I am trying to use Mathematica to do some data analysis. I am trying to label points in a plot of a dataset. I know that Callout should work, but can't for the life of me figure out how to make it when I'm plotting b against a and then labeling the points with c.

Here is a subset of the dataset:

dataset = 
   <|"a" -> 0.0, "b" -> -1.26, "c" -> 1|>,
   <|"a" -> 0.3, "b" -> -1.25, "c" -> 2|>,
   <|"a" -> 0.6, "b" -> -1.26, "c" -> 3|>,
   <|"a" -> 0.9, "b" -> -1.19, "c" -> 4|>,
   <|"a" -> 1.2, "b" -> -0.8, "c" -> 5|>,
   <|"a" -> 1.5, "b" -> -0.98, "c" -> 6|>}]

ListLinePlot[Callout[dataset[All, {1, 2}], dataset[All, 3]]]

enter image description here

This doesn't give me what I want as I want to label each point.

I realize I'm missing something fundamental in terms of associating each point. Any assistance is appreciated.

Thanks for the two solutions. I want to modify the plot by assigning the plot to "g" and then using Show:

Show[g, ScalingFunctions -> "Reverse", PlotTheme -> "Detailed"]

but do not see how to do it. I tried using Show without success.

  • 2
    $\begingroup$ Does dataset[Map[Callout[{#a, #b}, #c] &] /* ListLinePlot] do what you want? $\endgroup$ – J. M.'s torpor Feb 12 at 14:43
  • $\begingroup$ Yes. Thanks. Hard to get my head around some of Mathematica's cryptic code. $\endgroup$ – reo Feb 12 at 14:58
  • 1
    $\begingroup$ Alright, I'll post an answer later... $\endgroup$ – J. M.'s torpor Feb 12 at 15:10
  • 6
    $\begingroup$ @J.M.'sennui That could also be written as dataset[ListLinePlot, Callout[{#a, #b}, #c] &]. $\endgroup$ – WReach Feb 12 at 15:39
  • $\begingroup$ @WReach, that's way less awkward; please write an answer so I can upvote it. :) $\endgroup$ – J. M.'s torpor Feb 12 at 15:42

As suggested by @J.M.~~___, here is one way to do it:

dataset[ListLinePlot, Callout[{#a, #b}, #c] &]

Additional options can be added to ListLinePlot like this:

  ListLinePlot[#, ScalingFunctions->"Reverse", PlotTheme->"Detailed"]&
, Callout[{#a, #b}, #c] &

ListLinePlot image


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