2. Function
$$
y=\tan 2013 x-\tan 2014 x+\tan 2015 x
$$

The number of zeros of the function in the interval $[0, \pi]$ is $\qquad$.
2. 2014 .
$$
\begin{aligned}
y & =\tan 2013 x-\tan 2014 x+\tan 2015 x \\
& =\frac{\sin (2013 x+2015 x)}{\cos 2013 x \cdot \cos 2015 x}-\frac{\sin 2014 x}{\cos 2014 x} \\
& =\frac{\sin 4028 x}{\cos 2013 x \cdot \cos 2015 x}-\frac{\sin 2014 x}{\cos 2014 x} \\
& =\frac{2 \sin 2014 x \cdot \cos 2014 x}{\cos 2013 x \cdot \cos 2015 x}-\frac{\sin 2014 x}{\cos 2014 x} \\
& =\frac{\sin 2014 x \cdot(2 \cos 2014 x-\cos 2013 x \cdot \cos 2015 x)}{\cos 2013 x \cdot \cos 2014 x \cdot \cos 2015 x} \\
& =\frac{\sin 2014 x \cdot(1+\cos 4028 x-\cos 2013 x \cdot \cos 2015 x)}{\cos 2013 x \cdot \cos 2014 x \cdot \cos 2015 x} \\
& =\frac{\sin 2014 x \cdot(1-\sin 2013 x \cdot \sin 2015 x)}{\cos 2013 x \cdot \cos 2014 x \cdot \cos 2015 x} .
\end{aligned}
$$

From the domain of $y$, we know that
$$
\sin 2013 x \cdot \sin 2015 x \neq 1 \text{.}
$$

Therefore, the zeros of $y$ are
$$
x=\frac{k \pi}{2014}(k=0,1, \cdots, 2014) \text{.}
$$

However, when $k=1007$, the function is undefined.
Thus, $y$ has 2014 zeros in the interval $[0, \pi]$.