Plot different conditions & subjects at once in Mathematica

Question

Please Consider :

Needs["ErrorBarPlots`"];

fixNumberF1F6 = {{{7.11`, 7.51`, 11.14`, 8.19`, 6.58`}, 
                  {2.14`, 2.33`,2.25`, 1.53`,1.71`}},               
                  {{4.69`, 4.79`, 3.78,4.34`, 4.8`}, 
                   {2.22`, 2.71`, 3.18`, 2.29`, 1.93`}}}

fixNumberF1F6[[1,1]] refers to the mean fixation number for each subject for condition 1, fixNumberF1F6[[1,2]] the standard Deviation of those means. fixNumberF1F6[[2]] refers to condition 2.

plotOptionsXX[title_, yName_, xName_, colors_: Black] :=

  {Frame -> {{True, False}, {True, False}},
   PlotStyle -> colors,
   PlotLabel -> 
   Style[title, Bold, 14, Opacity[1], FontFamily -> "Helvetica"],
   PlotStyle -> Directive[Black, PointSize[Medium]],
   PlotRangePadding -> 0,
   Frame -> {{True, False}, {True, False}},
   LabelStyle -> Directive[Black, Bold, 12],
   FrameLabel -> {{Style[yName, Opacity[1]], 
   None}, {Style[xName, Opacity[1]], None}},
   FrameStyle -> Opacity[0],
   FrameTicksStyle -> Opacity[1],
   AxesStyle -> Directive[Black, 12],
   ImageSize -> laTaille};

ErrorListPlot[fixNumberF1F6[[#]] // Transpose,
              PlotRange -> {{0, 6}, {0, 15}},
              ImageSize -> 300,
              FrameTicks ->{{Range[1, 13, 2], None},{{1, 2, 3, 4, 5}, None}},
              PlotStyle -> Directive[Black, AbsolutePointSize[10], AbsoluteThickness[2]],
              plotOptionsXX["Mean Fixation number", "Fixation Nuber", "SubNo"]] & /@ Range[2]

The Plot On the Left represent each subject average fixation number along with the standard deviation for condition 1. On the right for condition 2.

How could I plot them both on 1 plot ?

If this :

fixNumberF1F6across = {{8.10, 1.99}, {4.48, 2.46}}

was the mean and SD across subject, How could I show both ?

-How can I show 2 conditions on a similar plot for different subjects -How could I show the group mean and SD on it to.

Answer 1

Edit:

I started from over. Given how the data are simple and clean, it may be easiest to use ListPlot and add the bars via an Epilog.

You can still tweak it a bit--e.g. put a slight space between the blue and red data points and bars, add a legend, etc, but the basic idea is there.

data = {{{7.11`, 7.51`, 11.14`, 8.19`, 6.58`}, {2.14`, 2.33`, 2.25`,   1.53`, 1.71`}}, {{4.69`, 4.79`, 3.78, 4.34`, 4.8`}, {2.22`,  2.71`, 3.18`, 2.29`, 1.93`}}};

ListPlot[{data[[1, 1]], data[[2, 1]]},
   PlotStyle -> {{PointSize[.025], Red}, {PointSize[0.025], Blue}},
   Frame -> True, 
   PlotRange -> {{0.5, 5.5}, {0, 14}},
   FrameTicks -> {{Automatic, Automatic}, {Range[5], None}},
   FrameLabel -> {{"Fixation (ms)", None}, {"Subject", None}},
   Epilog -> {{Red, Thickness[0.003], Dashed, 
      Line[{{0, m1 = Mean@data[[1, 1]]}, {5.5, m1}}],
      Blue, Line[{{0, m1 = Mean@data[[2, 1]]}, {5.5, m1}}]},
      Thickness[0.005], Red, 
      Line[{{#[[1]], #[[2, 1]]}, {#[[1]], #[[2, 2]]}}] & /@ 
      Transpose@{Range[5], ({#[[1]] + #[[2]], #[[1]] - #[[2]]} & /@ 
      Transpose@data[[1]])},
      Thickness[0.005], Blue, 
      Line[{{#[[1]], #[[2, 1]]}, {#[[1]], #[[2, 2]]}}] & /@ 
      Transpose@{Range[5], ({#[[1]] + #[[2]], #[[1]] - #[[2]]} & /@ 
      Transpose@data[[2]])},
      }]

The BoxWhiskerChart below is from your data. If this looks vaguely like something you are interested in, it could be modified so that the spread from the 25th percentile to the 75% percentile is altered to reflect the spread of one s.d. above and below the mean.

And, yes, it is easy to overlay the group means (N=5) onto the Chart.

[The reason there isn't perfect symmetry around the mean is that I used your means and standard deviations to generate raw data, assuming a normal distribution. I only used 100 data points per trial, so a little skewing is natural. This would not happen if we were to tweak the chart to reflect standard deviations, which are symmetrical.]