I'm building regression models. As a preprocessing step, I scale my feature values to have mean 0 and standard deviation 1. Is it necessary to normalize the target values also?

  • $\begingroup$ Scaling inputs help to save much more computation time. $\endgroup$
    – NickMa
    Commented Feb 2, 2021 at 17:18

8 Answers 8


Let's first analyse why feature scaling is performed. Feature scaling improves the convergence of steepest descent algorithms, which do not possess the property of scale invariance.

In stochastic gradient descent training examples inform the weight updates iteratively like so, $$w_{t+1} = w_t - \gamma\nabla_w \ell(f_w(x),y)$$

Where $w$ are the weights, $\gamma$ is a stepsize, $\nabla_w$ is the gradient wrt weights, $\ell$ is a loss function, $f_w$ is the function parameterized by $w$, $x$ is a training example, and $y$ is the response/label.

Compare the following convex functions, representing proper scaling and improper scaling.

Feature scaling

A step through one weight update of size $\gamma$ will yield much better reduction in the error in the properly scaled case than the improperly scaled case. Shown below is the direction of $\nabla_w \ell(f_w(x),y)$ of length $\gamma$.

Gradient update

Normalizing the output will not affect shape of $f$, so it's generally not necessary.

The only situation I can imagine scaling the outputs has an impact, is if your response variable is very large and/or you're using f32 variables (which is common with GPU linear algebra). In this case it is possible to get a floating point overflow of an element of the weights. The symptom is either an Inf value or it will wrap-around to the other extreme representation.

  • $\begingroup$ But if we do not scale the inputs and apply Gradient Descent, to solve for theta in something like y = theta0 + theta1 * x1 + theta2 * x2, if we are updating the values of X1 and X2 (by scaling them) while keeping Y (expected output) the same, won't the resulting predictions for theta1, theta2 be wrong when we apply them to the original equation? $\endgroup$
    – Prashant
    Commented Apr 24, 2018 at 8:17
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    $\begingroup$ my intuition tells me that if the target value has large values then choosing a step size should be trickier compared to target values in a range of say, -1 to +1. I can see a benefit in choosing hyperparameters for GD. $\endgroup$ Commented Jul 24, 2020 at 18:10
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    $\begingroup$ @Prashant Yes. That is why you also scale the future inputs to the model after training using the same parameters(mu, sigma) used to scale the training input. $\endgroup$ Commented Oct 27, 2021 at 12:07

Yes, you do need to scale the target variable. I will quote this reference:

A target variable with a large spread of values, in turn, may result in large error gradient values causing weight values to change dramatically, making the learning process unstable.

In the reference, there's also a demonstration on code where the model weights exploded during training given the very large errors and, in turn, error gradients calculated for weight updates also exploded. In short, if you don't scale the data and you have very large values, make sure to use very small learning rate values. This was mentioned by @drSpacy as well.

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    $\begingroup$ This answer is underrated. It's actually easy to come up with an example when gradient will explode due to unscaled target value. Take y(x) = x^2 and try to fit it in range [-20, 20] and you'll see overflow happenning at first couple of iterations due to big gap between y_hat and y. $\endgroup$
    – ptyshevs
    Commented Dec 3, 2019 at 15:31
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    $\begingroup$ @PavelTyshevskyi the reference, although valid, is more relevant for neural networks (or gradient descent). In a generalised model fitting situation, scaling the target should not add any benefit. $\endgroup$
    – baibo
    Commented Feb 1, 2020 at 21:41
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    $\begingroup$ @baibo great observation! But usually, regression models also use some optimization algorithm based on gradient descent. $\endgroup$ Commented Jun 23, 2021 at 15:03
  • $\begingroup$ Exploding gradient problem! Scaling might help here $\endgroup$ Commented Jun 10, 2023 at 7:24
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    $\begingroup$ @ptyshevs If I get it right, lets take $y = wx^2$. We want to estimate $w$ (which in your example is 1) and the loss is $\frac{1}{2} (wx^2 - y)^2$. The update will be: $w' = w - \eta (wx^2 - y) \cdot x^2$. If we have $(x, y) = (5, 25)$ and we start with $w=10$ then the update will be $w' = w - \eta 225 \cdot 25$ and we may overshoot if we don't carefully select a small learning rate. The problem becomes worse if the given $(x, y)$ are even bigger. $\endgroup$
    – ado sar
    Commented Jul 1, 2023 at 10:01

Generally, It is not necessary. Scaling inputs helps to avoid the situation, when one or several features dominate others in magnitude, as a result, the model hardly picks up the contribution of the smaller scale variables, even if they are strong. But if you scale the target, your mean squared error (MSE) is automatically scaled. Additionally, you need to look at the mean absolute scaled error (MASE). MASE>1 automatically means that you are doing worse than a constant (naive) prediction.

