I am unable to understand how decision boundary is calculated once we get the coefficients of the model.
Here is the link that I am referring: http://scikit-learn.org/stable/auto_examples/svm/plot_svm_margin.html
Here is the code
# get the separating hyperplane
w = clf.coef_[0]
a = -w[0] / w[1]
xx = np.linspace(-5, 5)
yy = a * xx - (clf.intercept_[0]) / w[1]
I didn't understand a = -w[0] / w[1] this line.
Why we are dividing one coefficient with another?
The separating hyperplane has the form w[0]*x+w[1]*y+intercept=0. So
w[1]*y=-w[0]*x-intercept
Now divide both sides by w[1], and you get
y=-(w[0]/w[1])*x-intercept/w[1].
This is exactly the equation that appears in your code.
Related
Question
Please forgive me asking Coursera ML course specific question. Hope someone who did the couser can answer.
In Coursera ML Week 4 Multi-class Classification and Neural Networks assignment, why the weight (theta) gradient is adding (plus) the derivative instead of subtracting?
% Calculate the gradients of Weight2
% Derivative at Loss function J=L(Z) : dJ/dZ = (oi-yi)/oi(1-oi)
% Derivative at Sigmoid activation function dZ/dY = oi(1-oi)
delta_theta2 = oi - yi; % <--- (dJ/dZ) * (dZ/dY)
# Using +/plus NOT -/minus
Theta2_grad = Theta2_grad + <-------- Why plus(+)?
bsxfun(#times, hi, transpose(delta_theta2));
Code Excerpt
for i = 1:m
% i is training set index of X (including bias). X(i, :) is 401 data.
xi = X(i, :);
yi = Y(i, :);
% hi is the i th output of the hidden layer. H(i, :) is 26 data.
hi = H(i, :);
% oi is the i th output layer. O(i, :) is 10 data.
oi = O(i, :);
%------------------------------------------------------------------------
% Calculate the gradients of Theta2
%------------------------------------------------------------------------
delta_theta2 = oi - yi;
Theta2_grad = Theta2_grad + bsxfun(#times, hi, transpose(delta_theta2));
%------------------------------------------------------------------------
% Calculate the gradients of Theta1
%------------------------------------------------------------------------
% Derivative of g(z): g'(z)=g(z)(1-g(z)) where g(z) is sigmoid(H_NET).
dgz = (hi .* (1 - hi));
delta_theta1 = dgz .* sum(bsxfun(#times, Theta2, transpose(delta_theta2)));
% There is no input into H0, hence there is no theta for H0. Remove H0.
delta_theta1 = delta_theta1(2:end);
Theta1_grad = Theta1_grad + bsxfun(#times, xi, transpose(delta_theta1));
end
I thought it is subtracting the derivative.
Derivative of Binary Cross Entropy - why are my signs not right?
Since the gradients are calculated by averaging the gradients across all training examples, we first "accumulate" the gradients while looping across all the training examples. We do this by summing the gradient across all training examples. So the line you highlighted with the plus is not the gradient update step. (Notice that alpha is not there as well.) It might be somewhere else. It is most likely outside of the loop from 1 to m.
Also, I am not sure when you will learn about this (I'm sure it's somewhere in the course), but you could also vectorize the code :)
I've been implementing VAE and IWAE models on the caltech silhouettes dataset and am having an issue where the VAE outperforms IWAE by a modest margin (test LL ~120 for VAE, ~133 for IWAE!). I don't believe this should be the case, according to both theory and experiments produced here.
I'm hoping someone can find some issue in how I'm implementing that's causing this to be the case.
