Posted on Categories Coding, data science, math programming, Statistics, TutorialsLeave a comment on Y-Conditionally Regularized Neural Nets

## Y-Conditionally Regularized Neural Nets

Win Vector LLC’s Dr. Nina Zumel has had great success applying y-aware methods to machine learning problems, and working out the detailed cross-validation methods needed to make y-aware procedures safe. I thought I would try our hand at y-aware neural net or deep learning methods here.

Posted on 1 Comment on When Cross-Validation is More Powerful than Regularization

## When Cross-Validation is More Powerful than Regularization

Regularization is a way of avoiding overfit by restricting the magnitude of model coefficients (or in deep learning, node weights). A simple example of regularization is the use of ridge or lasso regression to fit linear models in the presence of collinear variables or (quasi-)separation. The intuition is that smaller coefficients are less sensitive to idiosyncracies in the training data, and hence, less likely to overfit.

Cross-validation is a way to safely reuse training data in nested model situations. This includes both the case of setting hyperparameters before fitting a model, and the case of fitting models (let’s call them base learners) that are then used as variables in downstream models, as shown in Figure 1. In either situation, using the same data twice can lead to models that are overtuned to idiosyncracies in the training data, and more likely to overfit.

In general, if any stage of your modeling pipeline involves looking at the outcome (we’ll call that a y-aware stage), you cannot directly use the same data in the following stage of the pipeline. If you have enough data, you can use separate data in each stage of the modeling process (for example, one set of data to learn hyperparameters, another set of data to train the model that uses those hyperparameters). Otherwise, you should use cross-validation to reduce the nested model bias.

Cross-validation is relatively computationally expensive; regularization is relatively cheap. Can you mitigate nested model bias by using regularization techniques instead of cross-validation?

The short answer: no, you shouldn’t. But as, we’ve written before, demonstrating this is more memorable than simply saying “Don’t do that.”

## A simple example

Suppose you have a system with two categorical variables. The variable x_s has 10 levels, and the variable x_n has 100 levels. The outcome y is a function of x_s, but not of x_n (but you, the analyst building the model, don’t know this). Here’s the head of the data.

##     x_s  x_n           y
## 2  s_10 n_72  0.34228110
## 3  s_01 n_09 -0.03805102
## 4  s_03 n_18 -0.92145960
## 9  s_08 n_43  1.77069352
## 10 s_08 n_17  0.51992928
## 11 s_01 n_78  1.04714355

With most modeling techniques, a categorical variable with K levels is equivalent to K or K-1 numerical (indicator or dummy) variables, so this system actually has around 110 variables. In real life situations where a data scientist is working with high-cardinality categorical variables, or with a lot of categorical variables, the number of actual variables can begin to swamp the size of training data, and/or bog down the machine learning algorithm.

One way to deal with these issues is to represent each categorical variable by a single variable model (or base learner), and then use the predictions of those base learners as the inputs to a bigger model. So instead of fitting a model with 110 indicator variables, you can fit a model with two numerical variables. This is a simple example of nested models.

We refer to this procedure as “impact coding,” and it is one of the data treatments available in the vtreat package, specifically for dealing with high-cardinality categorical variables. But for now, let’s go back to the original problem.

## The naive way

For this simple example, you might try representing each variable as the expected value of y - mean(y) in the training data, conditioned on the variable’s level. So the ith “coefficient” of the one-variable model would be given by:

vi = E[y|x = si] − E[y]

Where si is the ith level. Let’s show this with the variable x_s (the code for all the examples in this article is here):

##     x_s      meany      coeff
## 1  s_01  0.7998263  0.8503282
## 2  s_02 -1.3815640 -1.3310621
## 3  s_03 -0.7928449 -0.7423430
## 4  s_04 -0.8245088 -0.7740069
## 5  s_05  0.7547054  0.8052073
## 6  s_06  0.1564710  0.2069728
## 7  s_07 -1.1747557 -1.1242539
## 8  s_08  1.3520153  1.4025171
## 9  s_09  1.5789785  1.6294804
## 10 s_10 -0.7313895 -0.6808876

In other words, whenever the value of x_s is s_01, the one variable model vs returns the value 0.8503282, and so on. If you do this for both variables, you get a training set that looks like this:

##     x_s  x_n           y         vs         vn
## 2  s_10 n_72  0.34228110 -0.6808876 0.64754957
## 3  s_01 n_09 -0.03805102  0.8503282 0.54991135
## 4  s_03 n_18 -0.92145960 -0.7423430 0.01923877
## 9  s_08 n_43  1.77069352  1.4025171 1.90394159
## 10 s_08 n_17  0.51992928  1.4025171 0.26448341
## 11 s_01 n_78  1.04714355  0.8503282 0.70342961

