# How to plot graphs from multi-variable list by filtering results

Having 100,000 points stored for 6 variables in a list. Following filters are to be applied to the variables: m1>0, 10^(-6) < m1 < 10^(-1)

sine of delta and deltap has to be taken. This will reduce all the values between [-1,1]. thetas have to be between 0-90 degrees. So corresponding radian limits and take their sine as well. We will get values from [0,1].

Once done with filtered results, I need to plot any two variables of the resultant list against any other variables, so 6C2=15 cases possible. For instance, m1 vs sin(delta) or sin(theta12) vs sin(theta23). So how do I go about manipulating "result" to get the desired plots?

Do[dm12 = RandomReal[{Sqrt[7.03*10^-5], Sqrt[8.09*10^-5]}];
dm13 = RandomReal[{Sqrt[2.407*10^-3], Sqrt[2.643*10^-3]}];
sp12 = RandomReal[{Sqrt[0.271], Sqrt[0.345]}];
sp23 = RandomReal[{Sqrt[0.385], Sqrt[0.635]}];
sp13 = RandomReal[{Sqrt[0.01934], Sqrt[0.02392]}];
Assuming[{m1 \[Element] Reals, \[Delta] \[Element] Reals,
s12 \[Element] Reals, s23 \[Element] Reals,
s13 \[Element] Reals, \[Delta]p \[Element] Reals},
data[i] =
FindRoot[{Sqrt[dm12^2 + m1^2] s12 s13 sp12 sp13 +
Sqrt[dm13^2 + m1^2] s13 sp13 Cos[\[Delta] - \[Delta]p] +
m1 ((-1 + s12^2)^2)^(1/4) ((-1 + s13^2)^2)^(
1/4) ((-1 + sp12^2)^2)^(1/4) ((-1 + sp13^2)^2)^(1/4)
Cos[1/2 (Arg[1 - s12^2] + Arg[1 - s13^2] - Arg[1 - sp12^2] -
Arg[1 - sp13^2])] ==
0, -Sqrt[dm13^2 + m1^2] s13 sp13 Sin[\[Delta] - \[Delta]p] -
m1 ((-1 + s12^2)^2)^(1/4) ((-1 + s13^2)^2)^(
1/4) ((-1 + sp12^2)^2)^(1/4) ((-1 + sp13^2)^2)^(1/4)
Sin[1/2 (Arg[1 - s12^2] + Arg[1 - s13^2] - Arg[1 - sp12^2] -
Arg[1 - sp13^2])] ==
0, -m1 ((-1 + s12^2)^2)^(1/4)
s13 s23 sp12 sp23 Cos[\[Delta]p + 1/2 Arg[1 - s12^2]] -
m1 s12 ((-1 + s23^2)^2)^(1/4)
sp12 sp23 Cos[1/2 Arg[1 - s23^2]] -
Sqrt[dm12^2 + m1^2] ((-1 + s12^2)^2)^(1/4) ((-1 + s23^2)^2)^(
1/4) ((-1 + sp12^2)^2)^(1/4)
sp23 Cos[
1/2 (Arg[1 - s12^2] + Arg[1 - s23^2] - Arg[1 - sp12^2])] +
Sqrt[dm12^2 + m1^2] s12 s13 s23 ((-1 + sp12^2)^2)^(1/4)
sp23 Cos[\[Delta]p - 1/2 Arg[1 - sp12^2]] +
Sqrt[dm13^2 + m1^2] ((-1 + s13^2)^2)^(1/4)
s23 ((-1 + sp13^2)^2)^(1/4) ((-1 + sp23^2)^2)^(1/4)
Cos[1/2 (Arg[1 - s13^2] - Arg[1 - sp13^2] -
Arg[1 - sp23^2])] +
Sqrt[dm12^2 + m1^2] s12 s13 s23 sp12 sp13 ((-1 + sp23^2)^2)^(
1/4) Cos[\[Delta] - \[Delta]p + 1/2 Arg[1 - sp23^2]] -
Sqrt[dm12^2 + m1^2] ((-1 + s12^2)^2)^(1/4) ((-1 + s23^2)^2)^(
1/4) sp12 sp13 ((-1 + sp23^2)^2)^(1/4)
Cos[1/2 (2 \[Delta] - Arg[1 - s12^2] - Arg[1 - s23^2] +
Arg[1 - sp23^2])] +
m1 ((-1 + s12^2)^2)^(1/4) s13 s23 ((-1 + sp12^2)^2)^(1/4)
sp13 ((-1 + sp23^2)^2)^(1/4)
Cos[1/2 (2 \[Delta] - 2 \[Delta]p - Arg[1 - s12^2] +
Arg[1 - sp12^2] + Arg[1 - sp23^2])] +
m1 s12 ((-1 + s23^2)^2)^(1/4) ((-1 + sp12^2)^2)^(1/4)
sp13 ((-1 + sp23^2)^2)^(1/4)
Cos[1/2 (2 \[Delta] - Arg[1 - s23^2] + Arg[1 - sp12^2] +
Arg[1 - sp23^2])]
== 0,
m1 ((-1 + s12^2)^2)^(1/4)
s13 s23 sp12 sp23 Sin[\[Delta]p + 1/2 Arg[1 - s12^2]] +
m1 s12 ((-1 + s23^2)^2)^(1/4)
sp12 sp23 Sin[1/2 Arg[1 - s23^2]] +
Sqrt[dm12^2 + m1^2] ((-1 + s12^2)^2)^(1/4) ((-1 + s23^2)^2)^(
1/4) ((-1 + sp12^2)^2)^(1/4)
sp23 Sin[
1/2 (Arg[1 - s12^2] + Arg[1 - s23^2] - Arg[1 - sp12^2])] -
Sqrt[dm12^2 + m1^2] s12 s13 s23 ((-1 + sp12^2)^2)^(1/4)
sp23 Sin[\[Delta]p - 1/2 Arg[1 - sp12^2]] -
Sqrt[dm13^2 + m1^2] ((-1 + s13^2)^2)^(1/4)
