我正在 sci-kit learn 中构建 MLPClassifier 模型。我使用 gridSearchCV 和 roc_auc 对模型进行评分。训练和考试的平均成绩在 0.76 左右，还不错。 cv_results_ 的输出是:
Train set AUC: 0.553465272412
Grid best score (AUC): 0.757236688092
Grid best parameter (max. AUC): {'hidden_layer_sizes': 10}
{ 'mean_fit_time': array([63.54, 136.37, 136.32, 119.23, 121.38, 124.03]),
'mean_score_time': array([ 0.04, 0.04, 0.04, 0.05, 0.05, 0.06]),
'mean_test_score': array([ 0.76, 0.74, 0.75, 0.76, 0.76, 0.76]),
'mean_train_score': array([ 0.76, 0.76, 0.76, 0.77, 0.77, 0.77]),
'param_hidden_layer_sizes': masked_array(data = [5 (5, 5) (5, 10) 10 (10, 5) (10, 10)],
mask = [False False False False False False],
fill_value = ?)
,
'params': [ {'hidden_layer_sizes': 5},
{'hidden_layer_sizes': (5, 5)},
{'hidden_layer_sizes': (5, 10)},
{'hidden_layer_sizes': 10},
{'hidden_layer_sizes': (10, 5)},
{'hidden_layer_sizes': (10, 10)}],
'rank_test_score': array([ 2, 6, 5, 1, 4, 3]),
'split0_test_score': array([ 0.76, 0.75, 0.75, 0.76, 0.76, 0.76]),
'split0_train_score': array([ 0.76, 0.75, 0.75, 0.76, 0.76, 0.76]),
'split1_test_score': array([ 0.77, 0.76, 0.76, 0.77, 0.76, 0.76]),
'split1_train_score': array([ 0.76, 0.75, 0.75, 0.76, 0.76, 0.76]),
'split2_test_score': array([ 0.74, 0.72, 0.73, 0.74, 0.74, 0.75]),
'split2_train_score': array([ 0.77, 0.77, 0.77, 0.77, 0.77, 0.77]),
'std_fit_time': array([47.59, 1.29, 1.86, 3.43, 2.49, 9.22]),
'std_score_time': array([ 0.01, 0.01, 0.01, 0.00, 0.00, 0.01]),
'std_test_score': array([ 0.01, 0.01, 0.01, 0.01, 0.01, 0.01]),
'std_train_score': array([ 0.01, 0.01, 0.01, 0.01, 0.01, 0.00])}
def model_mlp (X_train, y_train, verbose=True, random_state = 42):
grid_values = {'hidden_layer_sizes': [(5), (5,5), (5, 10),
(10), (10, 5), (10, 10)]}
# MLP requires scaling of all predictors
scaler = StandardScaler()
X_train = scaler.fit_transform(X_train)
mlp = MLPClassifier(solver='adam', learning_rate_init=1e-4,
max_iter=200,
verbose=False,
random_state=random_state)
# perform the grid search
grid_auc = GridSearchCV(mlp,
param_grid=grid_values,
scoring='roc_auc',
verbose=2, n_jobs=-1)
grid_auc.fit(X_train, y_train)
y_hat = grid_auc.predict(X_train)
# print out the results
if verbose:
print('Train set AUC: ', roc_auc_score(y_train, y_hat))
print('Grid best score (AUC): ', grid_auc.best_score_)
print('Grid best parameter (max. AUC): ', grid_auc.best_params_)
print('')
pp = pprint.PrettyPrinter(indent=4)
pp.pprint (grid_auc.cv_results_)
print ('MLPClassifier fitted, {:.2f} seconds used'.format (time.time () - t))
return grid_auc.best_estimator_
由于这种差异，我决定“模拟”GridSearchCV 例程并得到以下结果:
Shape X_train: (107119, 15)
Shape y_train: (107119,)
Shape X_val: (52761, 15)
Shape y_val: (52761,)
layers roc-auc
Seq l1 l2 train test iters runtime
1 5 0 0.5522 0.5488 85 20.54
2 5 5 0.5542 0.5513 80 27.10
