I have 38 variables, like oxygen, temperature, pressure, etc and have a task to determine the total yield produced every day from these variables. When I calculate the regression coefficients and intercept value, they seem to be abnormal and very high (Impractical). For example, if 'temperature' coefficient was found to be +375.456, I could not give a meaning to them saying an increase in one unit in temperature would increase yield by 375.456g. That's impractical in my scenario. However, the prediction accuracy seems right. I would like to know, how to interpret these huge intercept( -5341.27355) and huge beta values shown below. One other important point is that I removed multicolinear columns and also, I am not scaling the variables/normalizing them because I need beta coefficients to have meaning such that I could say, increase in temperature by one unit increases yield by 10g or so. Your inputs are highly appreciated!
modl.intercept_
Out[375]: -5341.27354961415
modl.coef_
Out[376]:
array([ 1.38096017e+00, -7.62388829e+00, 5.64611255e+00, 2.26124164e-01,
4.21908571e-01, 4.50695302e-01, -8.15167717e-01, 1.82390184e+00,
-3.32849969e+02, 3.31942553e+02, 3.58830763e+02, -2.05076898e-01,
-3.06404757e+02, 7.86012402e+00, 3.21339318e+02, -7.00817205e-01,
-1.09676321e+04, 1.91481734e+00, 6.02929848e+01, 8.33731416e+00,
-6.23433431e+01, -1.88442804e+00, 6.86526274e+00, -6.76103795e+01,
-1.11406021e+02, 2.48270706e+02, 2.94836048e+01, 1.00279016e+02,
1.42906659e-02, -2.13019683e-03, -6.71427100e+02, -2.03158515e+02,
9.32094007e-03, 5.56457014e+01, -2.91724945e+00, 4.78691176e-01,
8.78121854e+00, -4.93696073e+00])
It's very unlikely that all of these variables are linearly correlated, so I would suggest that you have a look at simple non-linear regression techniques, such as Decision Trees or Kernel Ridge Regression. These are however more difficult to interpret.
Going back to your issue, these high weights might well be due to there being some high amount of correlation between the variables, or that you simply don't have very much training data.
If you instead of linear regression use Lasso Regression, the solution is biased away from high regression coefficients, and the fit will likely improve as well.
A small example on how to do this in scikit-learn, including cross validation of the regularization hyper-parameter:
from sklearn.linear_model LassoCV
# Make up some data
n_samples = 100
n_features = 5
X = np.random.random((n_samples, n_features))
# Make y linear dependent on the features
y = np.sum(np.random.random((1,n_features)) * X, axis=1)
model = LassoCV(cv=5, n_alphas=100, fit_intercept=True)
model.fit(X,y)
print(model.intercept_)
If you have a linear regression, the formula looks like this (y= target, x= features inputs):
y= x1*b1 +x2*b2 + x3*b3 + x4*b4...+ c
where b1,b2,b3,b4... are your modl.coef_. AS you already realized one of your bigges number is 3.319+02 = 331 and the intercept is also quite big with -5431.
As you already mentioned the coeffiecient variables means how much the target variable changes, if the coeffiecient feature changes with 1 unit and all others features are constant.
so for your interpretation, the higher the absoult coeffienct, the higher the influence of your analysis. But it is important to note that the model is using a lot of high coefficient, that means your model is not depending only of one variable
Related
Say for example, a dataset contains 60% instances for "Yes" class and 30% instances for "NO" class.
In this scenario, Precision, Recall for the random classifier are
Precision =60%
Recall =50%
Then, what will be the accuracy for random classifier in this scenario?
Some caution is required here, since the very definition of a random classifier is somewhat ambiguous; this is best illustrated in cases of imbalanced data.
By definition, the accuracy of a binary classifier is
acc = P(class=0) * P(prediction=0) + P(class=1) * P(prediction=1)
where P stands for probability.
Indeed, if we stick to the intuitive definition of a random binary classifier as giving
P(prediction=0) = P(prediction=1) = 0.5
then the accuracy computed by the above formula is always 0.5, irrespectively of the class distribution (i.e. the values of P(class=0) and P(class=1)).
However, in this definition, there is an implicit assumption, i.e. that our classes are balanced, each one consisting of 50% of our dataset.
