| Re: multiplication [message #19615 is a reply to message #19427] |
Tue, 28 March 2000 00:00   |
meron
Messages: 51
Registered: July 1995
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In article <38E0A379.34ADB7F7@wizard.net>, James Kuyper <kuyper@wizard.net> writes:
> meron@cars3.uchicago.edu wrote:
>>
>> In article <38E03BDC.868B8396@hotmail.com>, marc <m_schellens@hotmail.com> writes:
>>> Is there a function like TOTAL but for multiplication.
>>> Like the big PI symbol in mathematical notation.
>>> Or this really something for the for loop?
>>>
>>> I.E.
>>>
>>> a=[1,2,3,...]
>>>
>>> result=a[1]*a[2]*a[3]...
>>>
>> if all the elements of a are positive then you can simply do
>>
>> result = exp(total(alog(a)))
> ...
>> If some of the elements are negative, you can still handle it. do
>>
>> dum = where(a lt 0, ndum)
>> sig = (-1)^ndum
>> result = sig*exp(total(alog(abs(a))))
>
> You can't honestly be suggesting that this is a good technique?
Good? No, only not as bad as using "for".
> Ignore for a momement what happens if any element of 'a' is 0.
That's the easiest to deal with. You're already checking for presence
of negative elements, can check for zeroes as well. That should be
the first thing, in fact, since if even one of the elements is 0, then
the result is 0 and you can dispense with the rest of the evaluation.
> That code performs two transcendental function evaluations per element
> of 'a'.
Yep, indeed.
> IDL would have to be very badly engineered (which I suppose is possible),
> for a 'for' loop to execute more slowly than your code.
Well, I run a quick test, comparing the time it takes tto evaluate the
product using both methods (it run on an old Vms Alpha, somebody may
want to repeat it on a more modern platform. Being lazy, I'm simply
filling an array with a constant element, then doing the
multiplication. Here is the output
IDL> speed, 1.00001, 100, 10
"for" time = 0.0012000084 res = 1.00100
"exp-log" time = 0.00019999743 res = 1.00100
IDL> speed, 1.00001, 1000, 10
"for" time = 0.012699997 res = 1.01006
"exp-log" time = 0.0012000084 res = 1.01006
IDL> speed, 1.00001, 10000, 10
"for" time = 0.12589999 res = 1.10532
"exp-log" time = 0.011699998 res = 1.10532
IDL> speed, 1.00001, 100000, 10
"for" time = 1.2583000 res = 2.72191
"exp-log" time = 0.12850000 res = 2.72198
The first input to SPEED is the array element, the second is the
length of the array. the third is just telling SPEED how many times to
repeat the test. As you can see, the above was tried for arrays with
lengths ranging from 100 to 100000 and calculation using "for" loop is
consistently an order of magnitude slower.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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