Skip to main content

plotting - PlotLegends "Expressions" default for ListPlot?


gwr pointed out an apparent problem with setting a default Option value of PlotLegends for DateListPlot. I traced the problem to the existence of PlotLegends -> "Expressions" in the default PlotTheme. This is true for ListPlot as well. Since the explicit options within the Plot Theme override options set with SetOptions it appears that our option is ignored.


Actually "Expressions" is used across many Themes for ListPlot:


PlotLegends /. Charting`ResolvePlotTheme["Default", ListPlot] & /@
  {"Default", "Web", "Classic", "Detailed", "Scientific"}



{"Expressions", "Expressions", "Expressions", "Expressions", "Expressions"}

Why would this be? This option value doesn't seem to even work with these plot types, i.e.


dat = Table[{k, PDF[BinomialDistribution[50, p], k]}, {p, {0.3, 0.5, 0.8}}, {k, 0, 50}];

ListPlot[dat, Filling -> Axis, PlotLegends -> "Expressions"]

enter image description here



However it does cause the Themes to ignore a manually set default PlotLegends value:


SetOptions[ListPlot, PlotLegends -> PointLegend[Automatic, {"one", "two", "three"}]];

ListPlot[dat, Filling -> Axis]

enter image description here


Observe that turning off the Plot Theme causes the legend to suddenly appear:


 ListPlot[dat, Filling -> Axis, PlotTheme -> None]

enter image description here



Unless PlotLegends -> "Expressions" has some desirable effect I have failed to observe this seems like plainly undesirable behavior. Was the inclusion of PlotLegends -> "Expressions" in the Themes for ListPlot and related functions a mistake?


(All observations made in Mathematica 10.1.0 under Windows.)



Answer



"Expressions" for legend in ListPlot family is used because of Association. With Association input, the plot will automatically pick up the keys as legends. Here is an example:


ListPlot[<|"A"->{1,2,3},"B"->{2,3,4}|>]


enter image description here


SetOptions is to set the default values to options but it doesn't guarantee that the settings will be used by the plot since PlotTheme has been introduced.


The workaround would be to pack a set of options you would like to use through out the session in a variable and use it instead of using SetOptions.


Comments

Post a Comment

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...
Post a Comment
Read more

list manipulation - Selecting multiple columns from a matrix?

Sample data: data = { {{2013, 1, 1}, 24.13, 167.67, 231.82}, {{2013, 1, 2}, 32.15, 170.92, 225.99}, {{2013, 1, 3}, 35.43, 172.68, 221.67}, {{2013, 1, 4}, 36.73, 173.05, 218.32}, {{2013, 1, 5}, 58.19, 165.96, 197.05}, {{2013, 1, 6}, 69.99, 163.50, 187.52}, {{2013, 1, 7}, 71.37, 154.21, 175.58}, {{2013, 1, 8}, 72.51, 149.66, 163.25}}; I want a DateListPlot with three graphs, so for a matrix formed by columns 1 and 2, one for columns 1 and 3, and 1 for columns 1 and 4. At the moment I'm using this code: data2 = Transpose[{data[[All, 1]], data[[All, 2]]}]; data3 = Transpose[{data[[All, 1]], data[[All, 3]]}]; data4 = Transpose[{data[[All, 1]], data[[All, 4]]}]; DateListPlot[{data2, data3, data4}, Joined -> True, Filling -> {3 -> {1}}] but I have a hunch that this can be done more efficiently. I don't like the Transpose s in particular. Any ideas? edit (for extra credit) What if I need to multiply the second column by 2, which in my solution is simp...
Post a Comment
Read more

calculus and analysis - Integral of the Sinc product

Let us consider the following integral $$ B_n = \int_0^\infty \prod_{k=1,3,5,\dots}^n\frac{\sin (x/k)}{x/k}dx $$ By definition, Sinc[x] == Sin[x]/x , therefore B[n_?OddQ] := Integrate[Product[Sinc[x/k], {k, 1, n, 2}], {x, 0, Infinity}] B /@ Range[1, 13, 2] {Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, Pi/2} OK, everything is fine. But... B[15] 467807924713440738696537864469 Pi/935615849440640907310521750000 What's going on? Answer As Eckhard wrote in comments B[n] is the n-th Borwein integral . (The letter B was not accidental :) ) This funny properties of Borwein integrals is related to the Fourier transform of Sinc function FourierTransform[Sinc[x], x, k] 1/2 Sqrt[Pi/2] (Sign[1 - k] + Sign[1 + k]) Plot[%, {k, -2, 2}, Filling -> 0] which is the box function. The result is $\pi/2$ while the sum $$ 1/3+1/5+\dots+1/n $$ B_n = \frac{\pi}{2} - \pi \bigg(\sum_{k=3,5,\ldots}^n\frac{1}{k} -1\biggr)^\frac{n-1}{2}\prod_{k=3,5,\dots}^n\frac{k}{k-1}. $$ For $n=15$ it is equal to $$ \frac{46780...
Post a Comment
Read more