The network I'm using to approximate q and p is the same as that detailed in the appendix of the paper above. The calculation part of the model is below:
data_k_vec = data.repeat_interleave(K,0) # Generate K samples (in my case K=50 is producing this behavior)
mu, log_std = model.encode(data_k_vec)
z = model.reparameterize(mu, log_std) # z = mu + torch.exp(log_std)*epsilon (epsilon ~ N(0,1))
decoded = model.decode(z) # this is the sigmoid output of the model
log_prior_z = torch.sum(-0.5 * z ** 2, 1)-.5*z.shape[1]*T.log(torch.tensor(2*np.pi))
log_q_z = compute_log_probability_gaussian(z, mu, log_std) # Definitions below
log_p_x = compute_log_probability_bernoulli(decoded,data_k_vec)
if model_type == 'iwae':
log_w_matrix = (log_prior_z + log_p_x - log_q_z).view(-1, K)
elif model_type =='vae':
log_w_matrix = (log_prior_z + log_p_x - log_q_z).view(-1, 1)*1/K
log_w_minus_max = log_w_matrix - torch.max(log_w_matrix, 1, keepdim=True)[0]
ws_matrix = torch.exp(log_w_minus_max)
ws_norm = ws_matrix / torch.sum(ws_matrix, 1, keepdim=True)
ws_sum_per_datapoint = torch.sum(log_w_matrix * ws_norm, 1)
loss = -torch.sum(ws_sum_per_datapoint) # value of loss that gets returned to training function. loss.backward() will get called on this value
Here are the likelihood functions. I had to fuss with the bernoulli LL in order to not get nan during training
def compute_log_probability_gaussian(obs, mu, logstd, axis=1):
return torch.sum(-0.5 * ((obs-mu) / torch.exp(logstd)) ** 2 - logstd, axis)-.5*obs.shape[1]*T.log(torch.tensor(2*np.pi))
def compute_log_probability_bernoulli(theta, obs, axis=1): # Add 1e-18 to avoid nan appearances in training
return torch.sum(obs*torch.log(theta+1e-18) + (1-obs)*torch.log(1-theta+1e-18), axis)
In this code there's a "shortcut" being used in that the row-wise importance weights are being calculated in the model_type=='iwae' case for the K=50 samples in each row, while in the model_type=='vae' case the importance weights are being calculated for the single value left in each row, so that it just ends up calculating a weight of 1. Maybe this is the issue?
Any and all help is huge - I thought that addressing the nan issue would permanently get me out of the weeds but now I have this new problem.
EDIT:
Should add that the training scheme is the same as that in the paper linked above. That is, for each of i=0....7 rounds train for 2**i epochs with a learning rate of 1e-4 * 10**(-i/7)
The K-sample importance weighted ELBO is
$$ \textrm{IW-ELBO}(x,K) = \log \sum_{k=1}^K \frac{p(x \vert z_k) p(z_k)}{q(z_k;x)}$$
For the IWAE there are K samples originating from each datapoint x, so you want to have the same latent statistics mu_z, Sigma_z obtained through the amortized inference network, but sample multiple z K times for each x.
So its computationally wasteful to compute the forward pass for data_k_vec = data.repeat_interleave(K,0), you should compute the forward pass once for each original datapoint, then repeat the statistics output by the inference network for sampling:
mu = torch.repeat_interleave(mu,K,0)
log_std = torch.repeat_interleave(log_std,K,0)
Then sample z_k. And now repeat your datapoints data_k_vec = data.repeat_interleave(K,0), and use the resulting tensor to efficiently evaluate the conditional p(x |z_k) for each importance sample z_k.
Note you may also want to use the logsumexp operation when calculating the IW-ELBO for numerical stability. I can't quite figure out what's going on with the log_w_matrix calculation in your post, but this is what I would do:
log_pz = ...
log_qzCx = ....
log_pxCz = ...
log_iw = log_pxCz + log_pz - log_qzCx
log_iw = log_iw.reshape(-1, K)
iwelbo = torch.logsumexp(log_iw, dim=1) - np.log(K)
EDIT: Actually after thinking about it a bit and using the score function identity, you can interpret the IWAE gradient as an importance weighted estimate of the standard single-sample gradient, so the method in the OP for calculation of the importance weights is equivalent (if a bit wasteful), provided you place a stop_gradient operator around the normalized importance weights, which you call w_norm. So I the main problem is the absence of this stop_gradient operator.
