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python-3.x - 逻辑回归成本 = nan

转载 作者:行者123 更新时间:2023-11-30 09:58:57 26 4
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我正在尝试实现逻辑回归模型,但不断获得“nan”值作为成本。我用多个数据集尝试过,但给出了相同的结果。不同的来源给出的梯度下降的实现略有不同,所以我不确定这里梯度的实现是否正确。这是完整的代码:

import numpy as np
import matplotlib.pyplot as plt
from sklearn import datasets
from sklearn.model_selection import train_test_split

class LogisticRegression:
def __init__(self, lr=0.001, n_iter=8000):
self.lr = lr
self.n_iter = n_iter
self.weights = None

"""
z is dot product of features and weights, which is then mapped to discrete values, such as between 0 and 1
"""
def sigmoid(self, z):
return 1.0/(1+np.exp(-z))

def predict(self, x_features, weights):
"""Returns 1d array of probabilities that the class label == 1"""
z = np.dot(x_features, weights)
return self.sigmoid(z)

def cost(self, x_features, labels, weights):
"""
Using Mean Absolute Error

Cost = (labels*log(predictions) + (1-labels)*log(1-predictions) ) / len(labels)
"""
observation = len(labels)
predictions = self.predict(x_features, weights)
#take the error when label = 1
class1_cost = -labels*np.log(predictions)
#take the error when label = 0
class2_cost = (1-labels)*np.log(1-predictions)
#take sum of both the cost
cost = class1_cost+class2_cost
#take the average cost
cost = cost.sum()/observation
return cost

def update_weight(self, x_features, labels, weights):
"""
Vectorized Gradient Descent
"""
N = len(x_features)
#get predictions (approximation of y)
predictions = self.predict(x_features, weights)
gradient = np.dot(x_features.T, predictions-labels)
#take the average cost of derivative for each feature
gradient /= N
#multiply gradients by our learning rate
gradient *= self.lr
#subtract from our weights to minimize cost
weights -= gradient
return weights

def give_predictions(self, x_features, weights):
linear_model_prediction = self.predict(x_features, weights)
y_predicted_cls = [1 if i>0.5 else 0 for i in linear_model_prediction]
return y_predicted_cls

def train(self, features, labels):
n_samples, n_features = features.shape
self.weights = np.zeros((n_features,1)) #initialize the weight matrix
cost_history = []
for i in range(self.n_iter):
self.weights = self.update_weight(features, labels, self.weights)
#calculate error for auditing purposes
cost = self.cost(features, labels, self.weights)
cost_history.append(cost)
#Log process
if i%1000 == 0:
print("iter: {}, cost: {}".format(str(i),str(cost)))

return self.weights, cost_history

def generate_data():
bc = datasets.load_breast_cancer()
x_features, labels = bc.data, bc.target

x_train, x_test, y_train, y_test = train_test_split(x_features, labels, test_size=0.2, random_state=1234)
return x_train, x_test, y_train, y_test

x_train, x_test, y_train, y_test = generate_data()

model = LogisticRegression()
model.train(x_train, y_train)

最佳答案

在训练模型之前,我必须对 x_train 应用特征缩放。我使用了 sklearn StandardScaler 库

from sklearn.preprocessing import StandardScaler
sc_X = StandardScaler()
x_train = sc_X.fit_transform(x_train)

关于python-3.x - 逻辑回归成本 = nan,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/59850219/

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