我是python中机器学习和编码的完全初学者，我的任务是从头开始编写逻辑回归，以了解幕后发生的事情。到目前为止，我已经编码了假设函数，成本函数和梯度下降，然后编码进行逻辑回归。然而，在编码打印精度时，我得到低输出（0.69），这不会随着迭代次数的增加或学习速率的变化而变化。我的问题是，我的准确度代码存在问题吗？任何指向正确方向的帮助都将受到赞赏
X = data[['radius_mean', 'texture_mean', 'perimeter_mean',
'area_mean', 'smoothness_mean', 'compactness_mean', 'concavity_mean',
'concave points_mean', 'symmetry_mean', 'fractal_dimension_mean',
'radius_se', 'texture_se', 'perimeter_se', 'area_se', 'smoothness_se',
'compactness_se', 'concavity_se', 'concave points_se', 'symmetry_se',
'fractal_dimension_se', 'radius_worst', 'texture_worst',
'perimeter_worst', 'area_worst', 'smoothness_worst',
'compactness_worst', 'concavity_worst', 'concave points_worst',
'symmetry_worst', 'fractal_dimension_worst']]
X = np.array(X)
X = min_max_scaler.fit_transform(X)
Y = data["diagnosis"].map({'M':1,'B':0})
Y = np.array(Y)
X_train,X_test,Y_train,Y_test = train_test_split(X,Y,test_size=0.25)
X = data["diagnosis"].map(lambda x: float(x))
def Sigmoid(z):
if z < 0:
return 1 - 1/(1 + math.exp(z))
else:
return 1/(1 + math.exp(-z))
def Hypothesis(theta, x):
z = 0
for i in range(len(theta)):
z += x[i]*theta[i]
return Sigmoid(z)
def Cost_Function(X,Y,theta,m):
sumOfErrors = 0
for i in range(m):
xi = X[i]
hi = Hypothesis(theta,xi)
error = Y[i] * math.log(hi if hi >0 else 1)
if Y[i] == 1:
error = Y[i] * math.log(hi if hi >0 else 1)
elif Y[i] == 0:
error = (1-Y[i]) * math.log(1-hi if 1-hi >0 else 1)
sumOfErrors += error
constant = -1/m
J = constant * sumOfErrors
#print ('cost is: ', J )
return J
def Cost_Function_Derivative(X,Y,theta,j,m,alpha):
sumErrors = 0
for i in range(m):
xi = X[i]
xij = xi[j]
hi = Hypothesis(theta,X[i])
error = (hi - Y[i])*xij
sumErrors += error
m = len(Y)
constant = float(alpha)/float(m)
J = constant * sumErrors
return J
def Gradient_Descent(X,Y,theta,m,alpha):
new_theta = []
constant = alpha/m
for j in range(len(theta)):
CFDerivative = Cost_Function_Derivative(X,Y,theta,j,m,alpha)
new_theta_value = theta[j] - CFDerivative
new_theta.append(new_theta_value)
return new_theta
def Accuracy(theta):
correct = 0
length = len(X_test, Hypothesis(X,theta))
for i in range(length):
prediction = round(Hypothesis(X[i],theta))
answer = Y[i]
if prediction == answer.all():
correct += 1
my_accuracy = (correct / length)*100
print ('LR Accuracy %: ', my_accuracy)
def Logistic_Regression(X,Y,alpha,theta,num_iters):
theta = np.zeros(X.shape[1])
m = len(Y)
for x in range(num_iters):
new_theta = Gradient_Descent(X,Y,theta,m,alpha)
theta = new_theta
if x % 100 == 0:
Cost_Function(X,Y,theta,m)
print ('theta: ', theta)
print ('cost: ', Cost_Function(X,Y,theta,m))
Accuracy(theta)
initial_theta = [0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1]
alpha = 0.0001
iterations = 1000
Logistic_Regression(X,Y,alpha,initial_theta,iterations)
这是使用来自威斯康星乳腺癌数据集（https://www.kaggle.com/uciml/breast-cancer-wisconsin-data）的数据，其中我正在权衡30个特征 - 尽管将特征更改为已知相关的特征也不会改变我的准确性。
答案 0 :(得分：1)
我不确定你是如何得出0.0001 alpha的值，但我觉得它太低了。将您的代码与癌症数据结合使用表明，每次迭代都会降低成本 - 它只是冰川消失。
当我将其提高到0.5时，我仍然会降低成本，但处于更合理的水平。经过1000次迭代后，它会报告：
cost: 0.23668000993020666
在修复Accuracy功能后，我在数据的测试段上得到92％。