3 5 10 0.5544 0.5521 83 28.56
4 10 0 0.5532 0.5516 61 15.24
5 10 5 0.5540 0.5518 54 19.86
6 10 10 0.5507 0.5474 56 21.09
分数都在0.55左右，与上面代码中手动计算的结果一致。令我惊讶的是结果没有变化。看来我犯了一些错误，但我找不到错误，请参阅代码:
def simple_mlp (X, y, verbose=True, random_state = 42):
def do_mlp (X_t, X_v, y_t, y_v, n, l1, l2=None):
if l2 is None:
layers = (l1)
l2 = 0
else:
layers = (l1, l2)
t = time.time ()
mlp = MLPClassifier(solver='adam', learning_rate_init=1e-4,
hidden_layer_sizes=layers,
max_iter=200,
verbose=False,
random_state=random_state)
mlp.fit(X_t, y_t)
y_hat_train = mlp.predict(X_t)
y_hat_val = mlp.predict(X_v)
if verbose:
av = 'samples'
acc_trn = roc_auc_score(y_train, y_hat_train, average=av)
acc_tst = roc_auc_score(y_val, y_hat_val, average=av)
print ("{:5d}{:4d}{:4d}{:7.4f}{:7.4f}{:9d}{:8.2f}"
.format(n, l1, l2, acc_trn, acc_tst, mlp.n_iter_, time.time() - t))
return mlp, n + 1
X_train, X_val, y_train, y_val = train_test_split (X, y, test_size=0.33, random_state=random_state)
if verbose:
print('Shape X_train:', X_train.shape)
print('Shape y_train:', y_train.shape)
print('Shape X_val:', X_val.shape)
print('Shape y_val:', y_val.shape)
# MLP requires scaling of all predictors
scaler = StandardScaler()
X_train = scaler.fit_transform(X_train)
X_val = scaler.transform(X_val)
n = 1
layers1 = [5, 10]
layers2 = [5, 10]
if verbose:
print (" layers roc-auc")
print (" Seq l1 l2 train validation iters runtime")
for l1 in layers1:
mlp, n = do_mlp (X_train, X_val, y_train, y_val, n, l1)
for l2 in layers2:
mlp, n = do_mlp (X_train, X_val, y_train, y_val, n, l1, l2)
return mlp
我在这两种情况下使用完全相同的数据(159880 个观测值和 15 个预测变量)。我对 GridSearchCV 使用 cv=3 (默认值)，并在我的手工代码中使用相同比例的验证集。
在寻找可能的答案时，我发现 this post on SO它描述了同样的问题。没有人回答。也许有人明白到底发生了什么？
感谢您的宝贵时间。
编辑
我按照@Mohammed Kashif的建议检查了GridSearchCV和KFold的代码，确实发现了一个明确的注释，即KFold没有对数据进行混洗。因此，我在缩放器之前将以下代码添加到 model_mlp 中:
np.random.seed (random_state)
index = np.random.permutation (len(X_train))
X_train = X_train.iloc[index]
并进入 simple_mlp 作为 train_test_split 的替代品:
np.random.seed (random_state)
index = np.random.permutation (len(X))
X = X.iloc[index]
y = y.iloc[index]
train_size = int (2 * len(X) / 3.0) # sample of 2 third
X_train = X[:train_size]
X_val = X[train_size:]
y_train = y[:train_size]
y_val = y[train_size:]
这导致了以下输出:
Train set AUC: 0.5
Grid best score (AUC): 0.501410198106
Grid best parameter (max. AUC): {'hidden_layer_sizes': (5, 10)}
{ 'mean_fit_time': array([28.62, 46.00, 54.44, 46.74, 55.25, 53.33]),
'mean_score_time': array([ 0.04, 0.05, 0.05, 0.05, 0.05, 0.06]),
'mean_test_score': array([ 0.50, 0.50, 0.50, 0.50, 0.50, 0.50]),
'mean_train_score': array([ 0.50, 0.51, 0.51, 0.51, 0.50, 0.51]),
'param_hidden_layer_sizes': masked_array(data = [5 (5, 5) (5, 10) 10 (10, 5) (10, 10)],
mask = [False False False False False False],
fill_value = ?)
,
'params': [ {'hidden_layer_sizes': 5},
{'hidden_layer_sizes': (5, 5)},
{'hidden_layer_sizes': (5, 10)},
{'hidden_layer_sizes': 10},
{'hidden_layer_sizes': (10, 5)},
{'hidden_layer_sizes': (10, 10)}],
'rank_test_score': array([ 6, 2, 1, 4, 5, 3]),
'split0_test_score': array([ 0.50, 0.50, 0.51, 0.50, 0.50, 0.50]),
'split0_train_score': array([ 0.50, 0.51, 0.50, 0.51, 0.50, 0.51]),
'split1_test_score': array([ 0.50, 0.50, 0.50, 0.50, 0.49, 0.50]),
'split1_train_score': array([ 0.50, 0.50, 0.51, 0.50, 0.51, 0.51]),
'split2_test_score': array([ 0.49, 0.50, 0.49, 0.50, 0.50, 0.50]),
'split2_train_score': array([ 0.51, 0.51, 0.51, 0.51, 0.50, 0.51]),
'std_fit_time': array([19.74, 19.33, 0.55, 0.64, 2.36, 0.65]),
'std_score_time': array([ 0.01, 0.01, 0.00, 0.01, 0.00, 0.01]),
'std_test_score': array([ 0.01, 0.00, 0.01, 0.00, 0.00, 0.00]),
'std_train_score': array([ 0.00, 0.00, 0.00, 0.00, 0.00, 0.00])}
这似乎证实了穆罕默德的言论。我必须说，一开始我很怀疑，因为我无法想象随机化对如此大的数据集产生如此强烈的影响，而这个数据集看起来并不真正有序。
但是我有一些疑问。在原始设置中，GridSearchCV 始终过高约 0.20，现在始终过低约 0.05。这是一个进步，因为两种方法的偏差都减少了 4 倍。是否有对最后发现的解释，或者两种方法之间大约 0.05 的偏差仅仅是噪声的事实？我决定将此标记为正确答案，但我希望有人能够解答我的小疑问。
最佳答案
分数的差异主要是由于 GridSearchCV 分割数据集的方式不同以及模拟它的函数造成的。这样想吧。假设您的数据集中有 9 个数据点。现在 GridSearchCV 有 3 倍，假设分布如下:
train_cv_fold1_indices : 1 2 3 4 5 6
test_cv_fold1_indices : 7 8 9
train_cv_fold2_indices : 1 2 3 7 8 9
test_cv_fold2_indices : 4 5 6
train_cv_fold3_indices : 4 5 6 7 8 9
test_cv_fold3_indices : 1 2 3
但是，模拟 GridSearchCV 的函数可能会以不同的方式分割数据，例如:
train_indices : 1 3 5 7 8 9
test_indices : 2 4 6
现在正如您所看到的，这是数据集上的不同分割，因此在其上训练的分类器的行为可能会完全不同。 (它甚至可能表现相同，这完全取决于数据点和各种其他因素，例如它们的相关程度，它们是否有助于检查数据点之间的变化等)。
因此，为了完美模拟 GridSearchCV，您需要以相同的方式执行拆分。
检查GridSearchCV Source你会发现在第 592 行，为了执行 CV，他们调用了 check_cv 指定的另一个函数 at this link 。它实际上调用 Kfold CV或startified CV .
因此，根据您的实验，我建议使用固定随机种子和上述函数( Kfold CV 或 startified CV )在您的数据集上显式执行 CV。然后在模拟函数中使用相同的 CV 对象以获得更具可比性的分析。然后您可能会获得更多相关的值。
关于python - AUC 与 GridSearchCV AUC 有何不同？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/46940280/