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# Re: st:weave two matrices row by row

 From Amanda Fu To statalist@hsphsun2.harvard.edu Subject Re: st:weave two matrices row by row Date Mon, 27 Dec 2010 15:19:41 -0500

```Dear Mr. Chakravarty,

Thank you for the reply! It is really smart to use matrix multiplier
to deal with this. I appreciate it.

Amanda

On Sun, Dec 26, 2010 at 9:43 PM, Tirthankar Chakravarty
<tirthankar.chakravarty@gmail.com> wrote:
> Matrix multiplication is probably the most memory-efficient and
> generalisable way of doing this:
> *************************************
> clear*
> mata
> mA = runiform(10, 8)
> mB = runiform(10, 8)
> mD = (designmatrix(range(1, 19, 2))'\J(1, 10, 0))*mA +
> designmatrix(range(2, 20, 2))'*mB
> mD
> end
> *************************************
> T
>
> On Sun, Dec 26, 2010 at 4:34 PM, Amanda Fu <mandy.fu1@gmail.com> wrote:
>>
>> Hi All,
>>
>> I was wondered if there is any way to weave two matrix row by row.
>> Say, I have matrix A [10,8] and B[10,8]. I want to create a new matrix
>> C looks like as:
>>
>> matrix C [20,8]:
>> ---------------------------------
>> row 1 :        row 1 of A
>> row 2:         row 1 of B
>> row 3:         row 2 of A
>> row 4:         row 2 of B
>> ..........
>> row 19:      row 10 of A
>> row 20:      row 10 of B
>> ---------------------------------
>> Right now I think I could do the following:
>>
>> (1) matrix define C as a 20*8 matrix.
>> (2) use -mkmat- to change A and B into 8 variables respectively
>> A1,A2,..A8,B1,..,B8.
>> (3) Then define each row using the variables from (2).
>>
>> But  when actually conducting the above way,I find it requires to be
>> very careful and  much more work than I expected. This is why I think
>> it would be better that I ask the expert here to see how an expert
>> will deal with this.
>>
>>
>> Sincerely,
>> Amanda
>> *
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>
>
>
> --
> To every ω-consistent recursive class κ of formulae there correspond
> recursive class signs r, such that neither v Gen r nor Neg(v Gen r)
> belongs to Flg(κ) (where v is the free variable of r).
>
> *
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>

*
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```