Edit: Removing TransformedTargetRegressor and adding more info as requested.

Edit2: There were 18K rows where the relation did not hold. I'm sorry :(. After removing those rows and upon @Ben Reiniger's advice, I used LinearRegression and the metrics looked more saner. The new metrics are pasted below.

Original Question:

Given totalRevenue and costOfRevenue, I'm trying to predict grossProfit. Given that it's a simple formula totalRevenue - costOfRevenue = grossProfit, I was expecting that the following code would work. Is it a matter of hyperparameter optimization or have I missed some data cleaning. I have tried all the scalers and other regressions in sklearn but I don't see any big difference.

# X(107002 rows × 2 columns)
| totalRevenue | costOfRevenue |
| 2.256510e+05 | 2.333100e+04  |
| 1.183960e+05 | 2.857000e+04  |
| 2.500000e+05 | 1.693000e+05  |
| 1.750000e+05 | 8.307500e+04  |
| 3.905000e+09 | 1.240000e+09  |

# y
| 2.023200e+05 |
| 8.982600e+04 |
| 8.070000e+04 |
| 9.192500e+04 |
| 2.665000e+09 |
Name: grossProfit, Length: 107002, dtype: float64

# Training

import numpy as np
import sklearn

from sklearn.compose import TransformedTargetRegressor
from sklearn.preprocessing import StandardScaler
from sklearn.linear_model import Ridge
from sklearn.model_selection import train_test_split
from sklearn.pipeline import Pipeline

X_train, X_test, y_train, y_test = train_test_split(X, y, random_state=13)

x_scaler = StandardScaler()

pipe_l = Pipeline([
        ('scaler', x_scaler),
        ('regressor', Ridge())

regr = pipe_l

regr.fit(X_train, y_train)

y_pred = regr.predict(X_test)

print('R2 score: {0:.2f}'.format(sklearn.metrics.r2_score(y_test, y_pred)))
print('Mean Absolute Error:', sklearn.metrics.mean_absolute_error(y_test, y_pred))  
print('Mean Squared Error:', sklearn.metrics.mean_squared_error(y_test, y_pred))  
print('Root Mean Squared Error:', np.sqrt(sklearn.metrics.mean_squared_error(y_test, y_pred)))

print("Scaler Mean:",x_scaler.mean_)
print("Scaler Var:", x_scaler.var_)
print("Estimator Coefficient:",regr.steps[1][1].coef_)

Output of above metrics after training(Old Metrics with 18k rows which did not confirm to the relation)

R2 score: 0.69
Mean Absolute Error: 37216342513.01034
Mean Squared Error: 7.601569571667974e+23
Root Mean Squared Error: 871869805169.7842
Scaler Mean: [1.26326695e+13 2.14785735e+14]
Scaler Var: [1.24609190e+31 2.04306993e+32]
Estimator Coefficient: [1.16354874e+15 2.59046205e+09]

Ridge(After removing the 18k bad rows)

R2 score: 1.00
Mean Absolute Error: 15659273.260432156
Mean Squared Error: 8.539990125466045e+16
Root Mean Squared Error: 292232614.97420245
Scaler Mean: [1.57566809e+11 9.62274405e+10]
Scaler Var: [1.20924187e+25 5.95764210e+24]
Estimator Coefficient: [ 3.47663586e+12 -2.44005648e+12]

LinearRegression(After removing the 18K rows)

R2 score: 1.00
Mean Absolute Error: 0.00017393178061611583
Mean Squared Error: 4.68109129068828e-06
Root Mean Squared Error: 0.0021635829752261132
Scaler Mean: [1.57566809e+11 9.62274405e+10]
Scaler Var: [1.20924187e+25 5.95764210e+24]
Estimator Coefficient: [ 3.47741552e+12 -2.44082816e+12]
  • $\begingroup$ Could you provide the ridge regression's coefficients, and the scalers' mean/var? $\endgroup$ – Ben Reiniger Mar 17 '20 at 21:22
  • $\begingroup$ updated. I simplified the code. Thanks! $\endgroup$ – RAbraham Mar 17 '20 at 21:53
  • 1
    $\begingroup$ The scores you report are based on the unscaled labels and predictions, so the MSE and MAE make sense and are not relatively large. But the R2 score is worryingly low; it may be only due to the regularization penalty: try LinearRegression? $\endgroup$ – Ben Reiniger Mar 17 '20 at 22:13
  • $\begingroup$ It had 18k bad rows :(. Please see edit2. And LinearRegression worked well. I have posted the metrics above. Thanks so much! $\endgroup$ – RAbraham Mar 17 '20 at 23:20
  • $\begingroup$ @BenReiniger. would you care to put your above comment as the answer so that I can mark it as the answer, it helped me significantly. $\endgroup$ – RAbraham Mar 18 '20 at 11:58

(To summarize the comment thread into an answer)

Your original scores:

Mean Absolute Error: 37216342513.01034
Root Mean Squared Error: 871869805169.7842

are based on the original-scale target variable and are between $10^{10}$ and $10^{12}$, at least significantly smaller than the mean of the features (and the target)? So these aren't automatically bad scores, although for a perfect relationship we should hope for better. Furthermore, a 0.69 R2 values is pretty low, no scale-consciousness needed.

That both of the model's coefficients came out positive is the most worrisome point. I'm glad you identified the culprit rows; I don't know how I would have diagnosed that from here.

Your new ridge regression still has "large" errors, but significantly smaller than before, and quite small compared to the feature/target scale. And now the coefficients have different signs. (I think if you'd left the TransformedTargetRegressor in, you'd get largely the same results, but with less penalization.)

Finally, when such an exact relationship is the truth, it makes sense not to penalize the regression. Your coefficients here are a little bit larger, and the errors drop away to nearly nothing, especially considering the scale of the target.


It seems you are using a the standard scaler twice, once in your pipeline and once more in the TransformedTargetRegressor. Next to that, you are only fitting the scaler, never actually scaling the inputs (i.e. transforming the input).

  • 1
    $\begingroup$ Thanks! re: using the scaler twice. the first one is for X and the second one in TransformedTargetRegressor is for y. Even if I take it out, I get very similar results. Also, my understanding of pipelines is that fit will call fit_transform for all but the last. ref: iaml.it/blog/optimizing-sklearn-pipelines/images/… and scikit-learn.org/stable/modules/compose.html $\endgroup$ – RAbraham Mar 17 '20 at 20:00
  • $\begingroup$ I tried doing it manually but the same magnitude of error. python scaler = StandardScaler() X_train_transform = scaler.fit_transform(X_train) X_test_transform = scaler.transform(X_test) regr = LinearRegression() regr.fit(X_train_transform, y_train) y_pred = regr.predict(X_test_transform) $\endgroup$ – RAbraham Mar 17 '20 at 20:03
  • $\begingroup$ Looking at the image you linked it indeed would make sense that the fit method will fit_transform in all steps except the last one. However, when I quickly checked using your data it seemed that X_train was not properly scaled before the regr.fit was called. Can you check this on your side? $\endgroup$ – Oxbowerce Mar 17 '20 at 20:06
  • 1
    $\begingroup$ thank you for checking :). Do you mean that train_test_split should have scaled X_train for me? if I understand correctly X_train would be scaled after it is passed to regr.fit as the scaling would happen inside? $\endgroup$ – RAbraham Mar 17 '20 at 20:24
  • $\begingroup$ No, I meant that X_train should be getting scaled after being passes to regr.fit, however checking this on my end this did not seem the case. $\endgroup$ – Oxbowerce Mar 18 '20 at 7:39

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