For a complex of sheaves $\cal{F}^{\bullet}$ on a variety $X$, a useful fact is that the stalks of the cohomology sheaves of $\mathcal{F}^{\bullet}$ agree with the cohomology groups of the complex of stalks:

$$\mathcal{H}^{i}(\cal{F}^{\bullet})_{x} \cong \mathcal{H}^{i}(\cal{F}^{\bullet}_{x})$$

Consider now the category $D_{c}^{b}(X)$ of bounded constructible complexes on $X$, and the full abelian subcategory $\operatorname{Perv}(X)$ of perverse sheaves. Let us work with sheaves of $\mathbb{Q}$ or $\mathbb{C}$-modules. The *perverse* cohomology sheaves ${}^{\mathfrak{p}}\mathcal{H}^{i}(\cdot):D_{c}^{b}(X) \to \operatorname{Perv}(X)$ are themselves perverse sheaves.

I'm wondering, are there any nice results helping to compute stalks of ${}^{\mathfrak{p}}\mathcal{H}^{i}(\mathcal{F}^{\bullet})$? Like something analogous to the basic result I recall at the beginning? These stalks should be complexes of vector spaces.

The basic result above for ordinary sheaves can't work as it is. For example, if $X$ is smooth of dimension $n$, then $\mathbb{C}_{X}[n]$ is perverse. And ${}^{\mathfrak{p}}\mathcal{H}^{i}(\mathbb{C}_{X}[n])$ vanishes unless $i=0$ in which case it is just $\mathbb{C}_{X}[n]$. So at any point $q \in X$, the stalk is the skyscrapper sheaf:

$${}^{\mathfrak{p}}\mathcal{H}^{0}(\mathbb{C}_{X}[n])_{q} = \mathbb{C}_{q}[n]$$

But if we first pass to the stalk, and then take perverse cohomology ${}^{\mathfrak{p}}\mathcal{H}^{i}(\mathbb{C}_{q}[n])$, this vanishes unless $i=-n$, in which case it is just $\mathbb{C}_{q}$.

So are there any useful ways of computing stalks of perverse cohomology sheaves, or do you just have to do it directly?