如何在scikit-learn中计算相关的交叉验证分数？

Question

我正在做分类任务。 不过，我的结果略有不同：

#First Approach
kf = KFold(n=len(y), n_folds=10, shuffle=True, random_state=False)
pipe= make_pipeline(SVC())
for train_index, test_index in kf:
    X_train, X_test = X[train_index], X[test_index]
    y_train, y_test = y[train_index], y[test_index]

print ('Precision',np.mean(cross_val_score(pipe, X_train, y_train, scoring='precision')))



#Second Approach
clf.fit(X_train,y_train)
y_pred = clf.predict(X_test)
print ('Precision:', precision_score(y_test, y_pred,average='binary'))

#Third approach
pipe= make_pipeline(SCV())
print('Precision',np.mean(cross_val_score(pipe, X, y, cv=kf, scoring='precision')))

#Fourth approach

pipe= make_pipeline(SVC())
print('Precision',np.mean(cross_val_score(pipe, X_train, y_train, cv=kf, scoring='precision')))

日期：

Precision: 0.780422106837
Precision: 0.782051282051
Precision: 0.801544091998

/usr/local/lib/python3.5/site-packages/sklearn/cross_validation.py in cross_val_score(estimator, X, y, scoring, cv, n_jobs, verbose, fit_params, pre_dispatch)
   1431                                               train, test, verbose, None,
   1432                                               fit_params)
-> 1433                       for train, test in cv)
   1434     return np.array(scores)[:, 0]
   1435 

/usr/local/lib/python3.5/site-packages/sklearn/externals/joblib/parallel.py in __call__(self, iterable)
    798             # was dispatched. In particular this covers the edge
    799             # case of Parallel used with an exhausted iterator.
--> 800             while self.dispatch_one_batch(iterator):
    801                 self._iterating = True
    802             else:

/usr/local/lib/python3.5/site-packages/sklearn/externals/joblib/parallel.py in dispatch_one_batch(self, iterator)
    656                 return False
    657             else:
--> 658                 self._dispatch(tasks)
    659                 return True
    660 

/usr/local/lib/python3.5/site-packages/sklearn/externals/joblib/parallel.py in _dispatch(self, batch)
    564 
    565         if self._pool is None:
--> 566             job = ImmediateComputeBatch(batch)
    567             self._jobs.append(job)
    568             self.n_dispatched_batches += 1

/usr/local/lib/python3.5/site-packages/sklearn/externals/joblib/parallel.py in __init__(self, batch)
    178         # Don't delay the application, to avoid keeping the input
    179         # arguments in memory
--> 180         self.results = batch()
    181 
    182     def get(self):

/usr/local/lib/python3.5/site-packages/sklearn/externals/joblib/parallel.py in __call__(self)
     70 
     71     def __call__(self):
---> 72         return [func(*args, **kwargs) for func, args, kwargs in self.items]
     73 
     74     def __len__(self):

/usr/local/lib/python3.5/site-packages/sklearn/externals/joblib/parallel.py in <listcomp>(.0)
     70 
     71     def __call__(self):
---> 72         return [func(*args, **kwargs) for func, args, kwargs in self.items]
     73 
     74     def __len__(self):

/usr/local/lib/python3.5/site-packages/sklearn/cross_validation.py in _fit_and_score(estimator, X, y, scorer, train, test, verbose, parameters, fit_params, return_train_score, return_parameters, error_score)
   1522     start_time = time.time()
   1523 
-> 1524     X_train, y_train = _safe_split(estimator, X, y, train)
   1525     X_test, y_test = _safe_split(estimator, X, y, test, train)
   1526 

/usr/local/lib/python3.5/site-packages/sklearn/cross_validation.py in _safe_split(estimator, X, y, indices, train_indices)
   1589                 X_subset = X[np.ix_(indices, train_indices)]
   1590         else:
-> 1591             X_subset = safe_indexing(X, indices)
   1592 
   1593     if y is not None:

/usr/local/lib/python3.5/site-packages/sklearn/utils/__init__.py in safe_indexing(X, indices)
    161                                    indices.dtype.kind == 'i'):
    162             # This is often substantially faster than X[indices]
--> 163             return X.take(indices, axis=0)
    164         else:
    165             return X[indices]

IndexError: index 900 is out of bounds for size 900

所以，我的问题是上述哪种方法对于计算交叉验证的指标是正确的？ 我相信我的分数受到污染，因为我对何时执行交叉验证感到困惑。 那么，任何关于如何正确执行交叉验证分数的想法？

