hex.html")

# initialise figure/plot
fig = figure(title="SOM: Hexagonal Topology",
             plot_height=800, plot_width=800,
             match_aspect=True,
             tools="wheel_zoom,save,reset")

# create data stream for plotting
source_hex = ColumnDataSource(
    data = dict(
        x=hex_centre_col,
        y=hex_centre_row,
        c=hex_bokeh
    )
)

source_pages = ColumnDataSource(
    data=dict(
        wx=weight_x,
        wy=weight_y,
        species=label
    )
)

# define markers
MARKERS = ['diamond', 'cross', 'x']

# add shapes to plot
fig.hex(x='y', y='x', source=source_hex,
        size=100 * (.95 / np.sqrt(3)),
        alpha=.4,
        line_color='gray',
        fill_color='c')

fig.scatter(x='wy', y='wx', source=source_pages, 
            legend_field='species',
            size=20, 
            marker=factor_mark(field_name='species', markers=MARKERS, factors=SPECIES),
            color=factor_cmap(field_name='species', palette='Category10_3', factors=SPECIES))

# add hover-over tooltip
fig.add_tools(HoverTool(
    tooltips=[
        ("label", '@species'),
        ("(x,y)", '($x, $y)')],
    mode="mouse", 
    point_policy="follow_mouse"
))

show(fig)

如何将六边形 matplotlib 图转换为交互式散景图？

How to translate hexagon matplotlib plot to an interactive bokeh plot?

python-3.x

matplotlib

plot

bokeh

hexagonal-tiles