I am looking at the equation PPV and Sensitivity
and I got this
PPV = TP / (TF+FN)
and
Sensitivity = TP / (TF+FN)
Which means both are the same !!
So do we have them in 2 names?
and how come F1 Score is
F1 Score = 2*PPV*S / (PPV+S)
Can we rewrite F1 Score to be
F1 Score = 2*PPV*PPV / (PPV+PPV) = 2*PPV*PPV / (2*PPV) = PPV !!
They all the same?
It seems there is some condition or something I am missing here!
can someone please explain to me what am I missing?
Using medical diagnosis as an example:
sensitivity is the proportion testing positive among all those who actually have the disease.
Sensitivity = TP/(TP+FN) = TPR
While,
PPV is how likely a patient has a predicted specific disease given the test results.
PPV = TP/(TP+FP) which is definitely NOT equal to TP/(TP+FN)!
Regarding F1:
F1 is the harmonic mean of precision and sensitivity. One is normalized by column and the other normalized by row. Precision is synonymous with PPV while sensitivity is synonymous with TPR.
F1 = 2*PPV*TPR / (PPV+TPR)
Related
I have looked at the questions already asked about how to calculate precision and recall. However, I am still confused. There are specific questions that I face during the interviews that I am not sure how to approach them.
Question-1:
A binary classifier correctly filters 90% of spam emails but misclassifies 5% of non-spam emails as spam. Give the classifier's:
- recall
- precision
- accuracy,
or explain why there's not enough information to know.
For the above example, my solution would be:
TP = 90%
FP = 5%
Recall = TP / TP + FN => cannot be calculated without knowing FN
Precision = TP / TP + FP => 0.90/0.90 + 0.05 = 0.94
Accuracy = TP + TN / FN + FP + TP + TN => cannot be calculated without knowing TN & FN
Question-2:
1 percent of Uber transactions are fraud
We have a model that:
When a transaction is fraud, 99% classifies as fraud
When a transaction is not fraud, 99% classifies not fraud
What is the precision and recall of this model?
For the above example, my solution would be:
Let's say we have 1,000 daily transactions
number of positive classes (number of frauds): 10
number of negative classes (number of non-frauds): 990
TP = 10 * 99% = 9.9
TN = 990 * 99% = 980.1
FN = 990 * 1% = 9.9
FP = 10 * 1% = 0.1
Then use these in the below formula.
recall = TP / (TP + FN)
precision = TP / (FP + TP)
Can you please help me clarify an approach for these kinds of questions?
Thank you all!
I am trying to do a multiclass classification problem (containing 3 labels) with softmax regression.
This is my first rough implementation with gradient descent and back propagation (without using regularization and any advanced optimization algorithm) containing only 1 layer.
Also when learning-rate is big (>0.003) cost becomes NaN, on decreasing learning-rate the cost function works fine.
Can anyone explain what I'm doing wrong??
# X is (13,177) dimensional
# y is (3,177) dimensional with label 0/1
m = X.shape[1] # 177
W = np.random.randn(3,X.shape[0])*0.01 # (3,13)
b = 0
cost = 0
alpha = 0.0001 # seems too small to me but for bigger values cost becomes NaN
for i in range(100):
Z = np.dot(W,X) + b
t = np.exp(Z)
add = np.sum(t,axis=0)
A = t/add
loss = -np.multiply(y,np.log(A))
cost += np.sum(loss)/m
print('cost after iteration',i+1,'is',cost)
dZ = A-y
dW = np.dot(dZ,X.T)/m
db = np.sum(dZ)/m
W = W - alpha*dW
b = b - alpha*db
This is what I get :
cost after iteration 1 is 6.661713420377916
cost after iteration 2 is 23.58974203186562
cost after iteration 3 is 52.75811642877174
.............................................................
...............*upto 100 iterations*.................
.............................................................
cost after iteration 99 is 1413.555298639879
cost after iteration 100 is 1429.6533630169406
Well after some time i figured it out.
First of all the cost was increasing due to this :
cost += np.sum(loss)/m
Here plus sign is not needed as it will add all the previous cost computed on every epoch which is not what we want. This implementation is generally required during mini-batch gradient descent for computing cost over each epoch.