This assumption (and the corresponding intuition) breaks down in cases of class imbalance: if we have a dataset where, say, 90% of samples are of class 0 (i.e. P(class=0)=0.9), then it doesn't make much sense to use the above definition of a random binary classifier; instead, we should use the percentages of the class distributions themselves as the probabilities of our random classifier, i.e.:
P(prediction=0) = P(class=0) = 0.9
P(prediction=1) = P(class=1) = 0.1
Now, plugging these values to the formula defining the accuracy, we get:
acc = P(class=0) * P(prediction=0) + P(class=1) * P(prediction=1)
= (0.9 * 0.9) + (0.1 * 0.1)
= 0.82
which is nowhere close to the naive value of 0.5...
As I already said, AFAIK there are no clear-cut definitions of a random classifier in the literature. Sometimes the "naive" random classifier (always flip a fair coin) is referred to as a "random guess" classifier, while what I have described is referred to as a "weighted guess" one, but still this is far from being accepted as a standard...
The bottom line here is the following: since the main reason for using a random classifier is as a baseline, it makes sense to do so only in relatively balanced datasets. In your case of a 60-40 balance, the result turns out to be 0.52, which is admittedly not far from the naive one of 0.5; but for highly imbalanced datasets (e.g. 90-10), the usefulness itself of the random classifier as a baseline ceases to exist, since the correct baseline has become "always predict the majority class", which here would give an accuracy of 90%, in contrast to the random classifier accuracy of just 82% (let alone the 50% accuracy of the naive approach)...
As #desertnaut mentioned, if you're after a naïve benchmark for your model you're always better using "always predict the majority class" as your benchmark, achieving accuracy of %of_samples_in_majority_class (which is always better than either a random guess or a weighted guess).
In Deepchecks (a package I maintain) we have a check that automatically compares the performance of your model to a simple model (either weighted random, majority class or simple decision tree).
from deepchecks.checks import SimpleModelComparison
from deepchecks import Dataset
SimpleModelComparison().run(Dataset(train_df, label='target'), Dataset(test_df, label='target'), model)
I have a dataset X whose shape is (1741, 61). Using logistic regression with cross_validation I was getting around 62-65% for each split (cv =5).
I thought that if I made the data quadratic, the accuracy is supposed to increase. However, I'm getting the opposite effect (I'm getting each split of cross_validation to be in the 40's, percentage-wise) So,I'm presuming I'm doing something wrong when trying to make the data quadratic?
Here is the code I'm using,
from sklearn import preprocessing
X_scaled = preprocessing.scale(X)
from sklearn.preprocessing import PolynomialFeatures
poly = PolynomialFeatures(3)
poly_x =poly.fit_transform(X_scaled)
classifier = LogisticRegression(penalty ='l2', max_iter = 200)
from sklearn.cross_validation import cross_val_score
cross_val_score(classifier, poly_x, y, cv=5)
array([ 0.46418338, 0.4269341 , 0.49425287, 0.58908046, 0.60518732])
Which makes me suspect, I'm doing something wrong.
I tried transforming the raw data into quadratic, then using preprocessing.scale, to scale the data, but it was resulting in an error.
UserWarning: Numerical issues were encountered when centering the data and might not be solved. Dataset may contain too large values. You may need to prescale your features.
warnings.warn("Numerical issues were encountered "
So I didn't bother going this route.
The other thing that's bothering is the speed of the quadratic computations. cross_val_score is taking around a couple of hours to output the score when using polynomial features. Is there any way to speed this up? I have an intel i5-6500 CPU with 16 gigs of ram, Windows 7 OS.
Thank you.
Have you tried using the MinMaxScaler instead of the Scaler? Scaler will output values that are both above and below 0, so you will run into a situation where values with a scaled value of -0.1 and those with a value of 0.1 will have the same squared value, despite not really being similar at all. Intuitively this would seem to be something that would lower the score of a polynomial fit. That being said I haven't tested this, it's just my intuition. Furthermore, be careful with Polynomial fits. I suggest reading this answer to "Why use regularization in polynomial regression instead of lowering the degree?". It's a great explanation and will likely introduce you to some new techniques. As an aside #MatthewDrury is an excellent teacher and I recommend reading all of his answers and blog posts.
