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Re: Need speed partitioning WeatherData

  • To: mathgroup at smc.vnet.net
  • Subject: [mg112525] Re: Need speed partitioning WeatherData
  • From: "Hans Michel" <hmichel at cox.net>
  • Date: Sat, 18 Sep 2010 07:27:10 -0400 (EDT)

Try

Apply or Map AboluteTime to your list
In[1]:= AbsoluteTime[{1990,10,1,0,0,0}]
Out[1]= 2863728000


And used many of the grouping, select and partition functions available in
Mathematica. You can even use subtraction.

Hans
-----Original Message-----
From: P. Fonseca [mailto:public at fonseca.info] 
Sent: Friday, September 17, 2010 5:41 AM
To: mathgroup at smc.vnet.net
Subject: [mg112525] [mg112488] Need speed partitioning WeatherData

Hi,

Meanwhile I figured out that the problem comes from the time  
DateDifference takes.

One process of DateDifference costs 0.0045 seconds on my computer (ex.  
DateDifference[{1990, 10, 1, 0, 0, 0}, {1999, 10, 1, 0, 0, 0}]).

In all the different algorithms I implemented, each one of the 500 000  
samples, is "DateDifferenced" against around 3 of its neighbors. This  
means:

3 * 500 000 * 0.0045 = 6750 seconds !!!!!


I can now imagine a way of partitioning the data without applying so  
many times the DateDifference function: one first run to register the  
differences between every consecutive pair, and then I just work with  
these difference values.
Nevertheless, this still means 500 000*0.0045=2250 seconds!

Should I create my own DateDifference function (less options -> more  
fast)? Does someone has a simpler solution for a 100x speed up? (I  
already dropped the illusion of the couple of seconds...)

Thank you in advance,
Pedro




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