在正项等比数列 $\left\{a_{n}\right\}$ 中, $a_{3} a_{7}=4$, 数列 $\left\{\log _{2} a_{n}\right\}$ 的前 9 项之和为( )
$\because\left\{a_{n}\right\}$ 是正项等比数列, $\therefore a_{3} a_{7}=a_{1} a_{9}=a_{2} a_{8}=a_{4} a_{6}=a_{2}^{5}=4$,

$\therefore a_{5}=2$

$S_{9}=\log _{2} a_{1}+\log _{2} a_{2}+\log _{2} a_{3}+\log _{2} a_{4}+\log _{2} a_{5}+\log _{2} a_{6}+\log _{2} a_{7}+\log _{2} a_{8}+\log _{2} a_{9}$

$=\log _{2}\left(a_{1} a_{2} \ldots a_{9}\right)$

$=\log _{2}\left[\left(a_{3} a_{7}\right)\left(a_{1} a_{9}\right)\left(a_{2} a_{8}\right)\left(a_{4} a_{6}\right)\left(a_{5}\right)\right]$

$=\log _{2}\left(4^{4} \times 2\right)=\log _{2} 2^{9}=9$

故答案为 B