$A≔\mathrm{Vector}\left['\mathrm{row}'\right]\left(\left[9\,0\,16\,4\,1\right]\right)$

$A≔\left[\begin{array}{ccccc}9& 0& 16& 4& 1\end{array}\right]$

$\mathrm{SortBy}\left(A\right)$

$\left[\begin{array}{ccccc}0& 1& 4& 9& 16\end{array}\right]$

$A≔\mathrm{Matrix}\left(\left[\left[8\,18\,9\right]\,\left[3\,16\,7\right]\,\left[11\,1\,15\right]\,\left[4\,13\,12\right]\right]\right)$

$A≔\left[\begin{array}{ccc}8& 18& 9\\ 3& 16& 7\\ 11& 1& 15\\ 4& 13& 12\end{array}\right]$

$\mathrm{SortBy}\left(A\,'\mathrm{column}'\,2\right)$

$\left[\begin{array}{ccc}11& 1& 15\\ 4& 13& 12\\ 3& 16& 7\\ 8& 18& 9\end{array}\right]$

$\mathrm{SortBy}\left(A\,'\mathrm{row}'\,-1\right)$

$\left[\begin{array}{ccc}8& 9& 18\\ 3& 7& 16\\ 11& 15& 1\\ 4& 12& 13\end{array}\right]$

Custom sorting options can be passed:

$F≔\mathrm{Matrix}\left(\left[\left[1\,-2\,3\right]\,\left[-4\,5\,-6\right]\,\left[7\,-8\,9\right]\right]\right)$

$F≔\left[\begin{array}{ccc}1& \mathrm{-2}& 3\\ \mathrm{-4}& 5& \mathrm{-6}\\ 7& \mathrm{-8}& 9\end{array}\right]$

$G≔\mathrm{SortBy}\left(F\,'\mathrm{row}'\,2\right)$

$G≔\left[\begin{array}{ccc}3& 1& \mathrm{-2}\\ \mathrm{-6}& \mathrm{-4}& 5\\ 9& 7& \mathrm{-8}\end{array}\right]$

$H≔\mathrm{SortBy}\left(F\,'\mathrm{row}'\,2\,'\mathrm{key}'=\mathrm{abs}\right)$

$H≔\left[\begin{array}{ccc}1& \mathrm{-2}& 3\\ \mathrm{-4}& 5& \mathrm{-6}\\ 7& \mathrm{-8}& 9\end{array}\right]$

The unsorted rtable can be updated with the sorted version:

$A≔\mathrm{Matrix}\left(\left[\left[16\,9\right]\,\left[5\,8\right]\,\left[13\,19\right]\,\left[7\,6\right]\,\left[17\,15\right]\right]\right)$

$A≔\left[\begin{array}{cc}16& 9\\ 5& 8\\ 13& 19\\ 7& 6\\ 17& 15\end{array}\right]$

$B≔\mathrm{SortBy}\left(A\,'\mathrm{column}'\,-1\,'\mathrm{descending}'\,'\mathrm{inplace}'\right)$

$B≔\left[\begin{array}{cc}13& 19\\ 17& 15\\ 16& 9\\ 5& 8\\ 7& 6\end{array}\right]$

$'A'=A$

$A=\left[\begin{array}{cc}13& 19\\ 17& 15\\ 16& 9\\ 5& 8\\ 7& 6\end{array}\right]$

Arrays with starting indices different than $1$ are accepted:

$A≔\mathrm{Array}\left(0..2\,0..4\,\left[\left[4\,15\,8\,2\,18\right]\,\left[5\,18\,2\,3\,17\right]\,\left[11\,15\,20\,7\,1\right]\right]\right)$

For this Array, "row 1" refers to what would be "row 2" if the starting indices were shifted to all be $1$:

$B≔\mathrm{SortBy}\left(A\,'\mathrm{row}'\,1\right)$

Consider the following 3-D Array:

$X≔\mathrm{Array}\left(\left[\left[\left[0.223811939491137\,0.646313010111265\,0.823457828327293\,0.933993247757551\,0.970592781760616\right]\,\left[0.498364051982143\,0.795199901137063\,0.0318328463774207\,0.959492426392903\,0.546881519204984\right]\,\left[0.679702676853675\,0.0344460805029088\,0.392227019534168\,0.141886338627215\,0.913375856139019\right]\right]\,\left[\left[0.585267750979777\,0.445586200710899\,0.0971317812358475\,0.849129305868777\,0.157613081677548\right]\,\left[0.118997681558377\,0.765516788149002\,0.706046088019609\,0.792207329559554\,0.278498218867048\right]\,\left[0.276025076998578\,0.950222048838355\,0.743132468124916\,0.800280468888800\,0.126986816293506\right]\right]\,\left[\left[0.340385726666133\,0.489764395788231\,0.0461713906311539\,0.0357116785741896\,0.964888535199277\right]\,\left[0.162611735194631\,0.381558457093008\,0.171186687811562\,0.915735525189067\,0.0975404049994095\right]\,\left[0.754686681982361\,0.317099480060861\,0.757740130578333\,0.485375648722841\,0.905791937075619\right]\right]\,\left[\left[0.959743958516081\,0.186872604554379\,0.276922984960890\,0.655740699156587\,0.957506835434298\right]\,\left[0.655098003973841\,0.438744359656398\,0.655477890177557\,0.421761282626275\,0.632359246225410\right]\,\left[0.709364830858073\,0.694828622975817\,0.678735154857773\,0.957166948242946\,0.814723686393179\right]\right]\right]\right)$

The data can be sorted using any 1-D slice, and the sorting permutation can be returned in addition to or instead of the sorted Array. For example:

$P,Y≔\mathrm{SortBy}\left(X\,\left[2\,..\,1\right]\,'\mathrm{descending}'\,'\mathrm{output}'=\left['\mathrm{permutation}'\,'\mathrm{sorted}'\right]\right)$

The sorted Array $Y$ can be reproduced using this permutation $P$:

$Z≔X\left[..,P,..\right]$

$\mathrm{verify}\left(Y\,Z\,'\mathrm{Array}\left(\mathrm{float}\left(1\,\mathrm{digits}=15\right)\right)'\right)$

$\mathrm{true}$