您已安装Numpy，如X = np.array(X)所示。您应该考虑将它用于您的操作。对于像这样的工作，它将数量级更快。这是一个矢量化版本，可以立即给出结果而不是等待：
import math
import numpy as np
import matplotlib.pyplot as plt
import pandas as pd
from sklearn.preprocessing import MinMaxScaler
from sklearn.model_selection import train_test_split
df = pd.read_csv("cancerdata.csv")
X = df.values[:,2:-1].astype('float64')
X = (X - np.mean(X, axis =0)) / np.std(X, axis = 0)
## Add a bias column to the data
X = np.hstack([np.ones((X.shape[0], 1)),X])
X = MinMaxScaler().fit_transform(X)
Y = df["diagnosis"].map({'M':1,'B':0})
Y = np.array(Y)
X_train,X_test,Y_train,Y_test = train_test_split(X,Y,test_size=0.25)
def Sigmoid(z):
return 1/(1 + np.exp(-z))
def Hypothesis(theta, x):
return Sigmoid(x @ theta)
def Cost_Function(X,Y,theta,m):
hi = Hypothesis(theta, X)
_y = Y.reshape(-1, 1)
J = 1/float(m) * np.sum(-_y * np.log(hi) - (1-_y) * np.log(1-hi))
return J
def Cost_Function_Derivative(X,Y,theta,m,alpha):
hi = Hypothesis(theta,X)
_y = Y.reshape(-1, 1)
J = alpha/float(m) * X.T @ (hi - _y)
return J
def Gradient_Descent(X,Y,theta,m,alpha):
new_theta = theta - Cost_Function_Derivative(X,Y,theta,m,alpha)
return new_theta
def Accuracy(theta):
correct = 0
length = len(X_test)
prediction = (Hypothesis(theta, X_test) > 0.5)
_y = Y_test.reshape(-1, 1)
correct = prediction == _y
my_accuracy = (np.sum(correct) / length)*100
print ('LR Accuracy %: ', my_accuracy)
def Logistic_Regression(X,Y,alpha,theta,num_iters):
m = len(Y)
for x in range(num_iters):
new_theta = Gradient_Descent(X,Y,theta,m,alpha)
theta = new_theta
if x % 100 == 0:
#print ('theta: ', theta)
print ('cost: ', Cost_Function(X,Y,theta,m))
Accuracy(theta)
ep = .012
initial_theta = np.random.rand(X_train.shape[1],1) * 2 * ep - ep
alpha = 0.5
iterations = 2000
Logistic_Regression(X_train,Y_train,alpha,initial_theta,iterations)
我想我可能有不同版本的scikit，因为我更改了MinMaxScaler行以使其正常工作。结果是我可以在眨眼间进行10K迭代，并且将模型应用到测试集的结果的准确度大约为97％。
答案 1 :(得分：0)
准确度是最直观的性能指标之一，它只是正确预测的观察与总观察的比率。更高的准确性意味着模型预先形成得更好。
Accuracy = TP+TN/TP+FP+FN+TN
TP = True positives
TN = True negatives
FN = False negatives
TN = True negatives
当您使用准确度测量时，您的误报和漏报应该具有相似的成本。更好的指标是F1分数，由
给出F1-score = 2*(Recall*Precision)/Recall+Precision where,
Precision = TP/TP+FP
Recall = TP/TP+FN
在这里阅读更多内容
https://en.wikipedia.org/wiki/Precision_and_recall
python中关于机器学习的美妙之处在于像scikit-learn这样的重要模块是开源的，所以你总能看到实际的代码。 请使用以下链接scikit学习指标源代码，这将让您了解scikit-learn如何计算准确度得分
from sklearn.metrics import accuracy_score
accuracy_score(y_true, y_pred)
https://github.com/scikit-learn/scikit-learn/tree/master/sklearn/metrics
答案 2 :(得分：0)
Python为我们提供了这个scikit-learn库，使我们的工作更加轻松， 这对我有用：
from sklearn.metrics import accuracy_score
y_pred = log.predict(x_test)
score =accuracy_score(y_test,y_pred)
答案 3 :(得分：0)
这也适用于使用矢量化来计算准确度 但是准确度不是推荐的指标，如上述答案所述（如果数据不平衡，则不应使用准确度，而应使用 F1-score）
clf = sklearn.linear_model.LogisticRegressionCV();
clf.fit(X.T, Y.T);
LR_predictions = clf.predict(X.T)
print ('Accuracy of logistic regression: %d ' % float((np.dot(Y,LR_predictions) + np.dot(1-Y,1-LR_predictions))/float(Y.size)*100) +
'% ' + "(percentage of correctly labelled datapoints)")