There is a statement that "the accuracy is supposed to increase" with polynomial features. That is true if the polynomial features brings the model closer to the original data generating process. Polynomial features, especially making every feature interact and polynomial, may move the model further from the data generating process; hence worse results may be appropriate.
By using a 3 degree polynomial in scikit, the X matrix went from (1741, 61) to (1741, 41664), which is significantly more columns than rows.
41k+ columns will take longer to solve. You should be looking at feature selection methods. As Grr says, investigate lowering the polynomial. Try L1, grouped lasso, RFE, Bayesian methods. Try SMEs (subject matter experts who may be able to identify specific features that may be polynomial). Plot the data to see which features may interact or be best in a polynomial.
I have not looked at it for a while but I recall discussions on hierarchically well-formulated models (can you remove x1 but keep the x1 * x2 interaction). That is probably worth investigating if your model behaves best with an ill-formulated hierarchical model.
While following the Coursera-Machine Learning class, I wanted to test what I learned on another dataset and plot the learning curve for different algorithms.
I (quite randomly) chose the Online News Popularity Data Set, and tried to apply a linear regression to it.
Note : I'm aware it's probably a bad choice but I wanted to start with linear reg to see later how other models would fit better.
I trained a linear regression and plotted the following learning curve :
This result is particularly surprising for me, so I have questions about it :
Is this curve even remotely possible or is my code necessarily flawed?
If it is correct, how can the training error grow so quickly when adding new training examples? How can the cross validation error be lower than the train error?
If it is not, any hint to where I made a mistake?
Here's my code (Octave / Matlab) just in case:
Plot :
lambda = 0;
startPoint = 5000;
stepSize = 500;
[error_train, error_val] = ...
learningCurve([ones(mTrain, 1) X_train], y_train, ...
[ones(size(X_val, 1), 1) X_val], y_val, ...
lambda, startPoint, stepSize);
plot(error_train(:,1),error_train(:,2),error_val(:,1),error_val(:,2))
title('Learning curve for linear regression')
legend('Train', 'Cross Validation')
xlabel('Number of training examples')
ylabel('Error')
Learning curve :
S = ['Reg with '];
for i = startPoint:stepSize:m
temp_X = X(1:i,:);
temp_y = y(1:i);
% Initialize Theta
initial_theta = zeros(size(X, 2), 1);
% Create "short hand" for the cost function to be minimized
costFunction = #(t) linearRegCostFunction(X, y, t, lambda);
% Now, costFunction is a function that takes in only one argument
options = optimset('MaxIter', 50, 'GradObj', 'on');
% Minimize using fmincg
theta = fmincg(costFunction, initial_theta, options);
[J, grad] = linearRegCostFunction(temp_X, temp_y, theta, 0);
error_train = [error_train; [i J]];
[J, grad] = linearRegCostFunction(Xval, yval, theta, 0);
error_val = [error_val; [i J]];
fprintf('%s %6i examples \r', S, i);
fflush(stdout);
end
Edit : if I shuffle the whole dataset before splitting train/validation and doing the learning curve, I have very different results, like the 3 following :
Note : the training set size is always around 24k examples, and validation set around 8k examples.
Is this curve even remotely possible or is my code necessarily flawed?
It's possible, but not very likely. You might be picking the hard to predict instances for the training set and the easy ones for the test set all the time. Make sure you shuffle your data, and use 10 fold cross validation.
Even if you do all this, it is still possible for it to happen, without necessarily indicating a problem in the methodology or the implementation.
If it is correct, how can the training error grow so quickly when adding new training examples? How can the cross validation error be lower than the train error?
Let's assume that your data can only be properly fitted by a 3rd degree polynomial, and you're using linear regression. This means that the more data you add, the more obviously it will be that your model is inadequate (higher training error). Now, if you choose few instances for the test set, the error will be smaller, because linear vs 3rd degree might not show a big difference for too few test instances for this particular problem.
For example, if you do some regression on 2D points, and you always pick 2 points for your test set, you will always have 0 error for linear regression. An extreme example, but you get the idea.
How big is your test set?
Also, make sure that your test set remains constant throughout the plotting of the learning curves. Only the train set should increase.
If it is not, any hint to where I made a mistake?
