As the comments suggest, the tables enumerated knots are much larger than any table of "named" knots. So it might be easier to start with the "named" knots and identify them in the table. The snappy software (http://www.math.uic.edu/t3m/SnapPy/index.html) allows you to do this quite easily for hyperbolic knots. Here is a general outline of how to do that. (snappy: "text" are instructions to give snappy comments have a bullet in front of them.)

snappy: M = Manifold()

- If you have installed plink a window pops up where you can draw your knot

snappy: M.solution_type()

- This should be 'all tetrahedra positively oriented' if you have something hyperbolic

snappy: CK = CensusKnots()

- This loads the list of knots known to decompose into 8 or fewer tetrahedra

snappy: CK.identify(M)

- This will return the name of the manifold in the census you are looking at.

You can also get something similar to work if you want to look at the AlternatingKnotExteriors or NonalternatingKnotExteriors. Unfortunately, the identify function won't work on these list, but you can you can still get something to work.

snappy: M = Manifold()

snappy: M.solution_type()

- Again begin the same way and check that you have a hyperbolic knot that admits a diagram
with fewer than 16 crossings.

snappy: volM = M.volume()

snappy: for k in AlternatingKnotExteriors():

snappy: if k.volume() < volM + .002 and k.volume() > volM - .002:

snappy: print k

snappy: print k.is_isometric_to(M)

- This will give you a good list of what your knot could be. You could also do the same thing for the NonalternatingKnotExteriors(). Finally, the .002 is there to account for any error that might arise computationally. In principle, you could probably choose a number much smaller.