Re: [Jmol-users] why do I get avereaged values?

[Robert H. <ha...@st...>]

You're forgetting that

({1.53}[1][17]).elemno

is an average.

You are using array methods, so add ".all" to those:



print ({1.53}[1][17]).element.all.add("\t").add(({1.53}[1][17]).elemno.all
).add("\t").add(({1.53}[1][17]).x.all).add("\t").add(({1.53}[1][17]).y.all
).add("\t").add(({1.53}[1][17]).z.all)


but it might be simpler to use:

print {1.53}[1][17].label("%[element]\t%[elemno]\t%x\t%y\t%z")

in which case the formatting could be nicer:

print {1.53}[1][17].label("%[element]\t%[elemno]\t%5.3x\t%5.3y\t%5.3z")

or even

print {1.53}[1][17].label("%-3[element]%4.0[elemno]%10.3[xyz]")


Bob


On Wed, Dec 18, 2013 at 10:32 AM, Whitwell, George <GWh...@nc...>wrote:

>  when I use a command like:
>
>
>
> print
> ({1.53}[1][17]).element.add("\t").add(({1.53}[1][17]).elemno).add("\t").add(({1.53}[1][17]).x).add("\t").add(({1.53}[1][17]).y).add("\t").add(({1.53}[1][17]).z)
>
>
>
> result:
>
> C 2.4705882 -0.08696627 -0.8534151 1.8308194
> C 2.4705882 -0.08696627 -0.8534151 1.8308194
> C 2.4705882 -0.08696627 -0.8534151 1.8308194
> C 2.4705882 -0.08696627 -0.8534151 1.8308194
> C 2.4705882 -0.08696627 -0.8534151 1.8308194
> H 2.4705882 -0.08696627 -0.8534151 1.8308194
> H 2.4705882 -0.08696627 -0.8534151 1.8308194
> H 2.4705882 -0.08696627 -0.8534151 1.8308194
> H 2.4705882 -0.08696627 -0.8534151 1.8308194
> H 2.4705882 -0.08696627 -0.8534151 1.8308194
> H 2.4705882 -0.08696627 -0.8534151 1.8308194
> H 2.4705882 -0.08696627 -0.8534151 1.8308194
> H 2.4705882 -0.08696627 -0.8534151 1.8308194
> H 2.4705882 -0.08696627 -0.8534151 1.8308194
> H 2.4705882 -0.08696627 -0.8534151 1.8308194
> H 2.4705882 -0.08696627 -0.8534151 1.8308194
> H 2.4705882 -0.08696627 -0.8534151 1.8308194
>
>
>
> the atoms really are different, evidenced by C and H in the output or as
> produced by:
>
>  print
> ({1.53}[1]).element.add("\t").add(({1.53}[1]).elemno).add("\t").add(({1.53}[1]).x).add("\t").add(({1.53}[1]).y).add("\t").add(({1.53}[1]).z)
>
>
>
> result:
> C 6 0.5630599 -1.1161811 1.47227
>
>
>
> or:
>
> print data({1.53},"xyz")
>
>
>
> result:
> 17
>
> C     0.56306   -1.11618    1.47227
> C     0.57693   -2.34639    2.38065
> C    -0.82581   -0.78767    0.89831
> C     1.22154    0.08281    2.15912
> C    -1.84993   -0.20657    1.87432
> H     1.18245   -1.36436    0.61213
> H     1.59096   -2.59017    2.67957
> H     0.16333   -3.21266    1.87410
> H     0.00023   -2.18628    3.28660
> H     1.22370    0.95445    1.51206
> H     2.25210   -0.14288    2.41288
> H     0.71077    0.35029    3.07788
> H    -1.23407   -1.69083    0.45199
> H    -0.69344   -0.08274    0.08138
> H    -2.79985   -0.05883    1.37170
> H    -1.53391    0.75503    2.26115
> H    -2.02649   -0.86507    2.71782
>
>
>
> I've tried other approaches to producing data with the "symbol, atno, x,
> y, z" format, such as:
>
> modnum = 15
>
> atnum = 17
>
> for (var i=0;i<atnum;i++) {
>
> thisatom = "({1."+@modnum+"}["+@i+"])"
>
> print
> @thisatom.element.add("\t").add(thisatom.elemno).add("\t").add(thisatom.x).add("\t").add(thisatom.y).add("\t").add(thisatom.z)}
>
>
>
> with and w/o @ on thisatom, with no joy; however:
>
>
>
> print thisatom
>
>
>
> in the for loop does return a series of 17 ({1.53}[i]) style atom
> expressions.
>
>
>
> Hoping for new insight,
>
>
>
> George Whitwell
>
> NCWC Chemistry
>
>
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-- 
Robert M. Hanson
Larson-Anderson Professor of Chemistry
St. Olaf College
Northfield, MN
http://www.stolaf.edu/people/hansonr


If nature does not answer first what we want,
it is better to take what answer we get.

-- Josiah Willard Gibbs, Lecture XXX, Monday, February 5, 1900