Develop Reference

Marginal Density Probability using np

P = np.array(
[
[0.03607908, 0.03760034, 0.00503184, 0.0205082 , 0.01051408,
0.03776221, 0.00131325, 0.03760817, 0.01770659],
[0.03750162, 0.04317351, 0.03869997, 0.03069872, 0.02176718,
0.04778769, 0.01021053, 0.00324185, 0.02475319],
[0.03770951, 0.01053285, 0.01227089, 0.0339596 , 0.02296711,
0.02187814, 0.01925662, 0.0196836 , 0.01996279],
[0.02845139, 0.01209429, 0.02450163, 0.00874645, 0.03612603,
0.02352593, 0.00300314, 0.00103487, 0.04071951],
[0.00940187, 0.04633153, 0.01094094, 0.00172007, 0.00092633,
0.02032679, 0.02536328, 0.03552956, 0.01107725]
]
)
Here's the dataset where X corresponds to rows and Y corresponds to columns. I'm trying to figure out how to calculate the Covariance and Marginal Density Probability (MDP) for Y (columns). for the Covariance I have the following.
np.cov(P)
array([[ 2.247e-04, 6.999e-05, 2.571e-05, -2.822e-05, 1.061e-04],
[ 6.999e-05, 2.261e-04, 9.535e-07, 8.165e-05, -2.013e-05],
[ 2.571e-05, 9.535e-07, 7.924e-05, 1.357e-05, -8.118e-05],
[-2.822e-05, 8.165e-05, 1.357e-05, 2.039e-04, -1.267e-04],
[ 1.061e-04, -2.013e-05, -8.118e-05, -1.267e-04, 2.372e-04]])
How do I get the MDP? Also, is there a way to use numpy to select just the X and Y vals and assign them to variables where X= P's rows and Y=P's columns?

The data stored in P are a little ambiguous. In statistics, the X and Y have a very specific meaning. Usually, each row refers to one observation (i.e. datapoint) of some statistical object while the column represents a feature that is measured for each statistical object. In your case, there would be 9 observations with 5 features. This is referred to as a design matrix X, considered exogenous (independent), and serves as the foundation of most statistical learning algorithms. In supervised learning, there is additionally a vector Y; its length equals the number of rows in the X.
Your task at hand is of unsupervised nature as there is no true exogenous data Y and you are interested in the distribution of X alone. This opens up additional questions. Indeed, np.cov() computes the empirical covariance matrix measuring the pairwise covariance between each of these 5 features resulting in a symmetric 5x5 matrix as you indicated. Asking for the marginal probability density of each column (i.e. feature), however, refers to the univariate distribution of each feature alone. The covariance matrix in between features is irrelevant for this task.
There are several methodologies to obtain estimates of an unknown distribution given some data. Broadly speaking, they fall into two categories: parametric and non-parametric. I'll explain how each methodology works and can be implemented by leveraging NumPy exactly in the way you eluded to.
1. Parametric density estimation
In many cases, one assumes that the data stems from a particular parametric distribution. This distributional assumption is mostly based on convenience rather than prior knowledge. Then, estimating the unknown parameter values completely determines the distribution(s). In your case, for example, you could assume that each feature is univariate normal distributed.
import numpy as np
from matplotlib import pyplot as plt
from scipy import stats
# mere numerical grid to plot densities (no statistical significance)
x_ = np.linspace(0.0, 0.055, 1000)
# estimate mean (mu) and standard deviation (sigma) for each column
mean_vec = np.mean(P, axis=1)
std_vec = np.std(P, axis=1)
# histograms
for i in range(5):
plt.plot(x_, stats.norm.pdf(x_, loc=mean_vec[i], scale=std_vec[I]),
label='Col. {}'.format(i+1))
plt.suptitle('Marginal Distribution Estimates', fontsize=21, y=1.025)
plt.title('parametric via univariate Normal densities', fontsize=14)
plt.legend(loc='upper right')
plt.show()
2. Nonparametric density estimation
Alternatively, you can use a histogram as a non-parametric estimator of the unknown probability density functions (of each column/feature). Note, however, that you still have to choose a bandwidth h that determines the width of the bins. Additionally, non-parametric tools require a larger sample size to provide accurate estimates. Your sample size 9 is likely insufficient.
import numpy as np
from matplotlib import pyplot as plt
# endpoints on all bins: implies bandwith h=0.00229
bins = np.linspace(0.0, 0.055, 25)
h = np.diff(bins)[0]
# histograms
for i in range(5):
plt.hist(P[:,i], bins, alpha=0.5, label='Col. {}'.format(i+1))
plt.suptitle('Marginal Distribution Estimates', fontsize=21, y=1.025)
plt.title('nonparametric via histograms (h={})'.format(round(h, 4)), fontsize=14)
plt.legend(loc='upper right')
plt.show()

