       Re: Setting Negatives to Zero

• To: mathgroup at smc.vnet.net
• Subject: [mg82892] Re: Setting Negatives to Zero
• From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
• Date: Fri, 2 Nov 2007 03:34:46 -0500 (EST)
• Organization: The Open University, Milton Keynes, UK
• References: <fg6qha\$dj0\$1@smc.vnet.net> <fg9nla\$lc5\$1@smc.vnet.net>

```Jean-Marc Gulliet wrote:
> Kevin J. McCann wrote:
>
>> I have a very large data set (64000 x 583) in which negative values
>> indicate "no data", unfortunately these negatives are not all the same.
>> I would like to efficiently set all these negatives to zero. I know that
>> I will likely be embarrassed when I see how to do it, but I can't seem
>> to remember or figure it out. I should emphasize that because of the
>> size of the data set, this needs to be done efficiently. Another
>> programming language does it as follows:
>>
>> 		x(x < 0) = 0;
>
> Here is a couple of solutions. They works fine but speaking about
> efficiency they are about 70 times *slower* than the vectorization you
> used with the other product.
>
> First, we create a small set of data to show the principle.
>
> data = RandomReal[{-10, 100}, {6, 4}]
>
> {{90.6031, 16.644, 15.2568, 88.4432}, {95.3404, -0.391179, 22.6264,
>    41.0332}, {18.7866, 90.8717, 48.073, 59.3251}, {24.2224, 21.1771,
>    91.7082, 50.719}, {96.9408, 27.4581, 56.9265, 2.22925}, {31.6366,
>    0.266302, 68.7124, 7.80917}}
>
> Then we use a replacement rule,
>
> data /. x_ /; x < 0 -> 0.
>
> {{90.6031, 16.644, 15.2568, 88.4432}, {95.3404, 0., 22.6264,
>    41.0332}, {18.7866, 90.8717, 48.073, 59.3251}, {24.2224, 21.1771,
>    91.7082, 50.719}, {96.9408, 27.4581, 56.9265, 2.22925}, {31.6366,
>    0.266302, 68.7124, 7.80917}}
>
> We can also do it we *Cases*,
>
> Cases[data, x_ /; x < 0 -> 0., {-1}]
>
> {0.}
>
> Now we test both method on a matrix of doubles of the size you
> specified, and check the time spent in seconds.
>
> data = RandomReal[{-10, 100}, {64000, 583}];
> Timing[data /. x_ /; x < 0 -> 0.;][]
> Timing[Cases[data, x_ /; x < 0 -> 0., {-1}];][]
>
> 62.046
>
> 49.797
>
> In comparison, a similar replacement on a similar matrix done with the
> other product takes less than a second.
>
>   >> x = -10 + (100 - (-10)).*rand(64000,583);
>   >> tic; x(x < 0) = 0; toc
>   Elapsed time is 0.867847 seconds.
>   >> whos x
>     Name          Size                 Bytes  Class     Attributes
>
>     x         64000x583            298496000  double
>
> I am confident that we can improve the performances for Mathematica; but
> I draw a blank right now (though I suspect something is going on with
> the packed array technology used by Mathematica).

Thanks to Carl Woll's clipping method, Mathematica is now faster than
the other product.

In:= x =.
Timing[x = RandomReal[{-10, 100}, {64000, 583}];][]
Timing[x = Clip[x, {0, \[Infinity]}];][]

Out= 2.031

Out= 0.656

>> tic; x = -10 + (100 - (-10)).*rand(64000,583); toc
Elapsed time is 2.225936 seconds.
>> tic; x(x < 0) = 0; toc
Elapsed time is 0.897022 seconds.

(Tested on Windows with the latest version of both products as of today.)
--
Jean-Marc

```

• Prev by Date: Re: Re: get help info into my program
• Next by Date: Re: FindRoot and Bose-Einstein distribution
• Previous by thread: Re: Re: Setting Negatives to Zero
• Next by thread: Re: Re: Re: Setting Negatives to Zero