I am starting to use Octave and I am trying to understand how is the underlying calculation done for dividing a Scalar by vector ?
I am able to understand how ./ is operating to give us the results - dividing 1 by every element of the matrix column. However, I am not able to get my head around how we get the values in the second case ? 1 / (1 + a)
Example :
g = 1 ./ (1 + a)
g =
0.50000
0.25000
0.20000
>> g = 1 / (1 + a)
g =
0.044444 0.088889 0.111111
When you divide 1 by a vector, it gives you a vector that yields 1 when multiplied on the left by the first vector. In this sense, it is a sort of 'inverse' of the vector, although it will only be a one way inverse. In your example:
>> (1/(1+a))*(1+a)
ans = 1
>> (1+a)*(1/(1+a))
ans =
0.088889 0.177778 0.222222
0.177778 0.355556 0.444444
0.222222 0.444444 0.555556
You could say 1/(1+a) is the left inverse of 1+a. This would also explain why the dimensions of the vector are transposed. Another way to put it: given a vector v, 1/v is the solution (w) of the vector equation w*v=1.
I am working with a data-set of patient information and trying to calculate the Propensity Score from the data using MATLAB. After removing features with many missing values, I am still left with several missing (NaN) values.
I get errors due to these missing values, as the values of my cost-function and gradient vector become NaN, when I try to perform logistic regression using the following Matlab code (from Andrew Ng's Coursera Machine Learning class) :
[m, n] = size(X);
X = [ones(m, 1) X];
initial_theta = ones(n+1, 1);
[cost, grad] = costFunction(initial_theta, X, y);
options = optimset('GradObj', 'on', 'MaxIter', 400);
[theta, cost] = ...
fminunc(#(t)(costFunction(t, X, y)), initial_theta, options);
Note: sigmoid and costfunction are working functions I created for overall ease of use.
The calculations can be performed smoothly if I replace all NaN values with 1 or 0. However I am not sure if that is the best way to deal with this issue, and I was also wondering what replacement value I should pick (in general) to get the best results for performing logistic regression with missing data. Are there any benefits/drawbacks to using a particular number (0 or 1 or something else) for replacing the said missing values in my data?
Note: I have also normalized all feature values to be in the range of 0-1.
Any insight on this issue will be highly appreciated. Thank you
As pointed out earlier, this is a generic problem people deal with regardless of the programming platform. It is called "missing data imputation".
Enforcing all missing values to a particular number certainly has drawbacks. Depending on the distribution of your data it can be drastic, for example, setting all missing values to 1 in a binary sparse data having more zeroes than ones.
Fortunately, MATLAB has a function called knnimpute that estimates a missing data point by its closest neighbor.
From my experience, I often found knnimpute useful. However, it may fall short when there are too many missing sites as in your data; the neighbors of a missing site may be incomplete as well, thereby leading to inaccurate estimation. Below, I figured out a walk-around solution to that; it begins with imputing the least incomplete columns, (optionally) imposing a safe predefined distance for the neighbors. I hope this helps.
function data = dnnimpute(data,distCutoff,option,distMetric)
% data = dnnimpute(data,distCutoff,option,distMetric)
%
% Distance-based nearest neighbor imputation that impose a distance
% cutoff to determine nearest neighbors, i.e., avoids those samples
% that are more distant than the distCutoff argument.
%
% Imputes missing data coded by "NaN" starting from the covarites
% (columns) with the least number of missing data. Then it continues by
% including more (complete) covariates in the calculation of pair-wise
% distances.