Now fit a linear model for y as a function of vs and vn.

model_raw = lm(y ~ vs + vn,
data=dtrain_treated)
summary(model_raw)
##
## Call:
## lm(formula = y ~ vs + vn, data = dtrain_treated)
##
## Residuals:
##      Min       1Q   Median       3Q      Max
## -2.33068 -0.57106  0.00342  0.52488  2.25472
##
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)
## (Intercept) -0.05050    0.05597  -0.902    0.368
## vs           0.77259    0.05940  13.006   <2e-16 ***
## vn           0.61201    0.06906   8.862   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.8761 on 242 degrees of freedom
## Multiple R-squared:  0.6382, Adjusted R-squared:  0.6352
## F-statistic: 213.5 on 2 and 242 DF,  p-value: < 2.2e-16

Note that this model gives significant coefficients to both vs and vn, even though y is not a function of x_n (or vn). Because you used the same data to fit the one variable base learners and to fit the larger model, you have overfit.

## The right way: cross-validation

The correct way to impact code (or to nest models in general) is to use cross-validation techniques. Impact coding with cross-validation is already implemented in vtreat; note the similarity between this diagram and Figure 1 above.

The training data is used both to fit the base learners (as we did above) and to also to create a data frame of cross-validated base learner predictions (called a cross-frame in vtreat). This cross-frame is used to train the overall model. Let’s fit the correct nested model, using vtreat.

library(vtreat)
library(wrapr)
xframeResults = mkCrossFrameNExperiment(dtrain,
qc(x_s, x_n), "y",
codeRestriction = qc(catN),
verbose = FALSE)
# the plan uses the one-variable models to treat data
treatmentPlan = xframeResults$treatments # the cross-frame dtrain_treated = xframeResults$crossFrame

head(dtrain_treated)
##     x_s_catN   x_n_catN           y
## 1 -0.6337889 0.91241547  0.34228110
## 2  0.8342227 0.82874089 -0.03805102
## 3 -0.7020597 0.18198634 -0.92145960
## 4  1.3983175 1.99197404  1.77069352
## 5  1.3983175 0.11679580  0.51992928
## 6  0.8342227 0.06421659  1.04714355
variables = setdiff(colnames(dtrain_treated), "y")

model_X = lm(mk_formula("y", variables),
data=dtrain_treated)
summary(model_X)
##
## Call:
## lm(formula = mk_formula("y", variables), data = dtrain_treated)
##
## Residuals:
##     Min      1Q  Median      3Q     Max
## -3.2157 -0.7343  0.0225  0.7483  2.9639
##
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)
## (Intercept) -0.04169    0.06745  -0.618    0.537
## x_s_catN     0.92968    0.06344  14.656   <2e-16 ***
## x_n_catN     0.10204    0.06654   1.533    0.126
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.055 on 242 degrees of freedom
## Multiple R-squared:  0.4753, Adjusted R-squared:  0.471
## F-statistic: 109.6 on 2 and 242 DF,  p-value: < 2.2e-16

This model correctly determines that x_n (and its one-variable model x_n_catN) do not affect the outcome. We can compare the performance of this model to the naive model on holdout data.

rmse rsquared
ypred_naive 1.303778 0.2311538
ypred_crossval 1.093955 0.4587089

The correct model has a much smaller root-mean-squared error and a much larger R-squared than the naive model when applied to new data.

## An attempted alternative: regularized models.

But cross-validation is so complicated. Can’t we just regularize? As we’ll show in the appendix of this article, for a one-variable model, L2-regularization is simply Laplace smoothing. Again, we’ll represent each “coefficient” of the one-variable model as the Laplace smoothed value minus the grand mean.

vi = ∑xj = si yi/(counti + λ) − E[yi]

Where counti is the frequency of si in the training data, and λ is the smoothing parameter (usually 1). If λ = 1 then the first term on the right is just adding one to the frequency of the level and then taking the “adjusted conditional mean” of y.

Again, let’s show this for the variable x_s.

##     x_s      sum_y count_y   grandmean         vs
## 1  s_01  20.795484      26 -0.05050187  0.8207050
## 2  s_02 -37.302227      27 -0.05050187 -1.2817205
## 3  s_03 -22.199656      28 -0.05050187 -0.7150035
## 4  s_04 -14.016649      17 -0.05050187 -0.7282009
## 5  s_05  19.622340      26 -0.05050187  0.7772552
## 6  s_06   3.129419      20 -0.05050187  0.1995218
## 7  s_07 -35.242672      30 -0.05050187 -1.0863585
## 8  s_08  36.504412      27 -0.05050187  1.3542309
## 9  s_09  33.158549      21 -0.05050187  1.5577086
## 10 s_10 -16.821957      23 -0.05050187 -0.6504130

