Я пытался преобразовать код из Octave в Python, и в коде есть несколько случаев, когда они используют «лямбду», что-то равное, я спросил об этом моего учителя, и он сказал мне, что лямбда-константа, инасколько я знаю, они не являются константами в Python, поэтому мне было интересно, есть ли какой-нибудь способ, которым я могу преобразовать его или, возможно, создать константу.
%% Machine Learning Online Class - Exercise 2: Logistic Regression
%
% Instructions
% ------------
%
% This file contains code that helps you get started on the second part
% of the exercise which covers regularization with logistic regression.
%
% You will need to complete the following functions in this exericse:
%
% sigmoid.m
% costFunction.m
% predict.m
% costFunctionReg.m
%
% For this exercise, you will not need to change any code in this file,
% or any other files other than those mentioned above.
%
%% Initialization
clear ; close all; clc
%% Load Data
% The first two columns contains the X values and the third column
% contains the label (y).
data = load('ex2data2.txt');
X = data(:, [1, 2]); y = data(:, 3);
plotData(X, y);
% Put some labels
hold on;
% Labels and Legend
xlabel('Microchip Test 1')
ylabel('Microchip Test 2')
% Specified in plot order
legend('y = 1', 'y = 0')
hold off;
%% =========== Part 1: Regularized Logistic Regression ============
% In this part, you are given a dataset with data points that are not
% linearly separable. However, you would still like to use logistic
% regression to classify the data points.
%
% To do so, you introduce more features to use -- in particular, you add
% polynomial features to our data matrix (similar to polynomial
% regression).
%
% Add Polynomial Features
% Note that mapFeature also adds a column of ones for us, so the intercept
% term is handled
X = mapFeature(X(:,1), X(:,2));
% Initialize fitting parameters
initial_theta = zeros(size(X, 2), 1);
% Set regularization parameter lambda to 1
lambda = 1;
% Compute and display initial cost and gradient for regularized logistic
% regression
[cost, grad] = costFunctionReg(initial_theta, X, y, lambda);
fprintf('Cost at initial theta (zeros): %f\n', cost);
fprintf('Expected cost (approx): 0.693\n');
fprintf('Gradient at initial theta (zeros) - first five values only:\n');
fprintf(' %f \n', grad(1:5));
fprintf('Expected gradients (approx) - first five values only:\n');
fprintf(' 0.0085\n 0.0188\n 0.0001\n 0.0503\n 0.0115\n');
fprintf('\nProgram paused. Press enter to continue.\n');
pause;
% Compute and display cost and gradient
% with all-ones theta and lambda = 10
test_theta = ones(size(X,2),1);
[cost, grad] = costFunctionReg(test_theta, X, y, 10);
fprintf('\nCost at test theta (with lambda = 10): %f\n', cost);
fprintf('Expected cost (approx): 3.16\n');
fprintf('Gradient at test theta - first five values only:\n');
fprintf(' %f \n', grad(1:5));
fprintf('Expected gradients (approx) - first five values only:\n');
fprintf(' 0.3460\n 0.1614\n 0.1948\n 0.2269\n 0.0922\n');
fprintf('\nProgram paused. Press enter to continue.\n');
pause;
%% ============= Part 2: Regularization and Accuracies =============
% Optional Exercise:
% In this part, you will get to try different values of lambda and
% see how regularization affects the decision coundart
%
% Try the following values of lambda (0, 1, 10, 100).
%
% How does the decision boundary change when you vary lambda? How does
% the training set accuracy vary?
%
% Initialize fitting parameters
initial_theta = zeros(size(X, 2), 1);
% Set regularization parameter lambda to 1 (you should vary this)
lambda = 100;
% Set Options
options = optimset('GradObj', 'on', 'MaxIter', 400);
% Optimize
[theta, J, exit_flag] = ...
fminunc(@(t)(costFunctionReg(t, X, y, lambda)), initial_theta, options);
% Plot Boundary
plotDecisionBoundary(theta, X, y);
hold on;
title(sprintf('lambda = %g', lambda))
% Labels and Legend
xlabel('Microchip Test 1')
ylabel('Microchip Test 2')
legend('y = 1', 'y = 0', 'Decision boundary')
hold off;
% Compute accuracy on our training set
p = predict(theta, X);
fprintf('Train Accuracy: %f\n', mean(double(p == y)) * 100);
fprintf('Expected accuracy (with lambda = 1): 83.1 (approx)\n');
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