8-PSK baseband signal with real and imaginary part
You're making your life actively harder by considering I and Q separate instead of considering them to be the real and imaginary part of a complex signal $S = I + jQ$.

If you then do that, you'd simply get a table with 8 rows, one for every constellation point. Notice that for 8-PSK, you simply divide the full unit circle ($2\pi$) into 8 equal angles – hence, the complex value of all these points is simply $e^{j\angle(\text{constellation point})}= e^{j\frac{2\pi}{8}n}$; I arbitrarily chose that 000 should be mapped to $n=0$, 001 to $n=1$ and so on. There's no reason that this is the right mapping – all 8-PSK says is "there's 8 constellation points on equally distant points on the unit circle of the complex plane".  


000$_b\equiv 0$:&nbsp; $e^{j\frac{2\pi}8 \cdot 0}=1+0j$
001$_b\equiv 1$:&nbsp; $e^{j\frac{2\pi}8 \cdot 1}=\frac{\sqrt2}2+\frac{\sqrt2}2j$
010$_b\equiv 2$:&nbsp; $e^{j\frac{2\pi}8 \cdot 2}=0+1j$
011$_b\equiv 3$:&nbsp; $e^{j\frac{2\pi}8 \cdot 3}=-\frac{\sqrt2}2+\frac{\sqrt2}2j$
100$_b\equiv 4$:&nbsp; $e^{j\frac{2\pi}8 \cdot 4}=-1+0j$
101$_b\equiv 5$:&nbsp; $e^{j\frac{2\pi}8 \cdot 5}=-\frac{\sqrt2}2-\frac{\sqrt2}2j$
110$_b\equiv 6$:&nbsp; $e^{j\frac{2\pi}8 \cdot 6}=0-1j$
111$_b\equiv 7$:&nbsp; $e^{j\frac{2\pi}8 \cdot 7}=\frac{\sqrt2}2-\frac{\sqrt2}2j$


You'd then just look up the real and imaginary part from that table. The simple "bit maps to I (or Q)" simply doesn't work for anything but BPSK and QPSK.