From: "Shot (Piotr Szotkowski)" <shot@...>
Date: 2009-11-24T02:12:41+09:00
Subject: Re: Finding the closest value from a matrix

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Mahadev Ittina:

> thanks for all the methods. I never
> realised there were so many ways to do it!

> I decided to do the 1st method. I have repeat the method
> a couple of more times than shown here.. Its a good way.

> reinforcement =3D [6, 8, 10, 12, 16, 20, 25, 32, 40]
> number =3D [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
> diameters =3D [8, 10, 12, 16]
> spacings  =3D [85, 90, 100, 125, 150, 175, 200, 225, 250, 275, 300]
> @shear_links =3D diameters.product(spacings).map { |d, s| [0.5 * Math::PI=
 * d ** 2 / s, d, s] }.sort.find { |x,| x > @asw_s }
> @compression =3D reinforcement.product(number).map { |r, n| [0.25 *  Math=
::PI * r ** 2 * n, r, n] }.sort.find { |x,|x > @comp_steel}
> @tension =3D reinforcement.product(number).map { |r, n| [0.25 * Math::PI =
*  r ** 2 * n, r, n] }.sort.find { |x,|x > @ten_steel}

How about some DRY-ing up of the code (and
not recomputing the arrays every time)? :)



def values rows, cols, &formula
  rows.product(cols).map do |row, col|
    [formula.call(row, col), row, col]
  end.sort
end

reinforcement =3D [6, 8, 10, 12, 16, 20, 25, 32, 40]
number        =3D [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
diameters     =3D [8, 10, 12, 16]
spacings      =3D [85, 90, 100, 125, 150, 175, 200, 225, 250, 275, 300]

@ds ||=3D values(diameters, spacings)   { |d, s| 0.5  * Math::PI * d ** 2 /=
 s }
@rn ||=3D values(reinforcement, number) { |r, n| 0.25 * Math::PI * r ** 2 *=
 n }

@shear_links =3D @ds.find { |x,| x > @asw_s      }
@compression =3D @rn.find { |x,| x > @comp_steel }
@tension     =3D @rn.find { |x,| x > @ten_steel  }



=E2=80=94 Shot
--=20
Your lust for power doomed 700 men to a watery grave. Yes, you sank
my battleship =E2=80=93 but at what cost to your soul? Now go to your room.
                                               [Joshua Green Allen]

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