Here is the description of verb round: NB.*round v round y to nearest x (e.g. 1000 round 12345) round=: [ * [: <. 0.5 + %~
On Thu, Jan 14, 2016 at 3:41 PM, Raul Miller <[email protected]> wrote: > Did you try it? It seems to me that this just makes things worse: > > round=: [ * [: <. 0.5 + %~ > round 0.5 + 50.4 2e10 1e30 _ > |NaN error: round > > Thanks, > > -- > Raul > > On Thu, Jan 14, 2016 at 4:28 PM, Kip Murray <[email protected]> > wrote: > > Try > > > > round > > [ * [: <. 0.5 + %~ > > > > which I found in j602. > > > > --Kip Murray > > > > On Thu, Jan 14, 2016 at 1:56 PM, Marshall Lochbaum <[email protected] > > > > wrote: > > > >> Here's something I spent far too long on, and consequently thought was > >> worth sharing. I can turn it into an essay on the J wiki if people want > >> that. > >> > >> Recently I ran into the problem of rounding a J floating point number to > >> an integer, and forcing the result to have integer type. This seems like > >> a simple task: using a standard rounding function, we have > >> 0.5 <.@:+ 2.3 5.1 7.6 3.9 > >> 2 5 8 4 > >> But with numbers that are too large, the result still contains > >> floating-point numbers, and has type 8 (floating point) rather than 4 > >> (integer). > >> 0.5 <.@:+ _1e50 2e30 _ > >> _1e50 2e30 _ > >> 3!:0 ]0.5 <.@:+ _1e50 2e30 _ > >> 8 > >> When applied to a float, (<.) applies the C floor function, which yields > >> another float, and than casts the results to integers if all of them are > >> exactly representable as integers. They're not here, so they are left as > >> floating-point numbers. > >> > >> To give a more accurate problem statement, I want the 64-bit signed > >> integer which is closest to the function input. Thus numbers above the > >> maximum representable integer should round to that integer, and likewise > >> for numbers below the minimum representable integer. We define these two > >> bounds now. > >> MAX =: ->: MIN =: _2 <.@^ 63 > >> Note that since MAX is one less than 2^63, trying to take (2<.@^63) > >> would give us a float, and subtracting one would still leave us with a > >> floating point number, which is not actually equal to MAX since (>:MAX) > >> is representable as a float, while nearby integers are not. MIN on the > >> other hand is safely computed as an exponent. Note the negative base, > >> which works because 63 is odd. > >> > >> With these bounds, our problem should be easy: clamp to the integer > >> range, then use (<.). > >> ([: <. MIN>.MAX<.]) 0.5 + __ _1e30 _1e10 _100.3 > >> _9223372036854775808 _9223372036854775808 _10000000000 _100 > >> So far, so good... > >> ([: <. MIN>.MAX<.]) 0.5 + 50.4 2e10 1e30 _ > >> 50 2e10 9.22337e18 9.22337e18 > >> Oops. What happened? > >> MAX <. _ > >> 9.22337e18 > >> Since one of the arguments is a float, (<.) casts both to floats, and > >> takes the minimum. But the closest floating-point number to MAX is > >> (MAX+1), and that number's floor (MAX+1) isn't representable as an > >> integer--it's one too big. We didn't have this problem with MIN, since > >> it is exactly a negative power of two. > >> <. MAX+1 > >> 9.22337e18 > >> > >> We'll make a test case that contains numbers close to both bounds. I've > >> included the addition of 0.5 in t so I can focus on the floor function > >> from now on. The type of our result is a float, so we failed. > >> t =. 0.5 + __,_,~ (MIN,0,MAX) +/(,@:) i:1e5 > >> 3!:0 ([: <. MIN>.MAX<.]) t > >> 8 > >> > >> We can fix the problem by using exact integers, but it's extremely slow. > >> However, it serves as a good answer key. The ("0) is there for a > >> reason--otherwise the big array of exact integers tends to flood RAM. > >> fl_e =: (MIN>.MAX<.<.)&.:x:"0 NB. exact floor > >> 3!:0 key =. fl_e t > >> 4 > >> 10 (6!:2) 'fl_e t' > >> 3.62018 > >> > >> If we use a number small enough that its floating-point representation > >> is equal to a 64-bit integer, then we can force our answer to be a > >> float, but it's not correct since the results are sometimes too small. > >> If that doesn't matter and speed is critical, this is the right method > >> to use. > >> MAX1 =. MAX - 512 > >> fl_f =: [: <. MIN>.MAX1<.] NB. fast floor > >> 3!:0 fl_f t > >> 4 > >> key -: fl_f t > >> 0 > >> 10 (6!:2) 'fl_f t' > >> 0.0179205 > >> > >> Finally, my solution. It's not particularly elegant, but it is correct > >> and has good performance. We reduce all the values larger than MAX to > >> zero, then clamp on the minimum side and take the floor. For the values > >> that we removed, we add MAX back in. The comparison (<:&MAX) is only > >> computed once to save a little time. > >> fl =: ((MAX*-.@]) + [: <. MIN>.*) <:&MAX > >> key -: fl t > >> 1 > >> 10 (6!:2) 'fl t' > >> 0.0426181 > >> It's critical to use (<:) rather than (<) to test whether numbers are > >> acceptable even though it fails MAX, which wouldn't break (<.). That's > >> because comparisons cast their arguments to floats before comparing, so > >> MAX < MAX+1 > >> 0 > >> > >> Maybe there's a quicker solution to be found. The following rounds > >> towards zero quickly by negating all the positive numbers, and restoring > >> their signs later. However, adding in the cases to make it equal to (<.) > >> on small numbers removes its advantage. > >> fl_o =: (] * MIN <.@:>. *) -@:* NB. floor towards zero > >> fl_o _4.6 _3 _2.8 _1.2 3.4 5.8 9 > >> _5 _3 _3 _2 4 6 9 > >> 10 (6!:2) 'fl_o t' > >> 0.0324293 > >> > >> Any takers? > >> > >> Marshall > >> ---------------------------------------------------------------------- > >> For information about J forums see http://www.jsoftware.com/forums.htm > > ---------------------------------------------------------------------- > > For information about J forums see http://www.jsoftware.com/forums.htm > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm > ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