Logistic Regression vs predicting probability by splitting data into bin

So I am exploring using a logistic regression model to predict the probability of a shot resulting in a goal. I have two predictors but for simplicity lets assume I have one predictor: distance from the goal. When doing some data exploration I decided to investigate the relationship between distance and the result of a goal. I did this graphical by splitting the data into equal size bins and then taking the mean of all the results (0 for a miss and 1 for a goal) within each bin. Then I plotted the average distance from goal for each bin vs the probability of scoring. I did this in python
#use the seaborn library to inspect the distribution of the shots by result (goal or no goal)
fig, axes = plt.subplots(1, 2,figsize=(11, 5))
#first we want to create bins to calc our probability
#pandas has a function qcut that evenly distibutes the data
#into n bins based on a desired column value
df['Goal']=df['Goal'].astype(int)
df['Distance_Bins'] = pd.qcut(df['Distance'],q=50)
#now we want to find the mean of the Goal column(our prob density) for each bin
#and the mean of the distance for each bin
dist_prob = df.groupby('Distance_Bins',as_index=False)['Goal'].mean()['Goal']
dist_mean = df.groupby('Distance_Bins',as_index=False)['Distance'].mean()['Distance']
dist_trend = sns.scatterplot(x=dist_mean,y=dist_prob,ax=axes[0])
dist_trend.set(xlabel="Avg. Distance of Bin",
ylabel="Probabilty of Goal",
title="Probability of Scoring Based on Distance")
Probability of Scoring Based on Distance
So my question is why would we go through the process of creating a logistic regression model when I could fit a curve to the plot in the image? Would that not provide a function that would predict a probability for a shot with distance x.
I guess the problem would be that we are reducing say 40,000 data point into 50 but I'm not entirely sure why this would be a problem for predict future shot. Could we increase the number of bins or would that just add variability? Is this a case of bias-variance trade off? Im just a little confused about why this would not be as good as a logistic model.

The binning method is a bit more finicky than the logistic regression since you need to try different types of plots to fit the curve (e.g. inverse relationship, log, square, etc.), while for logistic regression you only need to adjust the learning rate to see results.
If you are using one feature (your "Distance" predictor), I wouldn't see much difference between the binning method and the logistic regression. However, when you are using two or more features (I see "Distance" and "Angle" in the image you provided), how would you plan to combine the probabilities for each to make a final 0/1 classification? It can be tricky. For one, perhaps "Distance" is more useful a predictor than "Angle". However, logistic regression does that for you because it can adjust the weights.
Regarding your binning method, if you use fewer bins you might see more bias since the data may be more complicated than you think, but this is not that likely because your data looks quite simple at first glance. However, if you use more bins that would not significantly increase variance, assuming that you fit the curve without varying the order of the curve. If you change the order of the curve you fit, then yes, it will increase variance. However, your data seems like it is amenable to a very simple fit if you go with this method.

Is there a way to get the probability of a prediction using XGBoostRegressor?

I have built a XGBoostRegressor model using around 200 categorical features predicting a countinous time variable.
But I would want to get both the actual prediction and the probability of that prediction as output. Is there any way to get this from the XGBoostRegressor model?
So I both want and P(Y|X) as output. Any idea how to do this?

There is no probability in regression, In regression the only output you will get is a predicted value thats why it is called regression, so for any regressor probability of a prediction is not possible. Its only there in classification.

As mentioned before, there is no probability associated with regression.
However, you could probably add a confidence interval on that regression, to see whether or not your regression can be trusted.
One thing to note though, is that the variance might not be the same along the data.
Let's assume that you study a time based phenomenon. Specifically, you have the temperature (y) after (x) time (in sec for instance) inside an oven. At x = 0s it is at 20°C, and you start heating it, and want to know the evolution in order to predict the temperature after x seconds. The variance could be the same after 20 seconds and after 5 minutes, or be completely different. This is called heteroscedasticity.
If you want to use a confidence interval, you probably want to make sure that you took care of heteroscedasticity, so your interval is the same for all the data.
You can probably try to get the distribution of your known outputs and compare the prediction on that curve, and check the pvalue. But that would only give you a measure of how realistic it is to get that output, without taking the input into consideration. If you know your inputs/outputs are in a specific interval, this could work.
EDIT
This is how I would do it. Obviously the outputs are your real outputs.
import numpy as np
import matplotlib.pyplot as plt
from scipy import integrate
from scipy.interpolate import interp1d
N = 1000 # The number of sample
mean = 0
std = 1
outputs = np.random.normal(loc=mean, scale=std, size=N)
# We want to get a normed histogram (since this is PDF, if we integrate
# it must be equal to 1)
nbins = N / 10
n = int(N / nbins)
p, x = np.histogram(outputs, bins=n, normed=True)
plt.hist(outputs, bins=n, normed=True)
x = x[:-1] + (x[ 1] - x[0])/2 # converting bin edges to centers
# Now we want to interpolate :
# f = CubicSpline(x=x, y=p, bc_type='not-a-knot')
f = interp1d(x=x, y=p, kind='quadratic', fill_value='extrapolate')
x = np.linspace(-2.9*std, 2.9*std, 10000)
plt.plot(x, f(x))
plt.show()
# To check :
area = integrate.quad(f, x[0], x[-1])
print(area) # (should be close to 1)
Now, the interpolate method is not great for outliers. if a predicted data is extremely far (more than 3 times the std) from your distribution, it wont work. Other than that, you can now use the PDF to get meaningful results.
It is not perfect, but it is the best I came up with in that time. I'm sure there are some better ways to do it. If your data follow a normal law, it becomes trivial.

