预测多个值作为scikit学习的模型结果

时间:2019-06-18 09:30:56

标签: python machine-learning scikit-learn prediction

我已经使用scikit学习算法创建了一个模型。

rf = RandomForestClassifier(n_estimators = 10,random_state=seed)
rf.fit(X_train,Y_train)

shift_id=2099.0
user_id=1402.0
status=['S']
shift_organisation_id=15.0
shift_department_id=20.0
open_positions=71.0
city=['taunton']
role_id=3.0
specialty_id=16.0
years_of_experience=10.0
nurse_zip=2780.0
shifts_zip=2021.0

status = status_encoder.transform(status)
city = city_encoder.transform(city)

X = np.array([shift_id, user_id, status, shift_organisation_id, shift_department_id, open_positions, city, role_id, specialty_id, years_of_experience, nurse_zip, shifts_zip])
location_id = rf.predict(X.reshape(1,-1))
print(location_id)

给出这样的结果

  

[25]

我了解的是25是此模型的最佳预测值。但我想获得最高的最佳3个值。我怎么能得到它?

在那种情况下,预测结果将是

  

[23,45,25]

2 个答案:

答案 0 :(得分:1)

您可以使用predict_proba方法返回类概率并从中获取前3个值ref

rf = RandomForestClassifier(n_estimators = 10,random_state=seed)
rf.fit(X_train,Y_train)

shift_id=2099.0
user_id=1402.0
status=['S']
shift_organisation_id=15.0
shift_department_id=20.0
open_positions=71.0
city=['taunton']
role_id=3.0
specialty_id=16.0
years_of_experience=10.0
nurse_zip=2780.0
shifts_zip=2021.0

status = status_encoder.transform(status)
city = city_encoder.transform(city)

X = np.array([shift_id, user_id, status, shift_organisation_id, shift_department_id, open_positions, city, role_id, specialty_id, years_of_experience, nurse_zip, shifts_zip])
location_id = rf.predict_proba(X.reshape(1,-1))
print(location_id)

答案 1 :(得分:1)

您可以使用predict_proba方法,该方法返回对类概率的预测。

让我们使用iris dataset来检查示例:

from sklearn import datasets
iris = datasets.load_iris()
X = iris.data[:, :2]  # we only take the first two features.
y = iris.target
# train/test split
X_train, X_test, y_train, y_test = train_test_split(X, y)

rf = RandomForestClassifier(n_estimators = 10, random_state=10)
rf.fit(x_train,y_train)

如果您现在按预期方式调用predict方法,则将获得最高的概率类别:

rf.predict(X_test)
# array([1, 2, 1, 0, 2, 0, 2, 0, 0, 1, 2, ...

无论如何致电predict_proba,您都会获得相应的概率:

rf.predict_proba(X_test)

array([[0.        , 1.        , 0.        ],
       [0.11      , 0.1       , 0.79      ],
       [0.        , 0.7       , 0.3       ],
       [0.5       , 0.4       , 0.1       ],
       [0.        , 0.3       , 0.7       ],
       [0.5       , 0.2       , 0.3       ],
       [0.4       , 0.        , 0.6       ],
       ...

为了获得最高的k概率,您可以使用argsort并索引相应的概率rf.classes_

k = 2
rf.classes_[rf.predict_proba(X_test).argsort()[:,-k:]]

array([[2, 1],
       [0, 2],
       [2, 1],
       [1, 0],
       [1, 2],
       [2, 0],
       [0, 2],
       [1, 0],
       [1, 0],
       [2, 1],
       ...

在上述情况中,可以使用argpartition进行改进,因为我们只对排名靠前的k概率感兴趣:

rf.classes_[rf.predict_proba(X_test).argpartition(range(k))[:,-k:]]