Question

我目前正在尝试使用GridSearchCV在scikit-learn中使用'Precision at k'评分指标来调整超参数，如果我将分类器得分的前k个百分位数分类为正类，我会给出精确度。我知道可以使用make_scorer创建自定义记分器并创建记分功能。这就是我现在所拥有的：

from sklearn import metrics
from sklearn.grid_search import GridSearchCV
from sklearn.linear_model import LogisticRegression

def precision_at_k(y_true, y_score, k):
    df = pd.DataFrame({'true': y_true, 'score': y_score}).sort('score')
    threshold = df.iloc[int(k*len(df)),1]
    y_pred = pd.Series([1 if i >= threshold else 0 for i in df['score']])
    return metrics.precision_score(y_true, y_pred)

custom_scorer = metrics.make_scorer(precision_at_k, needs_proba=True, k=0.1)

X = np.random.randn(100, 10)
Y = np.random.binomial(1, 0.3, 100)

train_index = range(0, 70)
test_index = range(70, 100)
train_x = X[train_index]
train_Y = Y[train_index]
test_x = X[test_index]
test_Y = Y[test_index]

clf = LogisticRegression()
params = {'C': [0.01, 0.1, 1, 10]}
clf_gs = GridSearchCV(clf, params, scoring=custom_scorer)
clf_gs.fit(train_x, train_Y)

但是，尝试致电fit会给我Exception: Data must be 1-dimensional而我不确定原因。有人可以帮忙吗？提前谢谢。

Answer 1

pd.DataFrame的参数应该是'list'而不是'numpy.arrays'

所以，只需尝试将y_true转换为python list ...

df = pd.DataFrame({'true': y_true.tolist(), 'score': y_score.tolist()}).sort('score')

在GridSearchCV的sklearn中自定义'精确在k'评分对象

1 个答案: