It can be simplified a bit, though, as the second operand in the multiplication
does not need to be a matrix:
y * rep(z,each=3)
On 19 Nov 2014, at 16:24 , Jeff Newmiller <jdnew...@dcn.davis.ca.us> wrote:
> When your matrices are the same size, the "*" operator does what you want.
> The problem is that you have to make a conforming version of z before you can
> use that operator.
>
> y*matrix(rep(z,3),ncol=2,byrow=TRUE)
>
> or
>
> y*matrix(rep(z,each=3),ncol=2)
>
> To interpret this, just keep in mind that matrices are folded vectors in R...
> every matrix can be thought of as a linear vector of columnwise data with
> dimension attributes.
>
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> Sent from my phone. Please excuse my brevity.
>
> On November 19, 2014 6:43:55 AM PST, "Ruima E." <ruimax...@gmail.com> wrote:
>> Thank you Chel Hee.
>>
>> Isn't there a simpler way to do so?
>>
>>
>> On Wed, Nov 19, 2014 at 3:35 PM, Chel Hee Lee <chl...@mail.usask.ca>
>> wrote:
>>>> y = matrix(cbind(c(0, 0.5, 1),c(0, 0.5, 1)),ncol=2)
>>>> z = matrix(c(12, -6),ncol=2)
>>>> t(apply(y, 1, function(x) x*z))
>>> [,1] [,2]
>>> [1,] 0 0
>>> [2,] 6 -3
>>> [3,] 12 -6
>>>
>>> I hope this helps.
>>>
>>> Chel Hee Lee
>>>
>>> On 14-11-19 08:22 AM, Ruima E. wrote:
>>>> Hi,
>>>>
>>>> I have this:
>>>>
>>>> y = matrix(cbind(c(0, 0.5, 1),c(0, 0.5, 1)),ncol=2)
>>>> z = matrix(c(12, -6),ncol=2)
>>>>
>>>> In matlab I would do this
>>>>
>>>>> y .* x
>>>> I would get this in matlab
>>>>
>>>>> ans
>>>> 0 -0
>>>> 6 -3
>>>> 12 -6
>>>>
>>>> What is the equivalent in R?
>>>>
>>>> Thanks
>>>>
>>>> [[alternative HTML version deleted]]
>>>>
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>>>
>>
>> ______________________________________________
>> R-help@r-project.org mailing list
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>
> ______________________________________________
> R-help@r-project.org mailing list
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> and provide commented, minimal, self-contained, reproducible code.
>
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