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31

Use the individual legends as tick labels: dates = Through[{First, Last}@#] & /@ res {{1769, 1821}, {1775, 1817}, {1770, 1831}, {1818, 1883}, {1777, 1855}, {1766, 1817}, {1870, 1924}} names={"Napoleon Bonaparte", "Jane Austen", "Hegel", "Marx", "Gauss", "Madame de Stael", "Lenin"}; llpd = MapIndexed[Thread@{#, First@#2} &, dates]; legends ...


14

Bob Hanlon's answer works very well, but in some ways it is the hard way of doing things. If you have v9 or v10, then it is arguably easier to use the legend constructs within it. Similar to his answer, we get the image and element names: img = Import["ExampleData/1PPT.pdb", "Rendering" -> "BallAndStick", ImageSize -> 500]; elements = ...


13

We may observe that the automatically generated legend limits the number of legend items to the number of available colors in the given color scheme. Using this utility function: plot[scheme_] := Plot[Evaluate[Range[20] + x], {x, -5, 5}, PlotLegends -> "Expressions", PlotStyle -> scheme] Observe the result for indexed color scheme #42 which has ...


12

This is an intentional change to make PlotLegends -> "Expressions" more consistent with PlotLegends -> Automatic. Both now do not produce legends when only one line is present. What you are looking for is PlotLegends -> "AllExpressions" which has the old behavior, e.g. Plot[x, {x, 0, 1}, PlotLegends -> "AllExpressions"] More generally, ...


12

Two work-arounds: (1) Use TickLabels BarLegend[{"Rainbow", {-0.015, 0.015}}, ImageSize->300, Charting`TickLabels -> (Style[NumberForm[#, {Infinity, 3}],Bold,Black,12] & /@ Range[-.015, .015, .005])] (2) Use LabelingFunction: BarLegend[{"Rainbow", {-0.015, 0.015}}, ImageSize->300, LabelingFunction -> (Style[NumberForm[#, ...


10

bas = Import["ExampleData/1PPT.pdb", "Rendering" -> "BallAndStick", ImageSize -> 500]; elements = Import["ExampleData/1PPT.pdb", "ResidueAtoms"] // Flatten // Union; legend = GraphicsColumn[{ {Graphics[{#[[1]], Disk[{0, 0}, 1]}, ImageSize -> 10], #[[2]]} & /@ Thread[{ ElementData[#, "IconColor"] & /@ elements, ...


10

ListPlot[Table[f, {f, {Sin[x], Cos[x]}}, {x, 0, 2 π, 0.1}], PlotLegends ->PointLegend[ (Style[#, 40] & /@ {"sin(x)", "cos(x)"}),Alignment->Center]] Update: As noted by rcollyer in the comments Alignment is an undocumented option for PointLegend. As an alternative, the following old Pane trick works without relying on an undocumented option: ...


10

You have to add the index for the colors to LineLegend Legended[Grid[{{Show[oniplot], Show[cpplot]}}], LineLegend[97, {"factor = 2", "factor = 3", "factor = 4", "factor = 5"}]] In order to have the legend above the plots and with markers: Legended[GraphicsGrid[{{oniplot, cpplot}}], Placed[LineLegend[97, Array["factor = " <> ToString@# ...


9

`AbsoluteThickne is another useful approach:e.g. op = Table[AbsoluteThickness[2], {3}]; leg = LineLegend[Automatic, {"Max", "Mea", "Min"}]; ListLinePlot[{historyobjetivomejor, historyobjetivomedia, historyobjetivopeor}, AxesLabel -> {"Generacion", "Objetivo"}, PlotStyle -> op, PlotLegends -> leg] Varying thickness: using: Manipulate[ ...


9

The following works in both v9 and v10: style = Directive[Thick, Black]; ContourPlot[y - x^2, {x, 0, 1}, {y, 0, 1}, ContourStyle -> Directive[Thick, Black, Opacity[1]], FrameStyle -> style, PlotLegends -> BarLegend[Automatic, Method -> {FrameStyle -> style}]] The idea to use the (undocumented) Method option comes from inspecting the ...


8

After a rather long debugging session in our chat we could determine the reason of the problem and come up with a workaround. In short, we first tried whether the issue appears for the most basic Graphics[], which it didn't. As it turned out the gray background is introduced by using PlotLegends as in the example above. We went further by comparing ...


7

You can do : LogLinearPlot[{Log[x], x Sin[x], x Cos[x]}, {x, 1, 100}, PlotLegends -> Placed[{"Log(x)", None, "x Cos(x)"}, After]]


7

points = Table[{x, .5 x^2 - 3 + RandomReal[]}, {x, -3, 3}]; Show[{ListPlot[points, PlotLegends -> {"Experiment"}], Plot[{.5 x^2 - 3}, {x, -3, 3}, PlotStyle -> Red, PlotLegends -> {"Model"}]}]


6

In version 10 the PointSize of the legend will automatically match the PointSize of the Plot: ListPlot[Table[RandomReal[NormalDistribution[], {20, 2}], {2}], PlotLegends -> {"a", "b"}, PlotStyle -> PointSize[0.02]] Edit: The answer by @eldo made me realize, that this is only true up to a PointSize that is equal to the default ...


6

As it often happens, when I was researching for writing a fine question I also digged through the documentation and found the answer. There is an option LegendMarkerkSize which in my opinion has an unintuitive name in the case of the BarLegend. Anyway it is exactly meant to do what I needed: testPlot = ArrayPlot[ Array[RandomInteger[100] &, {100, ...


