       Re: Assign new values to matrix using indices

• To: mathgroup at smc.vnet.net
• Subject: [mg99771] Re: Assign new values to matrix using indices
• From: "Sjoerd C. de Vries" <sjoerd.c.devries at gmail.com>
• Date: Thu, 14 May 2009 01:40:00 -0400 (EDT)
• References: <gue2k2\$7ll\$1@smc.vnet.net>

```Hi Mac,

(m - LowerTriangularize[m, -1]/2)

does the trick.

Cheers -- Sjoerd

On May 13, 11:07 am, Mac <mwjdavid... at googlemail.com> wrote:
> This simple problem has got me confounded (despite 2 years of
> Mathematica experience). Say I would like to divide all lower diagonal
> elements of a matrix by 2. It is fairly easy to generate a list of
> indices, here an example for 3 by 3 matrix
>
> In:=
> Table[{j, i}, {j, 2, cdim = 3}, {i, 1, j - 1}] // Flatten[#, 1] &
>
> Out= {{2, 1}, {3, 1}, {3, 2}}
>
> Now if I would like to divide the elements of a 3 x 3 matrix by 2, I
> don't know how to do this. The problem is the complex interplay of Part
> [], Set[] and Apply[]. To set the elements of hte matrix I can do
> this
>
> In:= t2 = {{0, 0, 0}, {2, 0, 0}, {2, 2, 0}};
> f = Function[{x}, Apply[Set[Part[t2, ##], 1] &, x]];
> Scan[f, {{2, 1}, {3, 1}, {3, 2}}]
> t2
>
> Out= {{0, 0, 0}, {1, 0, 0}, {1, 1, 0}}
>
> which seems complicated for a simple operation. However, to divide all
> elements with indices given in the form above i.e. {{2, 1}, {3, 1},
> {3, 2}} by 2 has got me stumped.
>
> Any help would be appreciated
>
> Mac

```

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