s23 ((-1 + sp13^2)^2)^(1/4) ((-1 + sp23^2)^2)^(1/4)
Sin[1/2 (Arg[1 - s13^2] - Arg[1 - sp13^2] -
Arg[1 - sp23^2])] +
Sqrt[dm12^2 + m1^2] s12 s13 s23 sp12 sp13 ((-1 + sp23^2)^2)^(
1/4) Sin[\[Delta] - \[Delta]p + 1/2 Arg[1 - sp23^2]] -
Sqrt[dm12^2 + m1^2] ((-1 + s12^2)^2)^(1/4) ((-1 + s23^2)^2)^(
1/4) sp12 sp13 ((-1 + sp23^2)^2)^(1/4)
Sin[1/2 (2 \[Delta] - Arg[1 - s12^2] - Arg[1 - s23^2] +
Arg[1 - sp23^2])] +
m1 ((-1 + s12^2)^2)^(1/4) s13 s23 ((-1 + sp12^2)^2)^(1/4)
sp13 ((-1 + sp23^2)^2)^(1/4)
Sin[1/2 (2 \[Delta] - 2 \[Delta]p - Arg[1 - s12^2] +
Arg[1 - sp12^2] + Arg[1 - sp23^2])] +
m1 s12 ((-1 + s23^2)^2)^(1/4) ((-1 + sp12^2)^2)^(1/4)
sp13 ((-1 + sp23^2)^2)^(1/4)
Sin[1/2 (2 \[Delta] - Arg[1 - s23^2] + Arg[1 - sp12^2] +
Arg[1 - sp23^2])] == 0,
m1 ((-1 + s12^2)^2)^(1/4) s13 ((-1 + s23^2)^2)^(1/4)
sp12 sp23 Sin[
1/2 (2 \[Delta]p + Arg[1 - s12^2] + Arg[1 - s23^2])] -
Sqrt[dm12^2 + m1^2] ((-1 + s12^2)^2)^(1/4)
s23 ((-1 + sp12^2)^2)^(1/4)
sp23 Sin[1/2 (Arg[1 - s12^2] - Arg[1 - sp12^2])] -
Sqrt[dm12^2 + m1^2] s12 s13 ((-1 + s23^2)^2)^(
1/4) ((-1 + sp12^2)^2)^(1/4)
sp23 Sin[\[Delta]p + 1/2 Arg[1 - s23^2] -
1/2 Arg[1 - sp12^2]] -
Sqrt[dm13^2 + m1^2] ((-1 + s13^2)^2)^(1/4) ((-1 + s23^2)^2)^(
1/4) ((-1 + sp13^2)^2)^(1/4) ((-1 + sp23^2)^2)^(1/4)
Sin[1/2 (Arg[1 - s13^2] + Arg[1 - s23^2] - Arg[1 - sp13^2] -
Arg[1 - sp23^2])] +
Sqrt[dm12^2 + m1^2] ((-1 + s12^2)^2)^(1/4)
s23 sp12 sp13 ((-1 + sp23^2)^2)^(1/4)
Sin[\[Delta] - 1/2 Arg[1 - s12^2] + 1/2 Arg[1 - sp23^2]] +
Sqrt[dm12^2 + m1^2] s12 s13 ((-1 + s23^2)^2)^(1/4)
sp12 sp13 ((-1 + sp23^2)^2)^(1/4)
Sin[\[Delta] - \[Delta]p - 1/2 Arg[1 - s23^2] +
1/2 Arg[1 - sp23^2]] -
m1 s12 s23 ((-1 + sp12^2)^2)^(1/4) sp13 ((-1 + sp23^2)^2)^(1/4)
Sin[1/2 (2 \[Delta] + Arg[1 - sp12^2] + Arg[1 - sp23^2])] +
m1 ((-1 + s12^2)^2)^(1/4) s13 ((-1 + s23^2)^2)^(
1/4) ((-1 + sp12^2)^2)^(1/4) sp13 ((-1 + sp23^2)^2)^(1/4)
Sin[1/2 (2 \[Delta] - 2 \[Delta]p - Arg[1 - s12^2] -
Arg[1 - s23^2] + Arg[1 - sp12^2] + Arg[1 - sp23^2])] == 0,
m1 s12 s23 sp12 sp23 -
m1 ((-1 + s12^2)^2)^(1/4) s13 ((-1 + s23^2)^2)^(1/4)
sp12 sp23 Cos[
1/2 (2 \[Delta]p + Arg[1 - s12^2] + Arg[1 - s23^2])] +
Sqrt[dm12^2 + m1^2] ((-1 + s12^2)^2)^(1/4)
s23 ((-1 + sp12^2)^2)^(1/4)
sp23 Cos[1/2 (Arg[1 - s12^2] - Arg[1 - sp12^2])] +
Sqrt[dm12^2 + m1^2] s12 s13 ((-1 + s23^2)^2)^(
1/4) ((-1 + sp12^2)^2)^(1/4)
sp23 Cos[\[Delta]p + 1/2 Arg[1 - s23^2] -
1/2 Arg[1 - sp12^2]] +
Sqrt[dm13^2 + m1^2] ((-1 + s13^2)^2)^(1/4) ((-1 + s23^2)^2)^(
1/4) ((-1 + sp13^2)^2)^(1/4) ((-1 + sp23^2)^2)^(1/4)
Cos[1/2 (Arg[1 - s13^2] + Arg[1 - s23^2] - Arg[1 - sp13^2] -
Arg[1 - sp23^2])] +
Sqrt[dm12^2 + m1^2] ((-1 + s12^2)^2)^(1/4)
s23 sp12 sp13 ((-1 + sp23^2)^2)^(1/4)
Cos[\[Delta] - 1/2 Arg[1 - s12^2] + 1/2 Arg[1 - sp23^2]] +
Sqrt[dm12^2 + m1^2] s12 s13 ((-1 + s23^2)^2)^(1/4)
sp12 sp13 ((-1 + sp23^2)^2)^(1/4)
Cos[\[Delta] - \[Delta]p - 1/2 Arg[1 - s23^2] +
1/2 Arg[1 - sp23^2]] -
m1 s12 s23 ((-1 + sp12^2)^2)^(1/4) sp13 ((-1 + sp23^2)^2)^(1/4)
Cos[1/2 (2 \[Delta] + Arg[1 - sp12^2] + Arg[1 - sp23^2])] +
m1 ((-1 + s12^2)^2)^(1/4) s13 ((-1 + s23^2)^2)^(
1/4) ((-1 + sp12^2)^2)^(1/4) sp13 ((-1 + sp23^2)^2)^(1/4)
Cos[1/2 (2 \[Delta] - 2 \[Delta]p - Arg[1 - s12^2] -
Arg[1 - s23^2] + Arg[1 - sp12^2] + Arg[1 - sp23^2])] ==
0}, {\[Delta], \[Pi]/6}, {m1, 0.001}, {s12, 0.2}, {s23,
0.6}, {s13, 0.2}, {\[Delta]p, \[Pi]/3}]], {i, 0, 100000, 1}]
results = data /@ Range[100000]

