Re-posted from: https://bkamins.github.io/julialang/2021/10/15/cv.html

Introduction

Today I decided to show how cross validation can be implemented using the
functions provided by DataFrames.jl. The objective is to discuss common
patterns used when working with data frames.

In this the post I use Julia 1.6.3, DataFrames.jl 1.2.2, and GLM.jl 1.5.1.

Preparing the data and initial analysis

Let us start by creating some synthetic data set we will work with.

julia> using DataFrames

julia> using GLM

julia> using Random

julia> using Statistics

julia> Random.seed!(1234);

julia> df = DataFrame(randn(50, 9), :auto);

julia> df.y = sum(eachcol(df)) .+ 1.0 .+ randn(50);

julia> df
50×10 DataFrame
 Row │ x1         x2          x3         x4          x5          x6          x7         x8         x9          y
     │ Float64    Float64     Float64    Float64     Float64     Float64     Float64    Float64    Float64     Float64
─────┼────────────────────────────────────────────────────────────────────────────────────────────────────────────────────
   1 │  0.867347  -1.22672     0.183976  -1.87215    -0.205782    0.295222    0.183203   0.358659  -0.833507   -2.20367
   2 │ -0.901744  -0.541716   -1.27635   -0.668331   -1.22338    -0.338215    0.975083   0.488578   0.0827196  -3.31337
  ⋮  │     ⋮          ⋮           ⋮          ⋮           ⋮           ⋮           ⋮          ⋮          ⋮           ⋮
  49 │ -1.00978   -1.66323     0.797165  -0.644069    0.127747    0.742054    0.196089  -1.05575    0.771387   -2.364
  50 │ -0.543805  -0.521229    0.103145  -1.37931     0.143105   -0.0665944  -2.34887   -1.37548    0.590872   -4.76852
                                                                                                           46 rows omitted

As you can see we have created data so that the target variable :y is a sum
of features :x1 to :x9 and an intercept equal to 1. The error term in the
model has a normal distribution with mean 0 and variance 1. Similarly all
features also have a normal distribution with mean 0 and variance 1.

Now we create a linear model on the whole data set to have a baseline:

julia> formula = Term(:y) ~ sum([Term(Symbol(:x, i)) for i in 1:9])
FormulaTerm
Response:
  y(unknown)
Predictors:
  x1(unknown)
  x2(unknown)
  x3(unknown)
  x4(unknown)
  x5(unknown)
  x6(unknown)
  x7(unknown)
  x8(unknown)
  x9(unknown)

julia> model = lm(formula, df)
StatsModels.TableRegressionModel{LinearModel{GLM.LmResp{Vector{Float64}}, GLM.DensePredChol{Float64, LinearAlgebra.CholeskyPivoted{Float64, Matrix{Float64}}}}, Matrix{Float64}}

y ~ 1 + x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9

Coefficients:
────────────────────────────────────────────────────────────────────────
                Coef.  Std. Error      t  Pr(>|t|)  Lower 95%  Upper 95%
────────────────────────────────────────────────────────────────────────
(Intercept)  1.01641     0.140685   7.22    <1e-08   0.732078    1.30075
x1           0.879802    0.141336   6.22    <1e-06   0.594151    1.16545
x2           1.14157     0.160897   7.10    <1e-07   0.81639     1.46676
x3           0.820527    0.137644   5.96    <1e-06   0.542338    1.09872
x4           1.28829     0.125645  10.25    <1e-12   1.03435     1.54223
x5           1.02118     0.156195   6.54    <1e-07   0.705503    1.33687
x6           0.844008    0.146142   5.78    <1e-06   0.548643    1.13937
x7           0.70256     0.15646    4.49    <1e-04   0.386343    1.01878
x8           0.925448    0.15797    5.86    <1e-06   0.60618     1.24472
x9           0.889749    0.167878   5.30    <1e-05   0.550456    1.22904
────────────────────────────────────────────────────────────────────────

We calculate mean square error of the model on the training data set:

julia> deviance(model) / nrow(df)
0.6689313350544223

julia> mse(model, df) = sum(x->x^2, predict(model, df) - df.y) / nrow(df);

julia> mse(model, df)
0.6689313350544223

It seems (and indeed is) a bit low given we know that the standard deviation of
the error term in our data generation process was 1. We will come back to
it in a second. However, first let us perfom 10-fold cross validation.

Doing cross validation

We start with generating a column holding the fold number for each row.
In the example we assume we do a 10-fold cross validation and will number folds
by integers starting with 0 and ending with 9. We need to make sure that
folds should be assigned to rows in a random order so we need to shuffle!
the initially generated fold numbers:

julia> df.fold = shuffle!((1:nrow(df)) .% 10)
50-element Vector{Int64}:
 4
 6
 5
 7
 ⋮
 8
 6
 7

Now we can quite easily extract out training and test data sets for each
individual fold using the following function:

get_fold_data(df, fold) =
    (train=view(df, df.fold .!= fold, :),
     test=view(df, df.fold .== fold, :))

Note that in this code we use views to avoid excessive copying of data.

