Convert the point $( -2, -2 \sqrt{3}, -1)$ in rectangular coordinates to cylindrical coordinates.  Enter your answer in the form $(r,\theta,z),$ where $r > 0$ and $0 \le \theta < 2 \pi.$
We have that $r = \sqrt{(-2)^2 + (-2 \sqrt{3})^2} = 4.$  We want $\theta$ to satisfy
\begin{align*}
-2 &= 4 \cos \theta, \\
-2 \sqrt{3} &= 4 \sin \theta.
\end{align*}Thus, $\theta = \frac{4 \pi}{3},$ so the cylindrical coordinates are $\boxed{\left( 4, \frac{4 \pi}{3}, -1 \right)}.$
We have that $r = \sqrt{(-2)^2 + (-2 \sqrt{3})^2} = 4.$  We want $\theta$ to satisfy
\begin{align*}
-2 &= 4 \cos \theta, \\
-2 \sqrt{3} &= 4 \sin \theta.
\end{align*}Thus, $\theta = \frac{4 \pi}{3},$ so the cylindrical coordinates are $\boxed{\left( 4, \frac{4 \pi}{3}, -1 \right)}.$
The final answer is $$\boxed{\left( 4, \frac{4 \pi}{3}, -1 \$. I hope it is correct.