绘制 SVM 决策边界

如何解决绘制 SVM 决策边界

以下代码拟合具有多项式核的 SVM 并绘制虹膜数据和决策边界。输入 X 使用数据的前 2 列，萼片长度和宽度。但是，我很难将第 3 列和第 4 列的输出再现为 X，即花瓣的长度和宽度。如何更改代码的绘图功能以使其正常工作？提前致谢。

import numpy as np
import matplotlib.pyplot as plt
from sklearn import datasets
from sklearn.svm import SVC

iris= datasets.load_iris()
y= iris.target
#X= iris.data[:,:2]  # sepal length and width
X= iris.data[:,2:]   # I tried a different X but it Failed.

# Ref: https://medium.com/all-things-ai/in-depth-parameter-tuning-for-svc-758215394769
def plotSVC(title):
    # create a mesh to plot in
    x_min,x_max = X[:,0].min() - 1,X[:,0].max() + 1
    y_min,y_max = X[:,1].min() - 1,1].max() + 1
    h = (x_max / x_min)/100
    xx,yy = np.meshgrid(np.arange(x_min,x_max,h),np.arange(y_min,y_max,h))
    plt.subplot(1,1,1)
    Z = svm_mod.predict(np.c_[xx.ravel(),yy.ravel()])
    Z = Z.reshape(xx.shape)
    plt.contourf(xx,yy,Z,cmap=plt.cm.Paired,alpha=0.8)
    plt.scatter(X[:,0],1],c=y,cmap=plt.cm.Paired)
    plt.xlabel('Sepal length')
    plt.ylabel('Sepal width')
    plt.xlim(xx.min(),xx.max())
    plt.title(title)
    plt.show()

svm= SVC(C= 10,kernel='poly',degree=2,coef0=1,max_iter=500000)
svm_mod= svm.fit(X,y)
plotSVC('kernel='+ str('polynomial'))

错误：

  import sys
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-57-4515f111e34d> in <module>()
      2 svm= SVC(C= 10,max_iter=500000)
      3 svm_mod= svm.fit(X,y)
----> 4 plotSVC('kernel='+ str('polynomial'))

<ipython-input-56-556d4a22026a> in plotSVC(title)
     10     Z = svm_mod.predict(np.c_[xx.ravel(),yy.ravel()])
     11     Z = Z.reshape(xx.shape)
---> 12     plt.contourf(xx,alpha=0.8)
     13     plt.scatter(X[:,cmap=plt.cm.Paired)
     14     plt.xlabel('Sepal length')

~/anaconda3/lib/python3.6/site-packages/matplotlib/pyplot.py in contourf(*args,**kwargs)
   2931                       mplDeprecation)
   2932     try:
-> 2933         ret = ax.contourf(*args,**kwargs)
   2934     finally:
   2935         ax._hold = washold

~/anaconda3/lib/python3.6/site-packages/matplotlib/__init__.py in inner(ax,*args,**kwargs)
   1853                         "the Matplotlib list!)" % (label_namer,func.__name__),1854                         RuntimeWarning,stacklevel=2)
-> 1855             return func(ax,**kwargs)
   1856 
   1857         inner.__doc__ = _add_data_doc(inner.__doc__,~/anaconda3/lib/python3.6/site-packages/matplotlib/axes/_axes.py in contourf(self,**kwargs)
   6179             self.cla()
   6180         kwargs['filled'] = True
-> 6181         contours = mcontour.QuadContourSet(self,**kwargs)
   6182         self.autoscale_view()
   6183         return contours

~/anaconda3/lib/python3.6/site-packages/matplotlib/contour.py in __init__(self,ax,**kwargs)
    844         self._transform = kwargs.pop('transform',None)
    845 
--> 846         kwargs = self._process_args(*args,**kwargs)
    847         self._process_levels()
    848 

~/anaconda3/lib/python3.6/site-packages/matplotlib/contour.py in _process_args(self,**kwargs)
   1414                 self._corner_mask = mpl.rcParams['contour.corner_mask']
   1415 
-> 1416             x,y,z = self._contour_args(args,kwargs)
   1417 
   1418             _mask = ma.getmask(z)

~/anaconda3/lib/python3.6/site-packages/matplotlib/contour.py in _contour_args(self,args,kwargs)
   1472             args = args[1:]
   1473         elif Nargs <= 4:
-> 1474             x,z = self._check_xyz(args[:3],kwargs)
   1475             args = args[3:]
   1476         else:

~/anaconda3/lib/python3.6/site-packages/matplotlib/contour.py in _check_xyz(self,kwargs)
   1508             raise TypeError("Input z must be a 2D array.")
   1509         elif z.shape[0] < 2 or z.shape[1] < 2:
-> 1510             raise TypeError("Input z must be at least a 2x2 array.")
   1511         else:
   1512             Ny,Nx = z.shape

TypeError: Input z must be at least a 2x2 array.

解决方法

工作代码

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出：

---

原因：

在 import numpy as np import matplotlib.pyplot as plt from sklearn import datasets from sklearn.svm import SVC iris= datasets.load_iris() y= iris.target #X= iris.data[:,:2] # sepal length and width X= iris.data[:,2:] # I tried a different X but it failed. # Ref: https://medium.com/all-things-ai/in-depth-parameter-tuning-for-svc-758215394769 def plotSVC(title): # create a mesh to plot in x_min,x_max = X[:,0].min() - 1,X[:,0].max() + 1 y_min,y_max = X[:,1].min() - 1,1].max() + 1 h = (x_max - x_min)/100 xx,yy = np.meshgrid(np.arange(x_min,x_max,h),np.arange(y_min,y_max,h)) plt.subplot(1,1,1) z = svm_mod.predict(np.c_[xx.ravel(),yy.ravel()]) z = z.reshape(xx.shape) plt.contourf(xx,yy,z,cmap=plt.cm.Paired,alpha=0.8) plt.scatter(X[:,0],1],c=y,cmap=plt.cm.Paired) plt.xlabel('Sepal length') plt.ylabel('Sepal width') plt.xlim(xx.min(),xx.max()) plt.title(title) plt.show() pass svm= SVC(C= 10,kernel='poly',degree=2,coef0=1,max_iter=500000) svm_mod= svm.fit(X,y) plotSVC('kernel='+ str('polynomial')) 行的 inf 中除以零得到 h

需要h = (x_max / x_min)/100

---

我通过阅读规定的例外发现了这一点

类型错误：输入 z 必须至少为 2x2 数组。

然后倒回去看到h = (x_max - x_min)/100的形状来自z的形状，它依赖于xx，它是h，这没有意义，这然后就轻松解决了。

我认为您应该学习如何更好地使用调试器。