As a continuation of my previous question, Simon's method to find the list product of a PackedArray is fast, but it does not work with negative values.
This can be "fixed" by Abs with minimal time penalty, but the sign is lost, so I will need to find the product sign separately.
The fastest method that I tried is EvenQ @ Total @ UnitStep[-lst]
lst = RandomReal[{-2, 2}, 5000000];
Do[
EvenQ@Total@UnitStep[-lst],
{30}
] // Timing
Out[]= {3.062, Null}
Is there a faster way?
This is a little over two times faster than your solution and apart from the nonsense of using Rule@@@ to extract the relevant term, I find it more clear - it simply counts the number elements with each sign.
EvenQ[-1 /. Rule@@@Tally@Sign[lst]]
To compare timings (and outputs)
In[1]:= lst=RandomReal[{-2,2},5000000];
s=t={};
Do[AppendTo[s,EvenQ@Total@UnitStep[-lst]],{10}];//Timing
Do[AppendTo[t,EvenQ[-1/.Rule@@@Tally@Sign[lst]]],{10}];//Timing
s==t
Out[3]= {2.11,Null}
Out[4]= {0.96,Null}
Out[5]= True
A bit late-to-the-party post: if you are ultimately interested in speed, Compile with the C compilation target seems to be about twice faster than the fastest solution posted so far (Tally - Sign based):
fn = Compile[{{l, _Real, 1}},
Module[{sumneg = 0},
Do[If[i < 0, sumneg++], {i, l}];
EvenQ[sumneg]], CompilationTarget -> "C",
RuntimeOptions -> "Speed"];
Here are the timings on my machine:
In[85]:= lst = RandomReal[{-2, 2}, 5000000];
s = t = q = {};
Do[AppendTo[s, EvenQ@Total@UnitStep[-lst]], {10}]; // Timing
Do[AppendTo[t, EvenQ[-1 /. Rule @@@ Tally@Sign[lst]]], {10}]; // Timing
Do[AppendTo[q, fn [lst]], {10}]; // Timing
s == t == q
Out[87]= {0.813, Null}
Out[88]= {0.515, Null}
Out[89]= {0.266, Null}
Out[90]= True
