Undecimated DWT vs. CWT. In what cases is one preferable over another?

Question

Talking about undecimated DWT let's focus here on SWT algorithm.

SWT is said to be:

numerically more stable due to Mallat algorithm. And the inverse operation is capable of returning the original signal which is not the case with CWT
SWT is computationally faster than CWT

On the other hand:

Are there any other factors supporting the choise of one or another alternative?

When difference do you make between "non-orthogonal" and CWT? In "DWT produces orthogonal decomposition", how do you link that to the title "In what cases (bi)orthogonal wavelet" — Laurent Duval, Feb 24 '17 at 07:02
sorry, I've mixed two questions here in a confusing manner. I'll separate them — Alexander Reshytko, Feb 26 '17 at 15:54
The questions are clearer now. I can answer them later in the week, if nobody else does that — Laurent Duval, Feb 26 '17 at 17:03
I should say that I do not fully understand the exact need, and some of the given assertions, eg "SWT works on dyadic scale", can be debated — Laurent Duval, Aug 16 '17 at 21:24

score 1 · Accepted Answer · answered Aug 16 '17 at 13:58

Are there any other factors supporting the choice of one or another alternative?

possibility to use almost any wavelet shape (and associated sampling) or signal-matched: in favor of the CWT
possible parallelization in cycle-spinning form: in favor of SWT
tunable redundancy factor: in favor of the CWT
possibility to derive sharp signal/noise estimators for denoising: in favor of SWT
geometrical extensions (direction, shear, anisotropy) in higher dimension: in favor of the CWT

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