After applying the one variable models for x_s and x_n to the data, the head of the resulting treated data looks like this:

##     x_s  x_n           y         vs         vn
## 2  s_10 n_72  0.34228110 -0.6504130 0.44853367
## 3  s_01 n_09 -0.03805102  0.8207050 0.42505898
## 4  s_03 n_18 -0.92145960 -0.7150035 0.02370493
## 9  s_08 n_43  1.77069352  1.3542309 1.28612835
## 10 s_08 n_17  0.51992928  1.3542309 0.21098803
## 11 s_01 n_78  1.04714355  0.8207050 0.61015422

Now fit the overall model:

##
## Call:
## lm(formula = y ~ vs + vn, data = dtrain_treated)
##
## Residuals:
##      Min       1Q   Median       3Q      Max
## -2.30354 -0.57688 -0.02224  0.56799  2.25723
##
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)
## (Intercept) -0.06665    0.05637  -1.182    0.238
## vs           0.81142    0.06203  13.082  < 2e-16 ***
## vn           0.85393    0.09905   8.621  8.8e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.8819 on 242 degrees of freedom
## Multiple R-squared:  0.6334, Adjusted R-squared:  0.6304
## F-statistic: 209.1 on 2 and 242 DF,  p-value: < 2.2e-16

Again, both variables look significant. Even with regularization, the model is still overfit. Comparing the performance of the models on holdout data, you see that the regularized model does a little better than the naive model, but not as well as the correctly cross-validated model.

rmse rsquared
ypred_naive 1.303778 0.2311538
ypred_crossval 1.093955 0.4587089
ypred_reg 1.267648 0.2731756

## The Moral of the Story

Unfortunately, regularization is not enough to overcome nested model bias. Whenever you apply a y-aware process to your data, you have to use cross-validation methods (or a separate data set) at the next stage of your modeling pipeline.

### Appendix: Derivation of Laplace Smoothing as L2-Regularization

Without regularization, the optimal one-variable model for y in terms of a categorical variable with K levels {sj} is a set of K coefficients v such that

$f(\mathbf{v}) := \sum\limits_{i=1}^N (y_i - v_i)^2$

is minimized (N is the number of data points). L2-regularization adds a penalty to the magnitude of v, so that the goal is to minimize

$f(\mathbf{v}) := \sum\limits_{i=1}^N (y_i - v_i)^2 + \lambda \sum\limits_{j=1}^K {v_j}^2$

where λ is a known smoothing hyperparameter, usually set (in this case) to 1.

To minimize the above expression for a single coefficient vj, take the deriviative with respect to vj and set it to zero:

$\sum\nolimits_{x_i = s_j} -2 (y_i - v_j) + 2 \lambda v_j = 0\\ \sum\nolimits_{x_i = s_j }-y_i + \sum\nolimits_{x_i = s_j} v_j + \lambda v_j = 0\\ \sum\nolimits_{x_i = s_j }-y_i + \text{count}_j v_j + \lambda v_j = 0$

Where countj is the number of times the level sj appears in the training data. Now solve for vj:

$v_j (\text{count}_j + \lambda) = \sum\nolimits_{x_i = s_j} y_i\\ v_j = \sum\nolimits_{x_i = s_i} y_i / (\text{count}_j + \lambda)$

This is Laplace smoothing. Note that it is also the one-variable equivalent of ridge regression.

Posted on 4 Comments on Unprincipled Component Analysis

## Unprincipled Component Analysis

As a data scientist I have seen variations of principal component analysis and factor analysis so often blindly misapplied and abused that I have come to think of the technique as unprincipled component analysis. PCA is a good technique often used to reduce sensitivity to overfitting. But this stated design intent leads many to (falsely) believe that any claimed use of PCA prevents overfit (which is not always the case). In this note we comment on the intent of PCA like techniques, common abuses and other options.

The idea is to illustrate what can quietly go wrong in an analysis and what tests to perform to make sure you see the issue. The main point is some analysis issues can not be fixed without going out and getting more domain knowledge, more variables or more data. You can’t always be sure that you have insufficient data in your analysis (there is always a worry that some clever technique will make the current data work), but it must be something you are prepared to consider. Continue reading Unprincipled Component Analysis

Posted on Categories Statistics3 Comments on A pathological glm() problem that doesn’t issue a warning

## A pathological glm() problem that doesn’t issue a warning

I know I have already written a lot about technicalities in logistic regression (see for example: How robust is logistic regression? and Newton-Raphson can compute an average). But I just ran into a simple case where R‘s glm() implementation of logistic regression seems to fail without issuing a warning message. Yes the data is a bit pathological, but one would hope for a diagnostic or warning message from the fitter. Continue reading A pathological glm() problem that doesn’t issue a warning