I suggest you to look into Ngboost (essentially a wrapper of Xgboost which provides eventually a probabilistic model.
Here you can find slides on the Ngboost functioning and the seminal Ngboost paper.
The basic idea is to assume a specific distribution for $P(Y|X=x)$ (by default is the Gaussian distribution) and fit an Xgboost model to estimate the best parameters of the distribution (for the Gaussian $\mu$ and $\sigma$. The model will split the variables' space into different regions with different distributions, i.e. same family (eg. Gaussian) but different parameters.
After training the model, you're provided with the method '''pred_dist''' which returns the estimated distribution $P(Y|X=x)$ for a given set of values $x$

Sklearn LDA-analysis won't generate 2 dimensions

I'm trying to plot a 3-feature dataset with a binary classification on a matplotlib plot. This worked with an example dataset provided in a guide (http://www.apnorton.com/blog/2016/12/19/Visualizing-Multidimensional-Data-in-Python/) but when I try to instead insert my own dataset, the LinearDiscriminantAnalysis will only output a one-dimensional series, no matter what number I put in "n_components". Why would this not work with my own code?
Data = pd.read_csv("DataFrame.csv", sep=";")
x = Data.iloc[:, [3, 5, 7]]
y = Data.iloc[:, 8]
lda = LDA(n_components=2)
lda_transformed = pd.DataFrame(lda.fit_transform(x, y))
plt.scatter(lda_transformed[y==0][0], lda_transformed[y==0][1], label='Loss', c='red')
plt.scatter(lda_transformed[y==1][0], lda_transformed[y==1][1], label='Win', c='blue')
plt.legend()
plt.show()

In the case when the number of different class labels, C, is less than the number of observations (almost always), then linear discriminant analysis will always produce C - 1 discriminating components. Using n_components from the sklearn API is only a means to choose possibly fewer components, e.g. in the case when you know what dimensionality you'd like to reduce down to. But you could never use n_components to get more components.
This is discussed in the Wikipedia section on Multiclass LDA. The definition of the between-class scatter is given as
\Sigma_{b} = (1 / C) \sum_{i}^{C}( (\mu_{i} - mu)(\mu_{i} - mu)^{T}
which is the empirical covariance matrix among the population of class means. By definition, such a covariance matrix has rank at most C - 1.
... the variability between features will be contained in the subspace spanned by the eigenvectors corresponding to the C − 1 largest eigenvalues ...
So because LDA uses a decomposition of the class mean covariance matrix, it means the dimensionality reduction it can provide is based on the number of class labels, and not on the sample size nor the feature dimensionality.
In the example you linked, it doesn't matter how many features there are. The point is that the example uses 3 simulated cluster centers, so there are 3 class labels. This means linear discriminant analysis could produce projection of the data onto either 1-dimensional or 2-dimensional discriminating subspaces.
But in your data, you start out with only 2 class labels, a binary problem. This means the dimensionality of the linear discriminant model can be at most 1-dimensional, literally a line that forms the decision boundary between the two classes. Dimensionality reduction with LDA in this case would simply be the projection of data points onto a particular normal vector of that separating line.
If you want to specifically reduce down to two dimensions, you can try many of the other algorithms that sklearn provides: t-SNE, ISOMAP, PCA and kernel PCA, random projection, multi-dimensional scaling, among others. Many of these allow you to choose the dimensionality of the projected space, up to the original feature dimensionality, or sometimes you can even project into larger spaces, like with kernel PCA.

In the example you give, dimension reduction by LDA reduces the data from 13 features to 2 features, however in your example it reduces from 3 to 1 (even though you wanted to get 2 features), thus it is not possible to plot in 2D.
If you really want to select 2 features out of 3, you can use feature_selection.SelectKBest to choose 2 best features and there won't be any problems plotting in 2D.
For more information, please read this fantastic answer for PCA:
https://stats.stackexchange.com/questions/2691/making-sense-of-principal-component-analysis-eigenvectors-eigenvalues

Probably it's because of sklearn implementation that won't allowed you to do so if you only have 2 class. The problem has been stated in here, https://github.com/scikit-learn/scikit-learn/issues/1967.

