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Re: simple array math question [message #33688 is a reply to message #33687] Fri, 17 January 2003 14:10 Go to previous messageGo to previous message
Sean Raffuse is currently offline  Sean Raffuse
Messages: 46
Registered: July 2001
Member
"JD Smith" <jdsmith@as.arizona.edu> wrote in message
news:pan.2003.01.17.21.46.58.749790.29842@as.arizona.edu...
> On Thu, 16 Jan 2003 20:47:46 -0700, Craig Markwardt wrote:
>
>
>> Heinz Stege <reply_to_posting@arcor.de> writes:
>>> On Thu, 16 Jan 2003 14:05:27 -0600, "Sean Raffuse" <sean@me.wustl.edu>
>>> wrote:
>>>
>>>> >> a=[[1,2,3],[4,5,6],[7,8,9]]
>>>>
>>>> >> b=[1,2,3]
>>>>
>>>> What is the best (read, fastest) way to multiply b by each individual
>>>> row of a? I would like to return a result of:
>>>>
>>>> [[1,4,9],[4,10,18],[7,14,27]]
>>>
>>>
>>> result=a*b(*,intarr(3))
>>
>> WOW! I've never seen that! It scares me how cool that is. :-)
>>
>> Craig
>
>
> I may have to add that to the REBIN/REFORM tutorial. I'll see how fast
> it is first. It's definitely one of the more readeable ways to add a new
> trailing dimension. Doesn't work for leading or in-the-middle
> dimensions, as far as I can tell.
>
> JD

Yes, it worked wonders for me. Plenty fast compared to what I was doing
before. I did have to use it on some leading dimensions and so am using
transpose(). Not sure if it is the optimal solution, but good enough for my
needs.
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