Your test set might not be large enough or your train and test sets might not be properly randomized. You should shuffle the data and use 10 fold cross validation.
You might want to also try to find other research regarding that data set. What results are other people getting?
Regarding the update
That makes a bit more sense, I think. Test error is generally higher now. However, those errors look huge to me. Probably the most important information this gives you is that linear regression is very bad at fitting this data.
Once more, I suggest you do 10 fold cross validation for learning curves. Think of it as averaging all of your current plots into one. Also shuffle the data before running the process.
I developed a image processing program that identifies what a number is given an image of numbers. Each image was 27x27 pixels = 729 pixels. I take each R, G and B value which means I have 2187 variables from each image (+1 for the intercept = total of 2188).
I used the below gradient descent formula:
Repeat {
θj = θj−α/m∑(hθ(x)−y)xj
}
Where θj is the coefficient on variable j; α is the learning rate; hθ(x) is the hypothesis; y is real value and xj is the value of variable j. m is the number of training sets. hθ(x), y are for each training set (i.e. that's what the summation sign is for). Further the hypothesis is defined as:
hθ(x) = 1/(1+ e^-z)
z= θo + θ1X1+θ2X2 +θ3X3...θnXn
With this, and 3000 training images, I was able to train my program in just over an hour and when tested on a cross validation set, it was able to identify the correct image ~ 67% of the time.
I wanted to improve that so I decided to attempt a polynomial of degree 2.
However the number of variables jumps from 2188 to 2,394,766 per image! It takes me an hour just to do 1 step of gradient descent.
So my question is, how is this vast number of variables handled in machine learning? On the one hand, I don't have enough space to even hold that many variables for each training set. On the other hand, I am currently storing 2188 variables per training sample, but I have to perform O(n^2) just to get the values of each variable multiplied by another variable (i.e. the polynomial to degree 2 values).
So any suggestions / advice is greatly appreciated.
try to use some dimensionality reduction first (PCA, kernel PCA, or LDA if you are classifying the images)
vectorize your gradient descent - with most math libraries or in matlab etc. it will run much faster
parallelize the algorithm and then run in on multiple CPUs (but maybe your library for multiplying vectors already supports parallel computations)
Along with Jirka-x1's answer, I would first say that this is one of the key differences in working with image data than say text data for ML: high dimensionality.
Second... this is a duplicate, see How to approach machine learning problems with high dimensional input space?
I have a scenario where I have several thousand instances of data. The data itself is represented as a single integer value. I want to be able to detect when an instance is an extreme outlier.
For example, with the following example data:
a = 10
b = 14
c = 25
d = 467
e = 12
d is clearly an anomaly, and I would want to perform a specific action based on this.
I was tempted to just try an use my knowledge of the particular domain to detect anomalies. For instance, figure out a distance from the mean value that is useful, and check for that, based on heuristics. However, I think it's probably better if I investigate more general, robust anomaly detection techniques, which have some theory behind them.
Since my working knowledge of mathematics is limited, I'm hoping to find a technique which is simple, such as using standard deviation. Hopefully the single-dimensioned nature of the data will make this quite a common problem, but if more information for the scenario is required please leave a comment and I will give more info.
Edit: thought I'd add more information about the data and what I've tried in case it makes one answer more correct than another.
The values are all positive and non-zero. I expect that the values will form a normal distribution. This expectation is based on an intuition of the domain rather than through analysis, if this is not a bad thing to assume, please let me know. In terms of clustering, unless there's also standard algorithms to choose a k-value, I would find it hard to provide this value to a k-Means algorithm.
The action I want to take for an outlier/anomaly is to present it to the user, and recommend that the data point is basically removed from the data set (I won't get in to how they would do that, but it makes sense for my domain), thus it will not be used as input to another function.
So far I have tried three-sigma, and the IQR outlier test on my limited data set. IQR flags values which are not extreme enough, three-sigma points out instances which better fit with my intuition of the domain.
Information on algorithms, techniques or links to resources to learn about this specific scenario are valid and welcome answers.
What is a recommended anomaly detection technique for simple, one-dimensional data?
