Do we have $A_{(\mathfrak{p})} \simeq (A_{(f)})_{\mathfrak{q}}$?

Question

Let $A = \oplus_{d \ge 0} A_d$ be a graded ring. Let $\mathfrak{p}$ a homogeneous prime of $A$. Let $f$ be a homogeneous of $A$ of degree $\require{cancel}\cancel{d}$ $1$ such that $f \not\in \mathfrak{p}$. Let $$ A_{(\mathfrak{p})} = \Bigl\{ \frac{a}{b} : b \not\in \mathfrak{p}, a, b \text{ homogeneous of same degree} \Bigr\} \text{ and } A_{(f)} = \Bigl\{ \frac{a}{f^r} : a \in A_{dr} \Bigr\}. $$ Let $\mathfrak{q} = \mathfrak{p} A_f \cap A_{(f)}$. Do we have $A_{(\mathfrak{p})} \simeq (A_{(f)})_{\mathfrak{q}}$? We have a well-defined map $$ \phi : A_{(\mathfrak{p})} \longrightarrow (A_{(f)})_{\mathfrak{q}}, \frac{a}{b} \longmapsto \frac{a/f^r}{b/f^r}$$ where $r = \deg a = \deg b$. Is this map an isomorphism ?