Calculate the coefficient of determination using correlation. For the traditional measure of R squared, see rsq_trad().

## Usage

rsq(data, ...)

# S3 method for data.frame
rsq(data, truth, estimate, na_rm = TRUE, case_weights = NULL, ...)

rsq_vec(truth, estimate, na_rm = TRUE, case_weights = NULL, ...)

## Arguments

data

A data.frame containing the columns specified by the truth and estimate arguments.

...

Not currently used.

truth

The column identifier for the true results (that is numeric). This should be an unquoted column name although this argument is passed by expression and supports quasiquotation (you can unquote column names). For _vec() functions, a numeric vector.

estimate

The column identifier for the predicted results (that is also numeric). As with truth this can be specified different ways but the primary method is to use an unquoted variable name. For _vec() functions, a numeric vector.

na_rm

A logical value indicating whether NA values should be stripped before the computation proceeds.

case_weights

The optional column identifier for case weights. This should be an unquoted column name that evaluates to a numeric column in data. For _vec() functions, a numeric vector, hardhat::importance_weights(), or hardhat::frequency_weights().

## Value

A tibble with columns .metric, .estimator, and .estimate and 1 row of values.

For grouped data frames, the number of rows returned will be the same as the number of groups.

For rsq_vec(), a single numeric value (or NA).

## Details

The two estimates for the coefficient of determination, rsq() and rsq_trad(), differ by their formula. The former guarantees a value on (0, 1) while the latter can generate inaccurate values when the model is non-informative (see the examples). Both are measures of consistency/correlation and not of accuracy.

rsq() is simply the squared correlation between truth and estimate.

Because rsq() internally computes a correlation, if either truth or estimate are constant it can result in a divide by zero error. In these cases, a warning is thrown and NA is returned. This can occur when a model predicts a single value for all samples. For example, a regularized model that eliminates all predictors except for the intercept would do this. Another example would be a CART model that contains no splits.

## References

Kvalseth. Cautionary note about $$R^2$$. American Statistician (1985) vol. 39 (4) pp. 279-285.

Other numeric metrics: ccc(), huber_loss_pseudo(), huber_loss(), iic(), mae(), mape(), mase(), mpe(), msd(), poisson_log_loss(), rmse(), rpd(), rpiq(), rsq_trad(), smape()

Other consistency metrics: ccc(), rpd(), rpiq(), rsq_trad()

Max Kuhn

## Examples

# Supply truth and predictions as bare column names
rsq(solubility_test, solubility, prediction)
#> # A tibble: 1 × 3
#>   .metric .estimator .estimate
#>   <chr>   <chr>          <dbl>
#> 1 rsq     standard       0.879

set.seed(1234)
size <- 100
times <- 10

# create 10 resamples
solubility_resampled <- bind_rows(

n = times,
expr = sample_n(solubility_test, size, replace = TRUE),
simplify = FALSE
),
.id = "resample"
)

# Compute the metric by group
metric_results <- solubility_resampled %>%
group_by(resample) %>%
rsq(solubility, prediction)

metric_results
#> # A tibble: 10 × 4
#>    resample .metric .estimator .estimate
#>    <chr>    <chr>   <chr>          <dbl>
#>  1 1        rsq     standard       0.874
#>  2 10       rsq     standard       0.879
#>  3 2        rsq     standard       0.891
#>  4 3        rsq     standard       0.916
#>  5 4        rsq     standard       0.892
#>  6 5        rsq     standard       0.858
#>  7 6        rsq     standard       0.873
#>  8 7        rsq     standard       0.852
#>  9 8        rsq     standard       0.915
#> 10 9        rsq     standard       0.884

# Resampled mean estimate
metric_results %>%
summarise(avg_estimate = mean(.estimate))
#> # A tibble: 1 × 1
#>   avg_estimate
#>          <dbl>
#> 1        0.883
# With uninformitive data, the traditional version of R^2 can return
# negative values.
set.seed(2291)
solubility_test$randomized <- sample(solubility_test$prediction)
rsq(solubility_test, solubility, randomized)
#> # A tibble: 1 × 3
#>   .metric .estimator .estimate
#>   <chr>   <chr>          <dbl>
#> 1 rsq     standard     0.00199
#> # A tibble: 1 × 3
#>   .metric  .estimator .estimate
#>   <chr>    <chr>          <dbl>
# A constant truth or estimate vector results in a warning from
# NA will be returned in these cases.
#> Warning: A correlation computation is required, but estimate is constant and has
#> 0 standard deviation, resulting in a divide by 0 error. NA will be