Understanding Partial Dependence for Gradient Boosted Regression trees

I am looking at the tutorial for partial dependence plots in Python. No equation is given in the tutorial or in the documentation. The documentation of the R function gives the formula I expected:
This does not seem to make sense with the results given in the Python tutorial. If it is an average of the prediction of house prices, then how is it negative and small? I would expect values in the millions. Am I missing something?
Update:
For regression it seems the average is subtracted off of the above formula. How would this be added back? For my trained model I can get the partial dependence by
from sklearn.ensemble.partial_dependence import partial_dependence
partial_dependence, independent_value = partial_dependence(model, features.index(independent_feature),X=df2[features])
I want to add (?) back on the average. Would I get this by just using model.predict() on the df2 values with the independent_feature values changed?

how the R formula works
The r formula presented in the question applies to a randomForest. Each tree in a random forest tries to predict the target variable directly. Thus, prediction of each tree lies in the expected interval (in your case, all house prices are positive), and prediction of the ensemble is just the average of all the individual predictions.
ensemble_prediction = mean(tree_predictions)
This is what the formula tells you: just take predictions of all the trees x and average them.
why the Python PDP values are small
In sklearn, however, partial dependence is calculated for a GradientBoostingRegressor. In gradient boosting, each tree predicts the derivative of the loss function at current prediction, which is only indirectly related to the target variable. For GB regression, prediction is given as
ensemble_prediction = initial_prediction + sum(tree_predictions * learning_rate)
and for GB classification predicted probability is
ensemble_prediction = softmax(initial_prediction + sum(tree_predictions * learning_rate))
For both cases, partial dependency is reported as just
sum(tree_predictions * learning_rate)
Thus, initial_prediction (for GradientBoostingRegressor(loss='ls') it equals just the mean of the training y) is not included into the PDP, which makes the predictions negative.
As for the small range of its values, the y_train in your example is small: mean hous value is roughly 2, so house prices are probably expressed in millions.
how the sklearn formula actually works
I have already said that in sklearn the value of partial dependence function is an average of all trees. There is one more tweak: all irrelevant features are averaged away. To describe the actual way of averaging, I will just quote the documentation of sklearn:
For each value of the ‘target’ features in the grid the partial
dependence function need to marginalize the predictions of a tree over
all possible values of the ‘complement’ features. In decision trees
this function can be evaluated efficiently without reference to the
training data. For each grid point a weighted tree traversal is
performed: if a split node involves a ‘target’ feature, the
corresponding left or right branch is followed, otherwise both
branches are followed, each branch is weighted by the fraction of
training samples that entered that branch. Finally, the partial
dependence is given by a weighted average of all visited leaves. For
tree ensembles the results of each individual tree are again averaged.
And if you are still not satisfied, see the source code.
an example
To see that the prediction is already on the scale of the dependent variable (but is just centered), you can look at a very toy example:
import numpy as np
import matplotlib.pyplot as plt
from sklearn.ensemble import GradientBoostingRegressor
from sklearn.ensemble.partial_dependence import plot_partial_dependence
np.random.seed(1)
X = np.random.normal(size=[1000, 2])
# yes, I will try to fit a linear function!
y = X[:, 0] * 10 + 50 + np.random.normal(size=1000, scale=5)
# mean target is 50, range is from 20 to 80, that is +/- 30 standard deviations
model = GradientBoostingRegressor().fit(X, y)
fig, subplots = plot_partial_dependence(model, X, [0, 1], percentiles=(0.0, 1.0), n_cols=2)
subplots[0].scatter(X[:, 0], y - y.mean(), s=0.3)
subplots[1].scatter(X[:, 1], y - y.mean(), s=0.3)
plt.suptitle('Partial dependence plots and scatters of centered target')
plt.show()
You can see that partial dependence plots reflect the true distribution of the centered target variable pretty well.
If you want not only the units, but the mean to coincide with your y, you have to add the "lost" mean to the result of the partial_dependence function and then plot the results manually:
from sklearn.ensemble.partial_dependence import partial_dependence
pdp_y, [pdp_x] = partial_dependence(model, X=X, target_variables=[0], percentiles=(0.0, 1.0))
plt.scatter(X[:, 0], y, s=0.3)
plt.plot(pdp_x, pdp_y.ravel() + model.init_.mean)
plt.show()
plt.title('Partial dependence plot in the original coordinates');

You are looking at a Partial Dependence Plot. A PDP is a graph that represents
a set of variables/predictors and their effect on the target field (in this case price). Those graphs do not estimate actual prices.
It is important to realize that a PDP is not a representation of the dataset values or price. It is a representation of the variables effect on the target field. The negative numbers are logits of probabilities, not raw probabilities.