UPDATE

在培训步骤中进行评估？

X_train, X_test, y_train, y_test = train_test_split(X, y, random_state = False)
clf = make_pipeline(SVC())
# However, fot clf, you can use whatever estimator you like
kf = StratifiedKFold(y = y_train, n_folds=10, shuffle=True, random_state=False)
scores = cross_val_score(clf, X_train, y_train, cv = kf, scoring='precision')
print('Mean score : ', np.mean(scores))
print('Score variance : ', np.var(scores))

Answer 1

对于任何分类任务，使用StratifiedKFold交叉验证分割始终是好的。 在分层KFold中，每个类别的样本数量与您的分类问题相同。

那么这取决于您的分类问题类型。 总是很高兴看到精确度和召回分数。 如果是二元分类偏差，人们倾向于使用ROC AUC分数：

from sklearn import metrics
metrics.roc_auc_score(ytest, ypred)

让我们看看你的解决方案：

import numpy as np
from sklearn.cross_validation import cross_val_score
from sklearn.metrics import precision_score
from sklearn.cross_validation import KFold
from sklearn.pipeline import make_pipeline
from sklearn.svm import SVC

np.random.seed(1337)

X = np.random.rand(1000,5)

y = np.random.randint(0,2,1000)

kf = KFold(n=len(y), n_folds=10, shuffle=True, random_state=42)
pipe= make_pipeline(SVC(random_state=42))
for train_index, test_index in kf:
    X_train, X_test = X[train_index], X[test_index]
    y_train, y_test = y[train_index], y[test_index]

print ('Precision',np.mean(cross_val_score(pipe, X_train, y_train, scoring='precision')))
# Here you are evaluating precision score on X_train.

#Second Approach
clf = SVC(random_state=42)
clf.fit(X_train,y_train)
y_pred = clf.predict(X_test)
print ('Precision:', precision_score(y_test, y_pred, average='binary'))

# here you are evaluating precision score on X_test

#Third approach
pipe= make_pipeline(SVC())
print('Precision',np.mean(cross_val_score(pipe, X, y, cv=kf, scoring='precision')))

# Here you are splitting the data again and evaluating mean on each fold

因此，结果是不同的

Answer 2

首先，如文档中所述并在一些示例中显示 ， scikit-learn交叉验证cross_val_score执行以下操作：

1. 将数据集X拆分为N个折叠（根据参数cv ）。 它相应地分割标签y 。
2. 使用估算器（参数estimator ）在N-1之前的折叠上训练它。
3. 使用估算器预测最后一次折叠的标签。
4. 通过比较预测值和真实值返回分数（参数scoring ）
5. 通过更改测试折叠重复步骤2.到步骤4. 因此，您最终获得了一系列N分数。

让我们来看看你的每一个方法。

第一种方法：

为什么你会在cross_validation之前拆分训练集，因为scikit-learn函数会为你做这个？ 因此，您可以使用较少的数据训练模型，并以值得验证的分数结束

第二种方法：

在这里，您在数据上使用另一个指标而不是cross_validation_sore 。 因此，您无法将其与其他验证分数进行比较 - 因为它们是两个不同的东西。 一个是经典的误差百分比，而precision是用于校准二元分类器的度量（真或假）。 这是一个很好的指标（您可以检查ROC曲线，精确度和召回指标），但只比较这些指标。

第三种方法：

这个是更自然的一个。 这个分数很好 （我的意思是如果你想将它与其他分类器/估算器进行比较）。 但是，我会警告你不要直接采取平均值。 因为有两件事你可以比较：平均值，也可以是方差。 阵列的每个分数都与另一个分数不同，你可能想知道与其他估算者相比多少（你肯定希望你的方差尽可能小）

第四种方法：

Kfold似乎存在与Kfold无关的cross_val_score

最后：

仅使用第二种或第三种方法来比较估算器。 但他们肯定不会估计同样的事情 - 精确度与错误率。

clf = make_pipeline(SVC())
# However, fot clf, you can use whatever estimator you like
scores = cross_val_score(clf, X, y, cv = 10, scoring='precision')
print('Mean score : ', np.mean(scores))
print('Score variance : ', np.var(scores))

通过将clf更改为另一个估算器（或将其集成到循环中），您可以为每个eastimator分数并比较它们