Check out the three-sigma rule:
mu = mean of the data
std = standard deviation of the data
IF abs(x-mu) > 3*std THEN x is outlier
An alternative method is the IQR outlier test:
Q25 = 25th_percentile
Q75 = 75th_percentile
IQR = Q75 - Q25 // inter-quartile range
IF (x < Q25 - 1.5*IQR) OR (Q75 + 1.5*IQR < x) THEN x is a mild outlier
IF (x < Q25 - 3.0*IQR) OR (Q75 + 3.0*IQR < x) THEN x is an extreme outlier
this test is usually employed by Box plots (indicated by the whiskers):
EDIT:
For your case (simple 1D univariate data), I think my first answer is well suited.
That however isn't applicable to multivariate data.
#smaclell suggested using K-means to find the outliers. Beside the fact that it is mainly a clustering algorithm (not really an outlier detection technique), the problem with k-means is that it requires knowing in advance a good value for the number of clusters K.
A better suited technique is the DBSCAN: a density-based clustering algorithm. Basically it grows regions with sufficiently high density into clusters which will be maximal set of density-connected points.
DBSCAN requires two parameters: epsilon and minPoints. It starts with an arbitrary point that has not been visited. It then finds all the neighbor points within distance epsilon of the starting point.
If the number of neighbors is greater than or equal to minPoints, a cluster is formed. The starting point and its neighbors are added to this cluster and the starting point is marked as visited. The algorithm then repeats the evaluation process for all the neighbors recursively.
If the number of neighbors is less than minPoints, the point is marked as noise.
If a cluster is fully expanded (all points within reach are visited) then the algorithm proceeds to iterate through the remaining unvisited points until they are depleted.
Finally the set of all points marked as noise are considered outliers.
There are a variety of clustering techniques you could use to try to identify central tendencies within your data. One such algorithm we used heavily in my pattern recognition course was K-Means. This would allow you to identify whether there are more than one related sets of data, such as a bimodal distribution. This does require you having some knowledge of how many clusters to expect but is fairly efficient and easy to implement.
After you have the means you could then try to find out if any point is far from any of the means. You can define 'far' however you want but I would recommend the suggestions by #Amro as a good starting point.
For a more in-depth discussion of clustering algorithms refer to the wikipedia entry on clustering.
This is an old topic but still it lacks some information.
Evidently, this can be seen as a case of univariate outlier detection. The approaches presented above have several pros and cons. Here are some weak spots:
Detection of outliers with the mean and sigma has the obvious disadvantage of dependence of mean and sigma on the outliers themselves.
The case of the small sample limit (see question for example) is not adequately covered by, 3 sigma, K-Means, IQR etc.
And I could go on... However the statistical literature offers a simple metric: the median absolute deviation. (Medians are insensitive to outliers)
Details can be found here: https://www.sciencedirect.com/book/9780128047330/introduction-to-robust-estimation-and-hypothesis-testing
I think this problem can be solved in a few lines of python code like this:
import numpy as np
import scipy.stats as sts
x = np.array([10, 14, 25, 467, 12]) # your values
np.abs(x - np.median(x))/(sts.median_abs_deviation(x)/0.6745) #MAD criterion
Subsequently you reject values above a certain threshold (97.5 percentile of the distribution of data), in case of an assumed normal distribution the threshold is 2.24. Here it translates to:
array([ 0.6745 , 0. , 1.854875, 76.387125, 0.33725 ])
or the 467 entry being rejected.
Of course, one could argue, that the MAD (as presented) also assumes a normal dist. Therefore, why is it that argument 2 above (small sample) does not apply here? The answer is that MAD has a very high breakdown point. It is easy to choose different threshold points from different distributions and come to the same conclusion: 467 is the outlier.
Both three-sigma rule and IQR test are often used, and there are a couple of simple algorithms to detect anomalies.
The three-sigma rule is correct
mu = mean of the data
std = standard deviation of the data
IF abs(x-mu) > 3*std THEN x is outlier
The IQR test should be:
Q25 = 25th_percentile
Q75 = 75th_percentile
IQR = Q75 - Q25 // inter-quartile range
If x > Q75 + 1.5 * IQR or x < Q25 - 1.5 * IQR THEN x is a mild outlier
If x > Q75 + 3.0 * IQR or x < Q25 – 3.0 * IQR THEN x is a extreme outlier