Re: Re: Counting list elements above/below a given value
- To: mathgroup at smc.vnet.net
- Subject: [mg17432] Re: [mg17271] Re: Counting list elements above/below a given value
- From: "Allan Hayes" <hay at haystack.demon.co.uk>
- Date: Sun, 9 May 1999 04:43:46 -0400
- References: <7g0qv1$drj@smc.vnet.net> <199904300634.CAA21396@smc.vnet.net.> <7gfu2e$6je@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Carl,
Hello Carl,
Your finding that x>.5 is a slower test than x-.5 being positive is most
intriguing and not what I would have expected.
I don't know how reliable the following speed up it, but at least it's what
I would expect.
Do[Length[Select[Flatten[data-.5], Positive]],{10}]//Timing
{48.06 Second,Null}
Do[Length[Select[Flatten[data]-.5, Positive]],{10}]//Timing
{42.13 Second,Null}
Allan
---------------------
Allan Hayes
Mathematica Training and Consulting
Leicester UK
www.haystack.demon.co.uk
hay at haystack.demon.co.uk
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565
Carl Woll <carlw at u.washington.edu> wrote in message
news:7gfu2e$6je at smc.vnet.net...
> Hi Allan,
>
> Your results were somewhat surprising to me, since I didn't expect the
Select
> option to be the quickest. So, I played around with it, and I came up with
an
> idea using Count that was faster. Witness
>
> In[17]:=
> data= Table[Random[],{280},{280}];
> Count[data, _?(#>.5&),{2}]//Timing
> Count[data, x_/;x>.5,{2}]//Timing
> Length[Select[Flatten[data], #1>.5&]]//Timing
> Count[data-.5,_?Positive,{2}]//Timing
>
> Out[17]=
> {14.36 Second,39012}
>
> Out[18]=
> {12.84 Second,39012}
>
> Out[19]=
> {11.03 Second,39012}
>
> Out[20]=
> {8.71 Second,39012}
>
> So, changing the data and using a simpler test allows Count to beat
Select.
> However, this isn't fair to Select. So doing the same thing for Select
yields
>
> In[21]:=
> Length[Select[Flatten[data-.5], Positive]]//Timing
>
> Out[21]=
> {5.29 Second,39012}
>
> I guess the message here is that Length[Select[...]] is just faster than
> Count[...].
>
> Carl
>
> Allan Hayes wrote:
>
> > Bill
> >
> > Count needs a pattern; and in this case a level specification. But
pattern
> > matching is slower than the flattening and testing (with Select) that
you
> > have used. Here are some timings.
> >
> > data= Table[Random[],{280},{280}];
> >
> > Count[data, _?(#>.5&),{2}]//Timing
> >
> > {6.86 Second,39219}
> >
> > Count[data, x_/;x>.5,{2}]//Timing
> >
> > {6.26 Second,39219}
> >
> > Length[Select[Flatten[data], #1>.5&]]//Timing
> >
> > {5.82 Second,39219}
> >
> > Allan
> >
> > ---------------------
> > Allan Hayes
> > Mathematica Training and Consulting
> > Leicester UK
> > www.haystack.demon.co.uk
> > hay at haystack.demon.co.uk
> > Voice: +44 (0)116 271 4198
> > Fax: +44 (0)870 164 0565
> >
> > W.W. Sampson <w.sampson at umist.ac.uk> wrote in message
> > news:7g0qv1$drj at smc.vnet.net...
> > > I wish to count the number of elements within a list above
> > > a given value. I can do this using the following:
> > >
> > > Length[Select[Flatten[listname], value > #1&]]
> > >
> > > However, my list is a 280 x 280 matrix and consequently
> > > this takes a while to evaluate. My guess is that there
> > > must be something more efficient. I've tried:
> > >
> > > Count[listname, value > #1&]
> > >
> > > but the output given is 0, which I know to be incorrect.
> > >
> > > Any ideas appreciated.
> > >
> > > Bill
> > >
>
>
>
> --
> Carl Woll
> Dept of Physics
> U of Washington
>
>
>
>