TITLE: How does one Segal-subdivide a 2-category?
QUESTION [5 upvotes]: Let $\mathcal{C}$ be a small category. Then, its Segal subdivision $\text{sd }\mathcal{C}$ is a new category whose objects are morphisms of $\mathcal{C}$, and a morphism from $f:x \to y$ to $g: w \to z$ is a pair of morphisms $a: x \to w$ and $b: y \to z$ in $\mathcal{C}$ so that $f = b\circ g \circ a$. It is easy to see (by a contractible-fiber argument) that there is a homotopy equivalence of classifying spaces $B\text{Sd }\mathcal{C} \simeq B\mathcal{C}$. My first encounter with this construction was in the preprint on Morse theory and classifying spaces (link here).
Here's the question:

What is the analogue of Segal subdivision for small (strict) $2$-categories?

In particular, from a given small $2$-category $\mathcal{D}$, I would like to construct a new $2$-category $\text{sd}_2\mathcal{D}$ whose one-skeleton coincides with the ordinary Segal subdivision of the one-skeleton of $\mathcal{D}$. Some mysterious and powerful $2$-morphisms should exist, and their addition should magically make the classifying spaces of $\text{sd}_2\mathcal{D}$ and $\mathcal{D}$ homotopy-equivalent. 
What, if anything, is the precise collection of $2$-morphisms which achieve the homotopy equivalence, and where can I find this written down?

REPLY [2 votes]: Another, somewhat similar, suggestion, in a cubical setting, which produces a double category:

In the picture I wrote the one morphisms as subdivisions, but if you want to see it as a double category, then these should be the vertical morphisms, and the top and bottom faces are the horizontal morphisms.