This deals with the plotting part of the question.

results = RandomReal[100, {50, 6}];
labels = {"m1", "delta", "deltap", "theta12", "theta23", "theta13"};
plots = ListPlot[results[[All, #]], Frame -> True,
FrameLabel -> {labels[[#[[1]]]], labels[[#[[2]]]]}] & /@ Subsets[Range[6], {2}];
Grid[Partition[plots, 5], Dividers -> All]

A better alternative may be

Needs["StatisticalPlots`"]
PairwiseScatterPlot[results, DataTicks -> True, DataLabels -> labels]

** is there a way to plot one column versus the sine of another column?**

It is easy to modify first approach above:

plots2 = ListPlot[Transpose[{results[[All, #[[1]]]], Sin[results[[All, #[[1]]]]]}],
Frame -> True, FrameLabel -> {labels[[#[[1]]]], "sin(" <> labels[[#[[2]]]] <> ")"}] & /@
Subsets[Range[6], {2}];
Grid[Partition[plots2, 5], Dividers -> All]

TODO: Modify PairwiseScatterPlot to show Sin[y] on the vertical axes.

• Thank you! I have specific labels in place of Column 1/2/3/4/5/6. I'm unable to label them pairwise. As in the labels need to be "m1, delta, deltap, theta12, theta23, theta13". So that I obtain corresponding graphs for the gridwise plot. I was able to get the same for Scatterplot. Also, is there a way to plot one column versus the sine of another column?(This is totally optional). Commented Mar 20, 2018 at 4:37
• Thank you very much! I didn't get a notification update, so just saw this. Simplifies my work a lot! :) Commented Apr 5, 2018 at 11:25
• @EshanBhargava, my pleasure. Thank you for the accept.
– kglr
Commented Apr 5, 2018 at 11:29