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我一直在使用此代码为小型数据集生成 SOM。它会产生一个我无法弄清楚的错误。

from mvpa2.suite import *
data = np.array(
         [[1,0,0,1,1,0,1,0,1,0],
[1,0,0,0,1,0,1,0,1,1],
[1,0,1,0,1,0,0,0,1,0],
[0,1,1,1,0,1,0,1,0,0],
[1,1,0,0,1,0,1,1,1,1],
[0,0,1,0,1,1,0,1,0,0],
[1,1,0,1,0,1,0,0,1,0],
[1,0,0,0,1,0,1,1,1,1],
[0,1,1,0,1,0,1,0,0,0],
[1,1,0,1,0,1,0,1,1,0]])


data_names = \
        ['u1', 'u2', 'u3', 'u4',
         'u5', 'u6', 'u7', 'u8',
         'u9', 'u10']
som = SimpleSOMMapper((10, 10), 100, learning_rate=0.05)
som.train(data)
pl.imshow(som.K, origin='lower')
mapped = som(data)

pl.title('DATA SOM')

for i, m in enumerate(mapped):
    pl.text(m[1], m[0], data_names[i], ha='center', va='center',
           bbox=dict(facecolor='white', alpha=0.5, lw=0))
pl.show()

错误 ：

pl.imshow(som.K, origin='lower')
TypeError: Invalid dimensions for image data

任何帮助理解这一点？

4

1 回答 1

1

pl (pylab) 告诉您它不能显示编码为颜色的 10 维图像数据，这与这个示例不同，它是 pylab 编码为 rgb 的 3 维图像数据。

使用此代码，我能够对数据进行聚类并生成一些合理的东西（见下文）。

from mvpa2.suite import *
q = [[1,0,0,1,1,0,1,0,1,0], [1,0,0,0,1,0,1,0,1,1], [1,0,1,0,1,0,0,0,1,0], [0,1,1,1,0,1,0,1,0,0], [1,1,0,0,1,0,1,1,1,1], [0,0,1,0,1,1,0,1,0,0], [1,1,0,1,0,1,0,0,1,0], [1,0,0,0,1,0,1,1,1,1], [0,1,1,0,1,0,1,0,0,0], [1,1,0,1,0,1,0,1,1,0]]
q2 = []
for item in q:
    ri = []
    for x in item:
        ri.append(float(x))
    q2.append(ri)

data = np.array(q2)
data_names = ['u1', 'u2', 'u3', 'u4', 'u5', 'u6', 'u7', 'u8', 'u9', 'u10']
som = SimpleSOMMapper((10, 10), 100, learning_rate=0.05)
som.train(data)
mapped = som(data)

pl.title('DATA SOM')
pl.ylim([0,9])
pl.xlim([0,9])
for i, m in enumerate(mapped):
    print m[1],m[0],data_names[i]
    pl.text(m[1], m[0], data_names[i], ha='center', va='center',
           bbox=dict(facecolor='white', alpha=0.5, lw=0))
pl.savefig('b.png');

结果如下：

在此处输入图像描述

于 2014-03-19T02:30:18.867 回答