我试图用python绘制一个特性图(SOM)。 为了简单起见,设想一个二维绘图,其中每个单元都表示为一个六边形。在
如本主题所示:Hexagonal Self-Organizing map in Python六边形并排放置,格式为网格。在
我设法写了下面的一段代码,它完美地适用于一定数量的多边形,并且只适用于少数形状(例如6×6或10×4的六边形)。然而,这种方法的一个重要特性是支持3x3的任何网格形状。在
def plot_map(grid,
d_matrix,
w=10,
title='SOM Hit map'):
"""
Plot hexagon map where each neuron is represented by a hexagon. The hexagon
color is given by the distance between the neurons (D-Matrix) Scaled
hexagons will appear on top of the background image whether the hits array
is provided. They are scaled according to the number of hits on each
neuron.
Args:
- grid: Grid dictionary (keys: centers, x, y ),
- d_matrix: array contaning the distances between each neuron
- w: width of the map in inches
- title: map title
Returns the Matplotlib SubAxis instance
"""
n_centers = grid['centers']
x, y = grid['x'], grid['y']
fig = plt.figure(figsize=(1.05 * w, 0.85 * y * w / x), dpi=100)
ax = fig.add_subplot(111)
ax.axis('equal')
# Discover difference between centers
collection_bg = RegularPolyCollection(
numsides=6, # a hexagon
rotation=0,
sizes=(y * (1.3 * 2 * math.pi * w) ** 2 / x,),
edgecolors = (0, 0, 0, 1),
array= d_matrix,
cmap = cm.gray,
offsets = n_centers,
transOffset = ax.transData,
)
ax.add_collection(collection_bg, autolim=True)
ax.axis('off')
ax.autoscale_view()
ax.set_title(title)
divider = make_axes_locatable(ax)
cax = divider.append_axes("right", size="5%", pad=0.05)
plt.colorbar(collection_bg, cax=cax)
return ax
我试着做一些能自动理解网格形状的东西。它没用(我也不知道为什么)。它总是在六边形之间出现一个不希望看到的空间
总结:我想生成3x3或6x6或10x4(等等)网格使用六边形之间没有空格为给定点和设置绘图宽度。在
按照要求,这是六边形位置的数据。正如你所看到的,它总是相同的模式
3x3
^{pr2}$6x6
{'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]]),
'x': array([ 6.]),
'y': array([ 6.])}
11x4
{'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 ],
[ 7.5 , 0.8660254 ],
[ 8.5 , 0.8660254 ],
[ 9.5 , 0.8660254 ],
[ 10.5 , 0.8660254 ],
[ 11.5 , 0.8660254 ],
[ 1. , 1.73205081],
[ 2. , 1.73205081],
[ 3. , 1.73205081],
[ 4. , 1.73205081],
[ 5. , 1.73205081],
[ 6. , 1.73205081],
[ 7. , 1.73205081],
[ 8. , 1.73205081],
[ 9. , 1.73205081],
[ 10. , 1.73205081],
[ 11. , 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],
[ 7.5 , 2.59807621],
[ 8.5 , 2.59807621],
[ 9.5 , 2.59807621],
[ 10.5 , 2.59807621],
[ 11.5 , 2.59807621],
[ 1. , 3.46410162],
[ 2. , 3.46410162],
[ 3. , 3.46410162],
[ 4. , 3.46410162],
[ 5. , 3.46410162],
[ 6. , 3.46410162],
[ 7. , 3.46410162],
[ 8. , 3.46410162],
[ 9. , 3.46410162],
[ 10. , 3.46410162],
[ 11. , 3.46410162]]),
'x': array([ 11.]),
'y': array([ 4.])}
我已经设法找到一个解决办法,根据给定的dpi计算出英寸的大小。之后,我计算两个相邻点之间的像素距离(通过使用隐藏散点图绘制)。这样我就可以计算出六边形apotem并正确估计六边形内圈的大小(正如matplotlib所期望的那样)。在
最后没有差距!在
相关问题 更多 >
编程相关推荐