Secondly the learning rate is too big for this problem that's why cost was overshooting the minimum value and becoming NaN.
I looked in my code and find out that my features were of very different range (one was from -1 to 1 and other was -5000 to 5000) which was limiting my algorithm to use greater values for learning rate.
So I applied feature scaling :
var = np.var(X, axis=1)
X = X/var
Now learning rate can be much bigger (<=0.001).
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 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'm in a situation where I need to train a model to predict a scalar value, and it's important to have the predicted value be in the same direction as the true value, while the squared error being minimum.
What would be a good choice of loss function for that?
For example:
Let's say the predicted value is -1 and the true value is 1. The loss between the two should be a lot greater than the loss between 3 and 1, even though the squared error of (3, 1) and (-1, 1) is equal.
Thanks a lot!
This turned out to be a really interesting question - thanks for asking it! First, remember that you want your loss functions to be defined entirely of differential operations, so that you can back-propagation though it. This means that any old arbitrary logic won't necessarily do. To restate your problem: you want to find a differentiable function of two variables that increases sharply when the two variables take on values of different signs, and more slowly when they share the same sign. Additionally, you want some control over how sharply these values increase, relative to one another. Thus, we want something with two configurable constants. I started constructing a function that met these needs, but then remembered one you can find in any high school geometry text book: the elliptic paraboloid!
The standard formulation doesn't meet the requirement of sign agreement symmetry, so I had to introduce a rotation. The plot above is the result. Note that it increases more sharply when the signs don't agree, and less sharply when they do, and that the input constants controlling this behaviour are configurable. The code below is all that was needed to define and plot the loss function. I don't think I've ever used a geometric form as a loss function before - really neat.
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from matplotlib import cm
def elliptic_paraboloid_loss(x, y, c_diff_sign, c_same_sign):
# Compute a rotated elliptic parabaloid.
t = np.pi / 4
x_rot = (x * np.cos(t)) + (y * np.sin(t))
y_rot = (x * -np.sin(t)) + (y * np.cos(t))
z = ((x_rot**2) / c_diff_sign) + ((y_rot**2) / c_same_sign)
return(z)
c_diff_sign = 4
c_same_sign = 2
a = np.arange(-5, 5, 0.1)
b = np.arange(-5, 5, 0.1)
loss_map = np.zeros((len(a), len(b)))
for i, a_i in enumerate(a):
for j, b_j in enumerate(b):
loss_map[i, j] = elliptic_paraboloid_loss(a_i, b_j, c_diff_sign, c_same_sign)
fig = plt.figure()
ax = fig.gca(projection='3d')
X, Y = np.meshgrid(a, b)
surf = ax.plot_surface(X, Y, loss_map, cmap=cm.coolwarm,
linewidth=0, antialiased=False)
plt.show()
From what I understand, your current loss function is something like:
loss = mean_square_error(y, y_pred)
What you could do, is to add one other component to your loss, being this a component that penalizes negative numbers and does nothing with positive numbers. And you can choose a coefficient for how much you want to penalize it. For that, we can use like a negative shaped ReLU. Something like this:
Let's call "Neg_ReLU" to this component. Then, your loss function will be:
loss = mean_squared_error(y, y_pred) + Neg_ReLU(y_pred)
So for example, if your result is -1, then the total error would be:
mean_squared_error(1, -1) + 1
And if your result is 3, then the total error would be:
mean_squared_error(1, -1) + 0
(See in the above function how Neg_ReLU(3) = 0, and Neg_ReLU(-1) = 1.
If you want to penalize more the negative values, then you can add a coefficient:
coeff_negative_value = 2
loss = mean_squared_error(y, y_pred) + coeff_negative_value * Neg_ReLU
Now the negative values are more penalized.
The ReLU negative function we can build it like this:
tf.nn.relu(tf.math.negative(value))
So summarizing, in the end your total loss will be:
coeff = 1
Neg_ReLU = tf.nn.relu(tf.math.negative(y))
total_loss = mean_squared_error(y, y_pred) + coeff * Neg_ReLU