We can regularize a linear model with L1 or L2 regularization.

But we usually write L2 with a square: $\|x\|_2^2$ and L1 with $\|x\|_1$. It seems a little bit strange and inconsistent for me, because one has a square and another does not have.

Is the reasons that adding L2 with square can directly associate with the radius $r$?

EDIT: I was trying to ask convention but not why there is a square. I think Mark answers my question perfectly. Thanks

In fact, there's not even standardization on whether the "baseline" portion of a "least squares" objective should be squared. ... In the absence of the regularization term, it doesn't matter, but with a regularization term, it does. ... Bottom line: regularization tends to be somewhat of an ad hoc or arbitrary process, and there's more than one way to do it.

We can regularize a linear model with L1 or L2 regularization.

But we usually write L2 with a square: $\|x\|_2^2$ and L1 with $\|x\|_1$. It seems a little bit strange and inconsistent for me, because one has a square and another does not have.

Is the reasons that adding L2 with square can directly associate with the radius $r$?

EDIT: I was trying to ask convention but not why there is a square. I think Mark answers my question perfectly. Thanks

We can regularize a linear model with L1 or L2 regularization.

But we usually write L2 with a square: $\|x\|_2^2$ and L1 with $\|x\|_1$. It seems a little bit strange and inconsistent for me, because one has a square and another does not have.

Is the reasons that adding L2 with square can directly associate with the radius $r$?

EDIT: I was trying to ask convention but not why there is a square. I think Mark answers my question perfectly. Thanks

In fact, there's not even standardization on whether the "baseline" portion of a "least squares" objective should be squared. ... In the absence of the regularization term, it doesn't matter, but with a regularization term, it does.

We can regularize a linear model with L1 or L2 regularization.

But we usually write L2 with a square: $\|x\|_2^2$ and L1 with $\|x\|_1$. It seems a little bit strange and inconsistent for me, because one has a square and another does not have.

Is the reasons that adding L2 with square can directly associate with the radius $r$?

EDIT: I was trying to ask convention but not why there is a square. I think Mark answers my question perfectly. Thanks

In fact, there's not even standardization on whether the "baseline" portion of a "least squares" objective should be squared. ... In the absence of the regularization term, it doesn't matter, but with a regularization term, it does. ... Bottom line: regularization tends to be somewhat of an ad hoc or arbitrary process, and there's more than one way to do it.