5
You can wrap each element with Pane[#, BaselinePosition -> Top] &:
Panel[Row[
Pane[#, BaselinePosition -> Top] & /@ {grf, Spacer[50], tbl}],
Background -> LightBlue]
4
Something like this?
Grid[Partition[Graphics[{#, Rectangle[{0, 0}, {1, 1}]},
PlotRange -> {{0, 1}, {0, 1}},
ImageSize -> 20] & /@ Map[ColorData[64],
Range[100]], UpTo[16], 16], Spacings -> {0, 0}]
or
Grid[Partition[Graphics[{#, Rectangle[{0, 0}, {1, 1}],
White, Inset["a", Center]},
PlotRange -> {{0, 1}, {0, ...
4
A zero-order interpolation is a good way to in effect invert the function from positive integers to values in sortedlist.
Timing[
table = Table[N[n^2 + m^2], {n, 1, 3000},
{m, 1, 3000}];
sortedlist = Sort[Apply[Join,
table[[1 ;; 50]]]];
lastpairs =
SplitBy[Join[{{0, 0}},
Transpose[{sortedlist,
Range[Length[sortedlist]]}]],
First][...
3
Here are some suggestions for simpler methods to find the counts, and to make a plot.
Instead of Array[table[[#]]&, 50]] use table[[;; 50]]. Check the documentation for Part.
An easy way to count the number of values that are less or equal to $x$, is to count the values with Tally, and then total the tallies with Accumulate.
table = Table[N[n^2 + m^2], {...
3
It's faster and cleaner not to use any For loops and use list processing whenever possible:
nn = 6;
x = Cos[Subdivide[nn]*Pi]
(* {1, Sqrt[3]/2, 1/2, 0, -1/2, -Sqrt[3]/2, -1} *)
d1 = Outer[Subtract, x, x]
(* {{0, 1 - Sqrt[3]/2, 1/2, 1, 3/2, 1 + Sqrt[3]/2, 2},
{-1 + Sqrt[3]/2, 0, -(1/2) + Sqrt[3]/2, Sqrt[3]/2, 1/2 + Sqrt[3]/2, Sqrt[3], 1 + ...
3
crystalsystemnames = {"Cubic", "Tetragonal", "Orthorhombic",
"Hexagonal", "Monoclinic", "Rhombohedral", "Triclinic"};
metrictensors = Partition[{g1, g2, g3, g4, g5, g6, g7}, 1];
data = MapThread[Append, {metrictensors, crystalsystemnames}];
MatrixForm[MapAt[MatrixForm, data, {All, 1}...
2
Maybe this is a bit closer to what you want to do.
RF = Table[
Plus[
f[d, b, a, c],
Conjugate[f[c, a, b, d]],
If[b == d, -Sum[f[a, i, i, c], {i, 1, n}], 0],
If[a == c, -Conjugate[Sum[f[b, i, i, d], {i, 1, n}]], 0]
]
, {a, 1, n}, {b, 1, n}, {c, 1, n}, {d, 1, n}];
Note that I use Plus instead of + only because it is more legible of ...
2
If @Rom38's pure-function definition looks a bit esoteric, here's a more straightforward definition to the same effect:
rij[ri_, rj_, kij_] = ri/2 + dri + rj/2 + drj + kij + 10 Log[ss/(1 lij)];
Try it out:
rij[1, 2, 3]
(* 9/2 + dri + drj + 10 Log[ss/lij] *)
Of course you'll have to define the other parameters (dri, drj, ss, lij) as well.
2
You should make a function:
rij=#1 /2 + dri + #2 /2 + drj + #3 + 10 Log[ss/(1 lij)]&
And further call it by rij[ri,rj,kij], substituting the desired values of ri, rj and kij which are associated with numbered slots for arguments (#1,#2,#3) of the function
2
It suffices to Flatten the Table[...] :
Uexact[x_, t_] := Exp[x + t]
Uapprox[x_, t_] := 0.1*Exp[x + t]
Grid[Prepend[
Flatten[
Table[{x, t, Uexact[x, t], Uapprox[x, t],
Abs[Uapprox[x, t] - Uexact[x, t]]}, {x, -1, 1, 0.2}, {t, 0, 1,
0.1}]
, 1]
, {"x", "t", "Approx", "...
1
A possible solution using GraphicsRow:
x = N[2 Pi/24] Range[0, 5];
data = Transpose[{x, Chop@Sin[x]}];
grf = Plot[Sin[x], {x, 0, 2 Pi},
Frame -> True,
ImageSize -> 400,
AspectRatio -> 0.75,
ImagePadding -> {{80, 20}, {15, 5}}
];
title = Item[Style["sine data", 18, Blue], Alignment -> Center];
hdr = Item[Style[#, Bold,...
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