Let $\mathcal{D}$ be a triangulated category and a $t$-structure $(\mathcal{D}^{\leq 0},\mathcal{D}^{\geq 0})$ on $\mathcal{D}$. The heart of the $t$-structure, $\mathcal{A}=\mathcal{D}^{\leq 0} \cap \mathcal{D}^{\geq 0}$, is an abelian category. 


I know that in general there is not a natural functor from the derived category of the heart and the triangulated category .


But it's seem to be very linked ...


But if you consider stable infinite category or Grothendieck derivator is there such functor ?



The relation between t-structures and derived category
Assume $\mathcal D$ is a presentable stable $\infty$-category with a $\mathrm t$-structure (which is accessible and compatible with filtered colimits), and let $\mathcal A$ be its heart, $\mathcal{D(A)}$ its derived $\infty$-category.


Note that under those hypotheses, $\mathcal A$ is Grothendieck abelian (*Higher Algebra*, 1.3.5.23.).


You get a natural inclusion functor $\mathcal A\to \mathcal D$, which you can extend to $Fun(\Delta^{op},\mathcal A)\to \mathcal D$ (by geometric realization).


This functor preserves weak equivalences : indeed, (see *HA*, 1.2.4.4. and 1.2.4.5.), if $X$ is a simplicial object of $\mathcal A$ (and therefore $\mathcal D\_{\geq 0}$), there is a spectral sequence with $E^2\_{p,q}=\pi\_p\pi\_q(X)$ converging to $\pi\_{p+q}(|X|)$ in $\mathcal A$.


Since $\pi\_q(X) = 0$ for $q\neq 0$ in our situation (as $X$ takes values in $\mathcal A$), this spectral sequence degenerates, and $\pi\_p(|X|) = \pi\_p(X\_\bullet)$ (where the latter are homotopy groups as computed in $\mathcal A$ via the classical Dold-Kan correspondance)


It follows that this functor yields a (unique up to a contractible space of choices) functor $\mathcal D\_{\geq 0}(\mathcal A)\to \mathcal D$ (via, again the Dold-Kan correspondance).


Now $\mathcal D$ is presentable and stable, and this functor $\mathcal D\_{\geq 0}(\mathcal A)\to \mathcal D$ preserves colimits, so it extends (again, uniquely) to a functor $\mathcal{D(A) \to D}$ which preserves colimits and preserves $\mathcal A$ (*HA* 1.3.5.21. : $\mathcal{D(A)}$ is right complete with its classical $\mathrm t$-structure, which implies in particular that $\mathcal{D(A)} = \lim(\dots \overset{\Omega}\to \mathcal D\_{\geq 0}(\mathcal A)\overset{\Omega}\to \mathcal D\_{\geq 0}(\mathcal A))$, and then we use 1.4.4.5. which says that this precisely has the universal property of "presentable stabilization")


Now these hypotheses on $\mathcal D$ may look pretty strong but they're reasonable ; and in fact you can't really hope for much better : if $\mathcal D$ isn't presentable, then it could be something like $\mathcal D^{-}(\mathcal A)$ and then there's no hope to get a sensible functor $\mathcal{D(A)}\to \mathcal D^{-}(\mathcal A)$ (this is something that won't change whether you're in an $\infty$-categorical setting or not). 


This is probably already somewhere in *HA* but I couldn't find it written down completely. 


As I pointed out in the comments, under different hypotheses (which may look weaker), you can get away with a natural functor $\mathcal D^{-}(\mathcal A)\to \mathcal D$ (morally, this is because $\mathcal D^{-}(\mathcal A) \subset \mathcal{D(A)}$ is $\bigcup\_n \mathcal D\_{\geq n}(\mathcal A)$, and so it is determined by $\mathcal D\_{\geq 0}(\mathcal A)$ via finite limits, so you don't need a presentability hypothesis- you do, however need some hypothesis on $\mathcal D$ to be able to replace the first step, since you can't take colimits as easily).


I guess there may also be a more general statement about $\mathcal D^b(\mathcal A)$, the bounded derived $\infty$-category, which should require less hypotheses, since "everything is finite" but I couldn't tell you on the top of my head. 


