本文介绍了如何在 sklearn 中编写自定义估算器并对其使用交叉验证?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我想通过交叉验证检查新方法的预测误差.我想知道我是否可以将我的方法传递给 sklearn 的交叉验证函数,以及如何传递.

我想要sklearn.cross_validation(cv=10).mymethod之类的东西.

我还需要知道如何定义 mymethod 如果它是一个函数以及哪个输入元素和哪个输出

例如,我们可以将 mymethod 视为最小二乘估计器的实现(当然不是 sklearn 中的那些).

我找到了这个教程 link 但我不太清楚.

文档中,他们使用

>>>将 numpy 导入为 np>>>从 sklearn 导入 cross_validation>>>从 sklearn 导入数据集>>>从 sklearn 导入 svm>>>虹膜 = datasets.load_iris()>>>iris.data.shape, iris.target.shape((150, 4), (150,))>>>clf = svm.SVC(内核=线性",C=1)>>>分数 = cross_validation.cross_val_score(... clf, iris.data, iris.target, cv=5)...>>>分数

但问题是他们使用的是由 sklearn 中内置的函数获得的估计器 clf.我应该如何定义自己的估算器才能将其传递给 cross_validation.cross_val_score 函数?

例如,假设一个简单的估计器使用线性模型 $y=x\beta$,其中 beta 被估计为 X[1,:]+alpha,其中 alpha 是一个参数.我应该如何完成代码?

class my_estimator():定义适合(X,y):beta=X[1,:]+alpha #哪里可以将alpha传递给函数?返回测试版def scorer(estimator, X, y) #scorer 函数应该计算什么?返回 ?????

使用以下代码我收到一个错误:

class my_estimator():def fit(X, y, **kwargs):#alpha = kwargs['alpha']beta=X[1,:]#+alpha返回测试版

>>>cv=cross_validation.cross_val_score(my_estimator,x,y,scoring="mean_squared_error")回溯(最近一次调用最后一次):文件<stdin>",第 1 行,在 <module> 中文件C:\Python27\lib\site-packages\scikit_learn-0.14.1-py2.7-win32.egg\sklearn\cross_validation.py",第1152行,在cross_val_score对于火车,在 cv 中测试)文件C:\Python27\lib\site-packages\scikit_learn-0.14.1-py2.7-win32.egg\sklearn\externals\joblib\parallel.py",第516行,在__call__对于可迭代的函数、args、kwargs:文件C:\Python27\lib\site-packages\scikit_learn-0.14.1-py2.7-win32.egg\sklearn\cross_validation.py",第1152行,在<genexpr>对于火车,在 cv 中测试)文件C:\Python27\lib\site-packages\scikit_learn-0.14.1-py2.7-win32.egg\sklearn\base.py",第43行,克隆% (repr(estimator), type(estimator)))类型错误:无法克隆对象"(type <type 'classobj'>):它似乎不是 scikit-learn 估算器,因为它没有实现 'get_params' 方法.>>>

解决方案

答案也在于 sklearn 的 文档.

您需要定义两件事:

  • 一个实现 fit(X, y) 函数的估计器,X 是输入矩阵,y 是输出向量

  • 一个评分器函数,或可调用对象,可用于:scorer(estimator, X, y) 并返回给定模型的分数

参考你的例子:首先,scorer 不应该是估算器的一种方法,它是一个不同的概念.只需创建一个可调用的:

def scorer(estimator, X, y)返回 ?????# 计算任何你想要的,这由你来定义# 给定的估计量是好"还是坏"是什么意思

或者更简单的解决方案:您可以传递一个字符串 'mean_squared_error''accuracy'(完整列表可在 这部分文档) 到 cross_val_score 函数以使用预定义的评分器.

另一种可能是使用 make_scorer 工厂函数.

至于第二件事,您可以通过 fit_params dict 参数将参数传递给您的模型.org/stable/modules/generated/sklearn.model_selection.cross_val_score.html#sklearn.model_selection.cross_val_score" rel="noreferrer">cross_val_score 函数(如文档中所述).这些参数将传递给 fit 函数.

class my_estimator():def fit(X, y, **kwargs):alpha = kwargs['alpha']beta=X[1,:]+alpha返回测试版

在阅读了所有错误消息后,这些消息提供了对缺少什么的清晰概念,这里是一个简单的例子:

将 numpy 导入为 np从 sklearn.cross_validation 导入 cross_val_score类正则化回归器:def __init__(self, l = 0.01):self.l = ldef组合(自我,输入):return sum([i*w for (i,w) in zip([1] + inputs, self.weights)])定义预测(自我,X):返回 [self.combine(x) for x in X]定义分类(自我,输入):返回符号(self.predict(输入))def fit(self, X, y, **kwargs):self.l = kwargs['l']X = np.matrix(X)y = np.matrix(y)W = (X.transpose() * X).getI() * X.transpose() * yself.weights = [w[0] for w in W.tolist()]def get_params(self, deep = False):返回 {'l':self.l}X = np.matrix([[0, 0], [1, 0], [0, 1], [1, 1]])y = np.matrix([0, 1, 1, 0]).transpose()打印 cross_val_score(RegularizedRegressor(),X,是,fit_params={'l':0.1},评分 = 'mean_squared_error')

I would like to check the prediction error of a new method trough cross-validation.I would like to know if I can pass my method to the cross-validation function of sklearn and in case how.