Now we are ready to calculate the cross validation mean squared error (in the
code we take advantage of the fact that each fold has exactly the same size
in our example):

julia> mean(0:9) do fold
           train, test = get_fold_data(df, fold)
           model_cv = lm(formula, train)
           return mse(model_cv, test)
       end
0.9500293257502437

Indeed it is higher than the MSE we have calculated above as expected.

Let us calculate the expected square error for the new observation fed to
our model (I am using here an analytic formula that is suitable for our specific
way of input data generation):

julia> mset(model) = 1 + sum(x -> (1 - x) ^ 2, coef(model));

julia> mset(model)
1.2810475573651952

We can see it is greater than 1 which manes sense as this error is a sum of
two errors: random term in our data generation process plus the estimation error
of parameters of our model.

If you are not convinced that the proper formula is used to calculate this
expectation then it is easy to check it using simulation:

julia> dfn = DataFrame(randn(10^6, 9), :auto);

julia> dfn.y = sum(eachcol(dfn)) .+ 1.0 .+ randn(10^6);

julia> mse(model, dfn)
1.2817533442813103

And we get a roughly similar result.

It is crucial to note that we can calculate mse_t here accurately because we
know the data generation process. In practice we would not know it but would
want get some measure that would as accurately as possible order the competing
models in terms of their true, unobserved mse_t.

Now you might wonder if the cross validation mean squared error should be
expected to be lower than the actual prediction expected squared error as in
our single run and if it is a better predictor of mse_t than mean squared
error calculated on a training data set. We check this in the next section.

Testing the procedure using simulation

Let us wrap our calculations in the function. We will collect three mean
squared errors:

mse_whole: calculated on training data set;
mse_cv: calculated using cross validation;
mse_t: expected prediction squared error.

Here is the function definition:

function runtest()
    df = DataFrame(randn(50, 9), :auto)
    df.y = sum.(eachrow(df)) .+ 1.0 .+ randn(50)
    model = lm(formula, df)
    mse_whole = mse(model, df)

    mse_t = mset(model)

    df.fold = shuffle!((1:nrow(df)) .% 10)
    mse_cv = mean(0:9) do fold
        train, test = get_fold_data(df, fold)
        model_cv = lm(formula, train)
        return mse(model_cv, test)
    end

    return (mse_whole=mse_whole, mse_cv=mse_cv, mse_t=mse_t)
end

Let us now run 10,000 independent repetitions of our experiment and analyze
the results:

julia> Random.seed!(12);

julia> res = DataFrame([runtest() for _ in 1:10_000])
10000×3 DataFrame
   Row │ mse_whole  mse_cv    mse_t
       │ Float64    Float64   Float64
───────┼──────────────────────────────
     1 │  0.767821  1.20796   1.43191
     2 │  0.595431  1.11465   1.2233
     3 │  0.801025  1.41942   1.25682
   ⋮   │     ⋮         ⋮         ⋮
  9998 │  0.589998  0.930069  1.24527
  9999 │  0.444815  0.685732  1.36184
 10000 │  1.05747   1.68496   1.23118
                     9994 rows omitted

julia> describe(res, :all)
3×13 DataFrame
 Row │ variable   mean      std       min       q25       median    q75       max      nunique  nmissing  first     last     eltype
     │ Symbol     Float64   Float64   Float64   Float64   Float64   Float64   Float64  Nothing  Int64     Float64   Float64  DataType
─────┼────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────
   1 │ mse_whole  0.799776  0.177712  0.286417  0.672772  0.785994  0.912926  1.71692                  0  0.767821  1.05747  Float64
   2 │ mse_cv     1.29295   0.305371  0.419517  1.07337   1.26767   1.48104   3.0517                   0  1.20796   1.68496  Float64
   3 │ mse_t      1.25542   0.132023  1.01737   1.16157   1.22982   1.32126   2.07322                  0  1.43191   1.23118  Float64

julia> cor(Matrix(res))
3×3 Matrix{Float64}:
  1.0          0.947       -0.00555373
  0.947        1.0         -0.00844714
 -0.00555373  -0.00844714   1.0

What have we learned from our specific experiment setup:

mse_whole is significantly biased downwards;
mse_cv is slightly biased upwards on the average when multiple experiments
are considered, but its variance is much higher than the one of mse_t;
mse_cv and mse_whole are significantly correlated and they both do not
exhibit correlation with mse_t.

Conclusions

I hope the examples I gave of using DataFrames.jl in the context of some typical
data science workflow will be useful. A next step – that I leave as an
exercise – would be to e.g. check how model selection using cross validation
works.

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DataFrames.jl training: implementing cross validation

Introduction

Preparing the data and initial analysis

Doing cross validation

Testing the procedure using simulation

Conclusions

Related