下面的代码正在工作。我有一个例程运行交叉验证方案,使用一个线性模型,先前定义的学习。我对此没有异议。我的问题是:如果我将代码model=linear_model.LinearRegression()替换为model=RBF('multiquadric') (请参阅__main__中的第14和第15行),它将不再工作。因此,我的问题实际上是在RBF课堂上,我试图模仿一个滑雪模型。
如果我替换了上面描述的代码,我会得到以下错误:
FitFailedWarning)
/home/daniel/anaconda3/lib/python3.7/site-packages/sklearn/model_selection/_validation.py:536: FitFailedWarning: Estimator fit failed. The score on this train-test partition for these parameters will be set to nan. Details:
ValueError: All arrays must be equal length.
FitFailedWarning)1)我应该在RBF类中定义分数函数吗?
2)如何做到这一点?我迷路了。由于我继承了BaseEstimator和RegressorMixin,我希望这在内部得到解决。
3)还有什么东西不见了吗?
from sklearn import datasets
import numpy as np
import pandas as pd
from sklearn import linear_model
from sklearn import model_selection
from sklearn.pipeline import Pipeline
from sklearn.preprocessing import StandardScaler
from scipy.interpolate import Rbf
np.random.seed(0)
from sklearn.base import BaseEstimator, RegressorMixin
class RBF(BaseEstimator, RegressorMixin):
def __init__(self,function):
self.function=function
def fit(self,x,y):
self.rbf = Rbf(x, y,function=self.function)
def predict(self,x):
return self.rbf(x)
if __name__ == "__main__":
# Load Data
targetName='HousePrice'
data=datasets.load_boston()
featuresNames=list(data.feature_names)
featuresData=data.data
targetData = data.target
df=pd.DataFrame(featuresData,columns=featuresNames)
df[targetName]=targetData
independent_variable_list=featuresNames
dependent_variable=targetName
X=df[independent_variable_list].values
y=np.squeeze(df[[dependent_variable]].values)
# Model Definition
model=linear_model.LinearRegression()
#model=RBF('multiquadric')
# Cross validation routine
number_splits=5
score_list=['neg_mean_squared_error','neg_mean_absolute_error','r2']
kfold = model_selection.KFold(n_splits=number_splits,shuffle=True, random_state=0)
scalar = StandardScaler()
pipeline = Pipeline([('transformer', scalar), ('estimator', model)])
results = model_selection.cross_validate(pipeline, X, y, cv=kfold, scoring=score_list,return_train_score=True)
for score in score_list:
print(score+':')
print('Train: '+'Mean',np.mean(results['train_'+score]),'Standard Error',np.std(results['train_'+score]))
print('Test: '+'Mean',np.mean(results['test_'+score]),'Standard Error',np.std(results['test_'+score]))发布于 2020-02-18 05:46:53
让我们看一下文档这里
*args :数组 x,y,z,…,d,其中x,y,z,…是节点的坐标,d是节点上的值数组。
因此,它采用可变长度参数,最后一个参数是值,在您的情况下是y。参数k是所有数据点(所有其他参数z, y, z, …相同)的kth坐标。
在文档之后,您的代码应该是
from sklearn import datasets
import numpy as np
import pandas as pd
from sklearn import linear_model
from sklearn import model_selection
from sklearn.pipeline import Pipeline
from sklearn.preprocessing import StandardScaler
from scipy.interpolate import Rbf
np.random.seed(0)
from sklearn.base import BaseEstimator, RegressorMixin
class RBF(BaseEstimator, RegressorMixin):
def __init__(self,function):
self.function=function
def fit(self,X,y):
self.rbf = Rbf(*X.T, y,function=self.function)
def predict(self,X):
return self.rbf(*X.T)
# Load Data
data=datasets.load_boston()
X = data.data
y = data.target
number_splits=5
score_list=['neg_mean_squared_error','neg_mean_absolute_error','r2']
kfold = model_selection.KFold(n_splits=number_splits,shuffle=True, random_state=0)
scalar = StandardScaler()
model = RBF(function='multiquadric')
pipeline = Pipeline([('transformer', scalar), ('estimator', model)])
results = model_selection.cross_validate(pipeline, X, y, cv=kfold, scoring=score_list,return_train_score=True)
for score in score_list:
print(score+':')
print('Train: '+'Mean',np.mean(results['train_'+score]),'Standard Error',np.std(results['train_'+score]))
print('Test: '+'Mean',np.mean(results['test_'+score]),'Standard Error',np.std(results['test_'+score]))输出
neg_mean_squared_error:
Train: Mean -1.552450953914355e-20 Standard Error 7.932530906290208e-21
Test: Mean -23.007377210596463 Standard Error 4.254629143836107
neg_mean_absolute_error:
Train: Mean -9.398502208736061e-11 Standard Error 2.4673749061941226e-11
Test: Mean -3.1319779583728673 Standard Error 0.2162343985534446
r2:
Train: Mean 1.0 Standard Error 0.0
Test: Mean 0.7144217179633185 Standard Error 0.08526294242760363为什么*X.T:正如我们所看到的,每个参数对应于所有数据点的一个轴,因此我们转换它们,然后使用*操作符扩展和传递每个子数组作为变量长度函数的参数。
看起来,最新的实现有一个mode参数,在这里我们可以直接传递N-D数组。
https://stackoverflow.com/questions/60245240
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