Efficient way to remove empty lists from lists without evaluating held expressions?
- by Alexey Popkov
In previous thread an efficient way to remove empty lists ({}) from lists was suggested:
Replace[expr, x_List :> DeleteCases[x, {}], {0, Infinity}]
Using the Trott-Strzebonski in-place evaluation technique this method can be generalized for working also with held expressions:
f1[expr_] :=
Replace[expr,
x_List :> With[{eval = DeleteCases[x, {}]}, eval /; True], {0, Infinity}]
This solution is more efficient than the one based on ReplaceRepeated:
f2[expr_] := expr //. {left___, {}, right___} :> {left, right}
But it has one disadvantage: it evaluates held expressions if they are wrapped by List:
In[20]:= f1[Hold[{{}, 1 + 1}]]
Out[20]= Hold[{2}]
So my question is: what is the most efficient way to remove all empty lists ({}) from lists without evaluating held expressions? The empty List[] object should be removed only if it is an element of another List itself.
Here are some timings:
In[76]:= expr = Tuples[Tuples[{{}, {}}, 3], 4];
First@Timing[#[expr]] & /@ {f1, f2, f3}
pl = Plot3D[Sin[x y], {x, 0, Pi}, {y, 0, Pi}];
First@Timing[#[pl]] & /@ {f1, f2, f3}
Out[77]= {0.581, 0.901, 5.027}
Out[78]= {0.12, 0.21, 0.18}
Definitions:
Clear[f1, f2, f3];
f3[expr_] :=
FixedPoint[
Function[e, Replace[e, {a___, {}, b___} :> {a, b}, {0, Infinity}]], expr];
f1[expr_] :=
Replace[expr,
x_List :> With[{eval = DeleteCases[x, {}]}, eval /; True], {0, Infinity}];
f2[expr_] := expr //. {left___, {}, right___} :> {left, right};