%
% option,
% 'median' - Median of the nearest neighboring values
% 'weighted' - Weighted average of the nearest neighboring values
% 'default' - Unweighted average of the nearest neighboring values
%
% distMetric,
% 'euclidean' - Euclidean distance (default)
% 'seuclidean' - Standardized Euclidean distance. Each coordinate
% difference between rows in X is scaled by dividing
% by the corresponding element of the standard
% deviation S=NANSTD(X). To specify another value for
% S, use D=pdist(X,'seuclidean',S).
% 'cityblock' - City Block distance
% 'minkowski' - Minkowski distance. The default exponent is 2. To
% specify a different exponent, use
% D = pdist(X,'minkowski',P), where the exponent P is
% a scalar positive value.
% 'chebychev' - Chebychev distance (maximum coordinate difference)
% 'mahalanobis' - Mahalanobis distance, using the sample covariance
% of X as computed by NANCOV. To compute the distance
% with a different covariance, use
% D = pdist(X,'mahalanobis',C), where the matrix C
% is symmetric and positive definite.
% 'cosine' - One minus the cosine of the included angle
% between observations (treated as vectors)
% 'correlation' - One minus the sample linear correlation between
% observations (treated as sequences of values).
% 'spearman' - One minus the sample Spearman's rank correlation
% between observations (treated as sequences of values).
% 'hamming' - Hamming distance, percentage of coordinates
% that differ
% 'jaccard' - One minus the Jaccard coefficient, the
% percentage of nonzero coordinates that differ
% function - A distance function specified using #, for
% example #DISTFUN.
%
if nargin < 3
option = 'mean';
end
if nargin < 4
distMetric = 'euclidean';
end
nanVals = isnan(data);
nanValsPerCov = sum(nanVals,1);
noNansCov = nanValsPerCov == 0;
if isempty(find(noNansCov, 1))
[~,leastNans] = min(nanValsPerCov);
noNansCov(leastNans) = true;
first = data(nanVals(:,noNansCov),:);
nanRows = find(nanVals(:,noNansCov)==true); i = 1;
for row = first'
data(nanRows(i),noNansCov) = mean(row(~isnan(row)));
i = i+1;
end
end
nSamples = size(data,1);
if nargin < 2
dataNoNans = data(:,noNansCov);
distances = pdist(dataNoNans);
distCutoff = min(distances);
end
[stdCovMissDat,idxCovMissDat] = sort(nanValsPerCov,'ascend');
imputeCols = idxCovMissDat(stdCovMissDat>0);
% Impute starting from the cols (covariates) with the least number of
% missing data.
for c = reshape(imputeCols,1,length(imputeCols))
imputeRows = 1:nSamples;
imputeRows = imputeRows(nanVals(:,c));
for r = reshape(imputeRows,1,length(imputeRows))
% Calculate distances
distR = inf(nSamples,1);
%
noNansCov_r = find(isnan(data(r,:))==0);
noNansCov_r = noNansCov_r(sum(isnan(data(nanVals(:,c)'==false,~isnan(data(r,:)))),1)==0);
%
for i = find(nanVals(:,c)'==false)
distR(i) = pdist([data(r,noNansCov_r); data(i,noNansCov_r)],distMetric);
end
tmp = min(distR(distR>0));
% Impute the missing data at sample r of covariate c
switch option
case 'weighted'
data(r,c) = (1./distR(distR<=max(distCutoff,tmp)))' * data(distR<=max(distCutoff,tmp),c) / sum(1./distR(distR<=max(distCutoff,tmp)));
case 'median'
data(r,c) = median(data(distR<=max(distCutoff,tmp),c),1);
case 'mean'
data(r,c) = mean(data(distR<=max(distCutoff,tmp),c),1);
end
% The missing data in sample r is imputed. Update the sample
% indices of c which are imputed.
nanVals(r,c) = false;
end
fprintf('%u/%u of the covariates are imputed.\n',find(c==imputeCols),length(imputeCols));
end
To deal with missing data you can use one of the following three options:
If there are not many instances with missing values, you can just delete the ones with missing values.
