Non-integer hybridization I've known that hybridization in distorted geometries is not exactly $sp^3$ or $sp^2$ or whatever. For example, $\ce{PH3}$ has nearly pure $p$ orbitals in the $\ce{P-H}$ bond, and the lone pair is in a nearly pure $s$ orbital.

Basically, since hybridisation is an addition of wavefunctions, instead of a perfect symmetrical addition of the kets, we get something else.

While answering this question, I realised that it's not that easy predicting the numbers. 

Take the same image of $\ce{B2H6}$:



At first, seeing the 97&deg;, I thought "well, the inner $\ce{B-H}$ bonds will be almost pure $p$". But, with that, I couldn't figure out where the 120&deg; came from, because that is for perfect $sp^2$.

I then realised that I was being stupid and it wasn't that simple--just because you have a 120&deg; doesn't mean pure $sp^2$. But, I was at a loss trying to find out approximate hybridizations for $\ce{B2H6}$.

How does one generally go about predicting such "nonuniform hybridizations" if one knows the bond angles?

Approximations are OK--I believe the exact mixture ratios will require some knowledge of the exact wavefunctions.
Classical hybridization theory does not allow for noninteger hybridizations. However, ab initio calculations can be interpreted using a bond order analysis method such as NBO, where the MO coefficients are used to provide the closest analogue possible to a classical hybridization picture.

For example, one of the pure $sp$ orbitals in Pauling's valence bond theory (where hybridization was first introduced) has wavefunction

$$ \phi_{sp} = \frac{1}{\sqrt 2} \phi_{s} + \frac{1}{\sqrt{2}} \phi_{p_x} $$

Taking the square of the coefficients, this orbital is $\frac 1 2$ s character and $\frac 1 2$ p character, i.e. it is an $sp$ orbital.

The basic idea of a bond order analysis method is to reexpress a molecular orbital into a form similar to

$$ \phi = c_1 \phi_{As} + c_2 \phi_{Ap_x} + c_3 \phi_{Ap_y} + c_4 \phi_{Ap_z} + \dots$$

If the other coefficients are very small, the ratio

$$ n = \frac {c_2^2 + c_3^2 + c_4^2}{c_1^2} $$

would yield a number that could be used interpret $\phi$ as a $sp^n$ orbital on atom $A$.

Edit: As for inferring hybridization states from a direct inspection of nuclear geometries, there is in principle no such direct relationship in electronic structure theories that are more sophisticated than VSEPR. The former is an electronic property whereas the spatial arrangement of atomic nuclei are not, and the relationship between the two becomes much more complicated.