如何为sklearn的线性回归预测增加范围 [英] How to add a range to sklearn's linear regression predictions

问题描述

我想知道是否有一种方法可以在拟合模型之前为预测添加范围.

I wonder if there is a way to add a range to the predictions prior to fitting the model.

火车数据中有问题的变量从技术上讲是百分比得分，但是当我预测测试集时，我会得到负值或> 100的值.

The variable in question in my train data is technically a percentage score, but when I predict my test set, I get negative values or values >100.

就目前而言，我正在手动对预测列表进行标准化.我也曾经剪掉负数和> 100，然后分配0和100.

For now, I am manually normalizing the predictions list. I also used to cut off negatives and >100 and assign then a 0 and 100.

但是，只有让fit函数知道此约束，才有意义，对吧?

However, it only makes sense if the fit function could be made aware of this constraint, right?

这是数据的示例行:

test_df = pd.DataFrame([[0, 40, 28, 30, 40, 22, 60, 40, 21, 0, 85, 29, 180, 85, 36, 741, 25.0]], columns=['theta_1', 'phi_1', 'value_1', 'theta_2', 'phi_2', 'value_2', 'theta_3', 'phi_3', 'value_3', 'theta_4', 'phi_4', 'value_4', 'theta_5', 'phi_5', 'value_5', 'sum_readings', 'estimated_volume'])

我一直在阅读，很多人认为这不是线性回归问题，但是他们的逻辑并不健全.另外，有人说可以应用对数刻度，但仅在与阈值进行比较的情况下才有效，即手动分类，即对逻辑回归问题使用线性回归！就我而言，我需要百分比，因为它们是必需的输出.

I have been reading and a lot of people consider this not a linear regression problem but their logic is not sound. Also, some say that one can apply a log scale but that only works in the case of comparison against a threshold, i.e., manual classification, i.e., using linear regression for a logistic regression problem! In my case, I need the percentages as they are the required output.

非常感谢您的反馈/想法.

Your feedbacks/thought are much appreciated.

推荐答案

某些算法不会提出超出范围的预测值，例如sklearn.neighbors.KNeighborsRegressor或sklearn.ensemble.RandomForestRegressor.

Some algorithms will not propose out of range predicted values such as sklearn.neighbors.KNeighborsRegressor or sklearn.ensemble.RandomForestRegressor.

线性回归器可以给出超出目标范围的值，此处为示例:

Linear Regressor can give out of target range values, here an example :

from sklearn.ensemble import RandomForestRegressor
import numpy as np
from sklearn.linear_model import LinearRegression

y = np.linspace(0,1,100)
X = 2* y
X = X.reshape(-1,1)

>>>> rf.predict(np.array([[4.]])), lr.predict(np.array([[4.]]))
# (array([0.9979798]), array([2.]))

但是您可以使用一个技巧:您可以将[0，1]空间映射到[-inf，inf]空间，并在预测后返回初始空间.

but you can use a trick : you can map your [0, 1] space to [-inf, inf] space and came back in the initial space after prediction.

以下是使用sigmoid的示例:

Here an example using sigmoid :

def sigmoid(x):
    return 1/(1+np.exp(-x))

def sigmoid_m1(x):
    return -np.log((1/x)-1)

rf = RandomForestRegressor()
lr = LinearRegression()
rf.fit(X,sigmoid_m1(y*0.9+0.05))
lr.fit(X,sigmoid_m1(y*0.9+0.05))
>>>> sigmoid(rf.predict(np.array([[4.]]))), sigmoid(lr.predict(np.array([[4.]])))
# (array([0.9457559]), array([0.99904361]))

使用这种解决方案要当心，因为您完全改变了数据的分布，并且会带来很多问题.

Take care using this kind of solution because you totally change the distribution of the data and it can create a lot of problems.

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