If you have many features and it is affordable to lose some information, delete the entire feature with missing values.
The best method is to fill some value (mean, median) in place of missing value. You can calculate the mean of the rest of the training examples for that feature and fill all the missing values with the mean. This works out pretty well as the mean value stays in the distribution of your data.
Note: When you replace the missing values with the mean, calculate the mean only using training set. Also, store that value and use it to change the missing values in the test set also.
If you use 0 or 1 to replace all the missing values then the data may get skewed so it is better to replace the missing values by an average of all the other values.
I've gone through few courses of Professor Andrew for machine Learning and viewed the transcript for Logistic Regression using Newton's method. However when implementing the logistic regression using gradient descent I face certain issue.
The graph generated is not convex.
My code goes as follows:
I am using the vectorized implementation of the equation.
%1. The below code would load the data present in your desktop to the octave memory
x=load('ex4x.dat');
y=load('ex4y.dat');
%2. Now we want to add a column x0 with all the rows as value 1 into the matrix.
%First take the length
m=length(y);
x=[ones(m,1),x];
alpha=0.1;
max_iter=100;
g=inline('1.0 ./ (1.0 + exp(-z))');
theta = zeros(size(x(1,:)))'; % the theta has to be a 3*1 matrix so that it can multiply by x that is m*3 matrix
j=zeros(max_iter,1); % j is a zero matrix that is used to store the theta cost function j(theta)
for num_iter=1:max_iter
% Now we calculate the hx or hypothetis, It is calculated here inside no. of iteration because the hupothesis has to be calculated for new theta for every iteration
z=x*theta;
h=g(z); % Here the effect of inline function we used earlier will reflect
j(num_iter)=(1/m)*(-y'* log(h) - (1 - y)'*log(1-h)) ; % This formula is the vectorized form of the cost function J(theta) This calculates the cost function
j
grad=(1/m) * x' * (h-y); % This formula is the gradient descent formula that calculates the theta value.
theta=theta - alpha .* grad; % Actual Calculation for theta
theta
end
The code per say doesn't give any error but does not produce proper convex graph.
I shall be glad if any body could point out the mistake or share insight on what's causing the problem.
thanks
2 things you need to look into:
Machine Learning involves learning patterns from data. If your files ex4x.dat and ex4y.dat are randomly generated, it won't have patterns that you can learn.
You have used variables like g, h, i, j which make debugging difficult. Since it's a very small program, it might be a better idea to rewrite it.
Here's my code that gives the convex plot
clc; clear; close all;
load q1x.dat;
load q1y.dat;
X = [ones(size(q1x, 1),1) q1x];
Y = q1y;
m = size(X,1);
n = size(X,2)-1;
%initialize
theta = zeros(n+1,1);
thetaold = ones(n+1,1);
while ( ((theta-thetaold)'*(theta-thetaold)) > 0.0000001 )
%calculate dellltheta
dellltheta = zeros(n+1,1);
for j=1:n+1,
for i=1:m,
dellltheta(j,1) = dellltheta(j,1) + [Y(i,1) - (1/(1 + exp(-theta'*X(i,:)')))]*X(i,j);
end;
end;
%calculate hessian
H = zeros(n+1, n+1);
for j=1:n+1,
for k=1:n+1,
for i=1:m,
H(j,k) = H(j,k) -[1/(1 + exp(-theta'*X(i,:)'))]*[1-(1/(1 + exp(-theta'*X(i,:)')))]*[X(i,j)]*[X(i,k)];
end;
end;
end;
thetaold = theta;
theta = theta - inv(H)*dellltheta;
(theta-thetaold)'*(theta-thetaold)
end
I get the following values of error after iterations:
2.8553
0.6596
0.1532
0.0057
5.9152e-06
6.1469e-12
Which when plotted looks like: