Forest plot with Mathematica 9

Question

Does anyone have experience with or a package for the creation of forest plots in Mathematica 9?

For instance for subgroup analyses of cox model or meta-analysis (e.g rmeta package in R)?

Answer 1

You can work with RLink to create native R graphics - this and this examples. I guess forestplot is no different if you install related R packages. I may be far off here, but can't you replicate these plots with built-in Mathemtica functionality? - like these examples on "Compare the distribution of salaries for several departments at a university:"

salaries =  ExampleData[{"Statistics", "UniversitySalaries"}, "DataElements"];
depts = {"Mathematics", "History", "English", "Chemistry", "Law", "Physics", "Statistics"};
data = Table[Cases[salaries, {d, _, salary_, "A"} :> salary], {d, depts}];
all = Cases[salaries, {_, _, salary_, "A"} :> salary];

DistributionChart[data, 
 ChartLabels -> Placed[{depts, Length /@ data}, {Axis, Center}], 
 ChartStyle -> 54, GridLines -> {{{Median[all], Gray}}, None}, BarOrigin -> Left]

BoxWhiskerChart[data, {"Outliers", {"MedianMarker", 1, Directive[Thick, White]}}, 
 ChartLabels -> Placed[{depts, Length /@ data}, {Axis, Before}], 
 ChartStyle -> 10, GridLines -> {{{Median[all], Gray}}, None}, BarOrigin -> Left]

Answer 2

The plotting is a little more cumbersome than the computation assuming you want relative risk shown in your forest plot. As you point out, one can use CoxModelFit to compute everything that needs computed.

The larynx cancer data features the variables age and four levels of cancer stage.

larynx = ExampleData[{"Statistics", "LarynxCancer"}];    
x = larynx[[All, {1, 3}]];
e = EventData[larynx[[All, 2]], larynx[[All, 5]]];
mod = CoxModelFit[{x, e}, {stage, age}, {stage, age}, NominalVariables -> stage];

Creating the plot is a little ugly if you want a lot of control over the appearance. Here is a first crack at it using graphics primitives that allows the diamonds to scale with the plot.

forestPlot[mod_CoxModel] := 
  Block[{rr, rrci, vars, p}, 
   rr = mod["RelativeRisk"]; 
   rrci = mod["RelativeRiskConfidenceIntervals"]; 
   vars = mod["BasisFunctions"] /.
                  HoldPattern[DiscreteIndicator[s_, i_, _]] :> s[i]; 
   p = Length[vars]; 

   Graphics[Join[Join[{{Dashed, Line[{{1, 0}, {1, p + 1}}]}}, 
    MapThread[{Thick, Line[{{#1[[1]], #2}, {#1[[2]], #2}}]} &, 
      {rrci,Range[p]}]], 
    MapThread[{EdgeForm[{Thick, Black}], FaceForm[White], 
       Polygon[{{-.2 + #1, #2}, {#1, .1 + #2}, {.2 + #1, #2}, 
        {#1, -.1 + #2}}]} &, {rr, Range[p]}]], 
    AspectRatio -> 1, Frame -> True, 
    PlotRangePadding -> {1, 0}, 
    FrameTicks -> {Automatic, Transpose[{Range[p], vars}], None, None},
    GridLines -> Automatic, GridLinesStyle -> White
   ]
]

With the larnyx cancer data we get...

forestPlot[mod]

Answer 3

I appreciate that the (somewhat critical) commentary regarding the lack of specificity of my question.

I have written my own function for the plotting (once the calculational aspects from the model or meta-analysis have been collected).

forestplot[x_List, tcks_, opts : OptionsPattern[]] := 
Module[{lg, ln, int, mx, lines}, lg = Length[x]; ln = Range[lg]; 
mx = Max[x[[All, 3]]]; 
int = {Transpose[{x[[All, 1]], ln}], Transpose[{x[[All, 3]], ln}]}; 
lines = MapThread[{Black, Line[{#1, #2}]} &, int]; 
ListPlot[Transpose[{x[[All, 2]], ln}], 
PlotMarkers -> Style[\[DiamondSuit], 20], 
PlotRange -> {{-2, 1.1 mx}, {0, Length[x] + 1}}, 
Epilog -> {{Red, Line[{{1, 0}, {1, lg + 1}}], lines}, 
MapThread[Text[#1, {-1, #2}] &, {tcks, ln}]}, 
Axes -> {True, False}, Ticks -> {Table[j, {j, 0, mx, 1}], None}, 
Evaluate[FilterRules[{opts}, Options[ListPlot]]]]]

The form of x is {lower bound, point estimate, upper bound} and tcks is list of labels. There are formatting issues and I am yet to make a summary statistic and formatting for this. Further, options for plot markers could be customized.

The following utilizes the laryngeal cancer dataset and functions for extracting the point estimates, standard errors and then calculation of 95% confidence intervals.

I would suggest that for new and inexperienced (but interested) users like me asking experts regarding functionalities is within the purview and purpose of this exchange. Further, despite the nominal release of Mathematica 9, I am certain that many contributors to this exchange had access to its functionality (as I saw at IMS2012 in May in London).

Finally, comments and commentary that are personally critical or make assumptions regarding the ignorance or other attributes are at the very least unhelpful. (Having surveyed this site there are a spectrum of users with diverse backgrounds and aims. This diversity would appear to something to be encouraged rather than crushed. I have previously fully accepted the NARQ of a previous question I submitted. In my naive opinion, this question is legitimate. ).