3. Define the function on $\mathbf{R}$
$$
f(x)=\left\{\begin{array}{ll}
\log _{2}(1-x), & x \leqslant 0 ; \\
f(x-1)-f(x-2), & x>0 .
\end{array}\right.
$$

Then $f(2014)=$
3. 1.

First, consider the method for finding the function values on the interval $(0,+\infty)$.
If $x>3$, then
$$
\begin{array}{l}
f(x)=f(x-1)-f(x-2) \\
=f(x-1)-(f(x-1)+f(x-3)) \\
=-f(x-3) .
\end{array}
$$

Thus, for $x>6$, we have
$$
f(x)=-f(x-3)=f(x-6) \text {. }
$$

Therefore, $f(x)$ is a periodic function with a period of 6 on the interval $(0,+\infty)$.
$$
\begin{array}{l}
\text { Hence } f(2014)=f(4)=f(3)-f(2) \\
=-f(0)+f(-1)=1 .
\end{array}
$$