When there is only one response (dependent) variable, what is the advantage of partial least squares (PLS) regression over principal component regression (PCR)?

My understanding is that PLS is only better when there are multiple response variables, correct?


1 Answer 1


They are different methods, independently of the number of response variables. Both methods combine PCA with ordinary multiple regression but it's done in a crucially different way. For a matrix of predictor variables X and one of dependent variables Y, principal component regression performs a PCA on predictor matrix X and then uses those principal components as regressors on Y. This technique removes multicolinearity but does not reduce the number of predictors down to the “best” subset. Picking out manually the most informative components won't work because these components were made from the variables in X and are therefore informative only to X, not Y.

On the other hand, PLS finds components which explain the covariance between X and Y (and calls them “latent vectors”). Hence, with PLS it's safer to assume that more informative components correspond to more relevant predictors.

  • 3
    $\begingroup$ +1. Just to be completely clear in response to the title question: with only one dependent variable, PLS is still different from PCR, they are not identical. $\endgroup$
    – amoeba
    Commented Jul 30, 2015 at 13:35
  • $\begingroup$ Yes they're different, even with one dependent variable. I edited my answer to reflect this better. $\endgroup$
    – Digio
    Commented Jul 30, 2015 at 13:45
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    $\begingroup$ I wouldn't say "PLS finds principal components [...]", by the way. "Principal components" is the term reserved for PCA. Perhaps you can say "PLS finds components [...]"? $\endgroup$
    – amoeba
    Commented Jul 30, 2015 at 14:15
  • $\begingroup$ I meant "the PLS method", abstractly speaking, but you're right. $\endgroup$
    – Digio
    Commented Jul 30, 2015 at 14:16
  • $\begingroup$ But if there is only one dependent variable, then $\mathbf{X}^T \mathbf{y}$ is a vector and we can't apply an SVD to a vector, so how do we find the component in this case? $\endgroup$
    – Confounded
    Commented Aug 4, 2022 at 14:56

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