The Categorize( 'f', 'L', 'opts' ) function categorizes the list L into lists $\mathrm{M1}\,\mathrm{M2}\,\dots$ such that f(Mi[1], j, opts) returns true for each element j in the ith list Mi. The procedure f must return a value of type truefalse, and it must be reflexive: f(x, x, opts) must return true for each x in L.

If f defines an equivalence relation, then the result is a partition of L into equivalence classes.

If f can be written in the form (x,y) -> g(x) = g(y), then the call ListTools[Classify](g, L, opts) is more efficient, and equivalent except in the form of the output (Classify returns a table of sets).

$\mathrm{with}\left(\mathrm{ListTools}\right)\:$

$L≔\left[\mathrm{seq}\left(1..12\right)\right]$

$L≔\left[1\,2\,3\,4\,5\,6\,7\,8\,9\,10\,11\,12\right]$

$f≔\left(x\,y\right)\mapsto x-y\mathbf{mod}4=0$

$f≔\left(x\,y\right)\mapsto x-y\mathbf{mod}4=0$

$\mathrm{Categorize}\left(f\,L\right)$

$\left[1\,5\,9\right],\left[2\,6\,10\right],\left[3\,7\,11\right],\left[4\,8\,12\right]$

$L≔\left[x^2-1\,x^2+1\,\left(x-1\right)\left(x+1\right)\,x^2-2x+1\,{\left(x-1\right)}^2\right]$

$L≔\left[x^2-1\,x^2+1\,\left(x-1\right)\left(x+1\right)\,x^2-2x+1\,{\left(x-1\right)}^2\right]$

$\mathrm{Categorize}\left(\mathrm{verify}\,L\,\mathrm{expand}\right)$

$\left[x^2-1\,\left(x-1\right)\left(x+1\right)\right],\left[x^2+1\right],\left[x^2-2x+1\,{\left(x-1\right)}^2\right]$