  • $\begingroup$ Good point about the effect on MSE, when the target variable is scaled. Error metrics like MSE and MAE are easier to interpret when the target variable is also scaled. $\endgroup$
    – Hari
    Commented Nov 12, 2020 at 16:35

No, linear transformations of the response are never necessary. They may, however, be helpful to aid in interpretation of your model. For example, if your response is given in meters but is typically very small, it may be helpful to rescale to i.e. millimeters. Note also that centering and/or scaling the inputs can be useful for the same reason. For instance, you can roughly interpret a coefficient as the effect on the response per unit change in the predictor when all other predictors are set to 0. But 0 often won't be a valid or interesting value for those variables. Centering the inputs lets you interpret the coefficient as the effect per unit change when the other predictors assume their average values.

Other transformations (i.e. log or square root) may be helpful if the response is not linear in the predictors on the original scale. If this is the case, you can read about generalized linear models to see if they're suitable for you.


It does affect gradient descent in a bad way. check the formula for gradient descent:

$$ x_{n+1} = x_{n} - \gamma\Delta F(x_n) $$

lets say that $x_2$ is a feature that is 1000 times greater than $x_1$

for $ F(\vec{x})=\vec{x}^2 $ we have $ \Delta F(\vec{x})=2*\vec{x} $. The optimal way to reach (0,0) which is the global optimum is to move across the diagonal but if one of the features dominates the other in terms of scale that wont happen.

To illustrate: If you do the transformation $\vec{z}= (x_1,1000*x_1)$, assume a uniform learning rate $ \gamma $ for both coordinates and calculate the gradient then $$ \vec{z_{n+1}} = \vec{z_{n}} - \gamma\Delta F(z_1,z_2) .$$ The functional form is the same but the learning rate for the second coordinate has to be adjusted to 1/1000 of that for the first coordinate to match it. If not coordinate two will dominate and the $\Delta$ vector will point more towards that direction.

As a result it biases the delta to point across that direction only and makes the converge slower.

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    $\begingroup$ "lets say that x2 is a feature that is 1000 times greater than x1". Question is about scaling output, not scaling inputs. $\endgroup$ Commented Mar 20, 2020 at 17:04

It may be useful for some cases.

Even though not being a common error function, when L1 error used to calculate loss, a rather slow learning may occur.

Assume that we have a linear regression model, and also have a constant learning rate $n$. Say,

$ y = b_1x + b_0 $

$ n = 0.1 $

$b_1$ and $b_0$ are updated as follows:

$b_1{new} = b_1{old} - n* \frac{\hat{y}-y} {|\hat{y}-y|} *x$

$b_0{new} = b_0{old} - n* \frac{\hat{y}-y} {|\hat{y}-y|} $

$\frac{\hat{y}-y} {|\hat{y}-y|}$ evaluates to -1 or 1. Hence, $b_0$ will be incremented/decremented by $n$, and $b_1$ will be incremented/decremented by $n*x$.

Now, if the output value is in millions or billions, obviosuly $b_0$ will require so much iteration to approach the cost to zero.

If the input is normalized (or standardized), $b_1$ will also be changed by similar and close values to $b_0$ (e.g. 0.1), and it will require too much iteration too.

Actually this is why a factor of the actual loss is desired in the derivative of the cost at a certain point (such as $\hat{y}-y$).

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    $\begingroup$ Thanks for the explanation. I experienced a similar behavior in a multivariate regression problem - (using keras.callbacks.EarlyStopping) - rescaling the model took 48 epochs to achieve the same results without scaling, which took 475 epochs. $\endgroup$
    – epifanio
    Commented Mar 31, 2021 at 22:00

I think the best way to know whether we should scale the output is to try both way, using scaler.inverse_transform in sklearn. Neural network is not robust to transformation, in general. Therefore, if you scale the output variables, train,then the MSE produced is for the scaled version. However, if you use that model to predict and use scaler.inverse_transform, and recompute MSE, it may be a different scence.


I'd like to stress that in some cases execution time seriously suffers when not scaling the target values. I experienced this with sklearn.neural_network.MLPRegressor, where execution time increased at least tenfold when not scaling the target. I'm sure this varies from case to case and also depends heavily on the exact regression method used but this experience seems to confirm that it can be very useful indeed to scale the target.

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    Commented Sep 10, 2023 at 16:38

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