I would like something like sklearn.cross_validation(cv=10).mymethod.

I need also to know how to define mymethod should it be a function and which input element and which output

For example we can consider as mymethod an implementation of the least square estimator (of course not the ones in sklearn) .

I found this tutorial link but it is not very clear to me.

In the documentation they use

>>> import numpy as np
>>> from sklearn import cross_validation
>>> from sklearn import datasets
>>> from sklearn import svm

>>> iris = datasets.load_iris()
>>> iris.data.shape, iris.target.shape
((150, 4), (150,))

 >>> clf = svm.SVC(kernel='linear', C=1)
 >>> scores = cross_validation.cross_val_score(
 ...    clf, iris.data, iris.target, cv=5)
 ...
 >>> scores

But the problem is that they are using as estimator clf that is obtained by a function built in sklearn. How should I define my own estimator in order that I can pass it to the cross_validation.cross_val_score function?

So for example suppose a simple estimator that use a linear model $y=x\beta$ where beta is estimated as X[1,:]+alpha where alpha is a parameter. How should I complete the code?

class my_estimator():
      def fit(X,y):
          beta=X[1,:]+alpha #where can I pass alpha to the function?
          return beta
      def scorer(estimator, X, y) #what should the scorer function compute?
          return ?????

With the following code I received an error:

class my_estimator():
    def fit(X, y, **kwargs):
        #alpha = kwargs['alpha']
        beta=X[1,:]#+alpha
        return beta


>>> cv=cross_validation.cross_val_score(my_estimator,x,y,scoring="mean_squared_error")
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "C:\Python27\lib\site-packages\scikit_learn-0.14.1-py2.7-win32.egg\sklearn\cross_validation.py", line 1152, in cross_val_score
    for train, test in cv)
  File "C:\Python27\lib\site-packages\scikit_learn-0.14.1-py2.7-win32.egg\sklearn\externals\joblib\parallel.py", line 516, in __call__
    for function, args, kwargs in iterable:
  File "C:\Python27\lib\site-packages\scikit_learn-0.14.1-py2.7-win32.egg\sklearn\cross_validation.py", line 1152, in <genexpr>
    for train, test in cv)
  File "C:\Python27\lib\site-packages\scikit_learn-0.14.1-py2.7-win32.egg\sklearn\base.py", line 43, in clone
    % (repr(estimator), type(estimator)))
TypeError: Cannot clone object '<class __main__.my_estimator at 0x05ACACA8>' (type <type 'classobj'>): it does not seem to be a scikit-learn estimator a it does not implement a 'get_params' methods.
>>>
解决方案

The answer also lies in sklearn's documentation.

You need to define two things:

  • an estimator that implements the fit(X, y) function, X being the matrix with inputs and y being the vector of outputs

  • a scorer function, or callable object that can be used with: scorer(estimator, X, y) and returns the score of given model

Referring to your example: first of all, scorer shouldn't be a method of the estimator, it's a different notion. Just create a callable:

def scorer(estimator, X, y)
    return ?????  # compute whatever you want, it's up to you to define
                  # what does it mean that the given estimator is "good" or "bad"

Or even a more simple solution: you can pass a string 'mean_squared_error' or 'accuracy' (full list available in this part of the documentation) to cross_val_score function to use a predefined scorer.

Another possibility is to use make_scorer factory function.

As for the second thing, you can pass parameters to your model through the fit_params dict parameter of the cross_val_score function (as mentioned in the documentation). These parameters will be passed to the fit function.

class my_estimator():
    def fit(X, y, **kwargs):
        alpha = kwargs['alpha']
        beta=X[1,:]+alpha
        return beta

After reading all the error messages, which provide quite clear idea of what's missing, here is a simple example:

import numpy as np
from sklearn.cross_validation import cross_val_score

class RegularizedRegressor:
    def __init__(self, l = 0.01):
        self.l = l

    def combine(self, inputs):
        return sum([i*w for (i,w) in zip([1] + inputs, self.weights)])

    def predict(self, X):
        return [self.combine(x) for x in X]

    def classify(self, inputs):
        return sign(self.predict(inputs))

    def fit(self, X, y, **kwargs):
        self.l = kwargs['l']
        X = np.matrix(X)
        y = np.matrix(y)
        W = (X.transpose() * X).getI() * X.transpose() * y

        self.weights = [w[0] for w in W.tolist()]

    def get_params(self, deep = False):
        return {'l':self.l}

X = np.matrix([[0, 0], [1, 0], [0, 1], [1, 1]])
y = np.matrix([0, 1, 1, 0]).transpose()

print cross_val_score(RegularizedRegressor(),
                      X,
                      y,
                      fit_params={'l':0.1},
                      scoring = 'mean_squared_error')

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08-13 19:39