6

I always do it like this: cm = 72/2.54; BarChart[{{1, 2, 3}, {1, 3, 2}}, ChartLegends -> SwatchLegend[{"a", "b", "c"}, LegendMarkerSize -> 20], ImageSize -> 4 cm] top: code from your question, bottom: this code The colors of the legend are inherited from the BarChart object: BarChart[{{1, 2, 3}, {1, 3, 2}}, ChartStyle -> {Red, Green, ...


6

For ChartLegends, although the documentation does not mention any values other than "Row" and "Column" for the option LegendAppearance, it turns out you can also use this option to set the LegendMarkerSize BarChart[{{1, 2, 3}, {1, 3, 2}}, ChartStyle -> {Red, Green, Blue}, LegendAppearance -> {LegendMarkerSize -> 20}, ChartLegends -> {"a", ...


6

A reasonable workaround is to use SwatchLegend with LegendMarkers set: ListPlot[ Table[f, {f, {Sin[x], Cos[x]}}, {x, 0, 2 π, 0.1}], PlotLegends -> SwatchLegend[{"sin(x)", "cos(x)"}, LabelStyle -> 40, LegendMarkers -> "Bubble" ] ]


6

Borrowing from kguler the legends legends = MapIndexed[ SwatchLegend[{ColorData[{"Rainbow", {1, 7}}][## & @@ #2]}, {#}, LegendMarkerSize -> {{10, 10}}] &, names]; p1 = ListLinePlot[par, ColorFunction -> "Rainbow", Ticks -> {Range[1750, 1980, 10], Automatic}, PlotRangePadding -> {{5, 50}, {0, 1}}, GridLines -> ...


6

f@x_ := ColorData["VisibleSpectrum"][Rescale[x, {0, 1}, {380, 750}]]; Plot3D[ Exp[-x^2 - y^2], {x, -2, 2}, {y, -2, 2}, ColorFunction -> f, PlotLegends -> BarLegend[{f@# &, {0, 1}}]]


6

This is a variation of Plot draws list of curves in same color when not using Evaluate. By using /. the head of the first argument is ReplaceAll rather than a List, and the styling subroutine of ParametricPlot does not know what to do with it. Make the substitutions part of the definitions and you get this: N1[a_] := (4 d)/(π a) ...


5

Adapting an example in the docs you could do it this way: table[pairs_] := Grid[pairs, BaseStyle -> {TextAlignment -> Left}, Alignment -> {Left, Automatic}] Then use LegendLayout -> table, to give you: Edit To reverse it just use Reverse: reversetable[pairs_] := Grid[Reverse@pairs, BaseStyle -> {TextAlignment -> Left}, ...


5

I am continuing to explore this as I can't yet demonstrate exactly what is going on, but first note that ArrayPlot does not experience this problem: tab = Table[Exp[-((i - 25)^2 + (j - 25)^2)], {i, 50}, {j, 50}]; ArrayPlot[500 * tab, PlotLegends -> Automatic, ColorFunction -> "ThermometerColors"] Note that I multiplied the original table by 500 ...


5

The style for labels in a legend is determined by the option LabelStyle. To get the legend markers to match the style of the plot without rescaling, one approach is to use AbsoluteDashing and AbsoluteThickness to style the plot lines. With[{d = AbsoluteDashing[8], t = AbsoluteThickness[4]}, Plot[{E^(4 x), E^(3 x), E^(2 x), E^x, E^(x/2), E^(x/3), E^(x/4)}, ...


5

I don't know if there's an elegant way. The colors are embedded in the plots, and changing them after the fact takes some work. I basically do what the OP alluded to, but as postprocessing. Collect the colors in the graph and remap them according to some color function. plots = Table[ Plot[Evaluate[Table[Sin[(4(3-j)-i)x], {i, 4-j}]], {x, 0, 2 Pi}], ...


5

This is straightforward. All of the legend functions accept the generic, top level arguments, e.g. LineLegend[Automatic] which allows you to modify the behavior. So, you would use BarLegend[Automatic, LabelStyle -> {FontSize -> 20}].


5

I was able to isolate the problem with BarLegend in v.10.0.0. Yes, it is clearly a bug. Let us see the how the thin grey lines are implemented: Cases[ ToBoxes[BarLegend[{"DeepSeaColors", {0, 1}}, LegendLayout -> "ReversedColumn"]], _LineBox, Infinity] {LineBox[ NCache[{{-(15/2), 225/2}, {15/2, 225/2}, {15/ 2, -(225/2)}, {-(15/2), ...


5

If it is to be printed and you have space (say, a page), then a grid can be effective, where each dataset is plotted in a highlight color over a monochromatic plot of all datasets: maxmin = {Min /@ Transpose[data], Max /@ Transpose[data]}; background = ListLinePlot[Join[maxmin, data], PlotStyle -> Directive[Thin, Gray], Filling -> {1 -> ...


5

You can use the (afaik undocumented) option "Ticks" ContourPlot[Sin[x] Cos[y], {x, 0, 2 Pi}, {y, 0, 2 Pi}, Contours -> {Automatic, 10}, ColorFunction -> "Rainbow", PlotLegends -> BarLegend[Automatic, None, "Ticks" -> {{-1, "cold"}, {0, "ok"}, {1, "hot"}}]] Note: You can also use Ticks instead of "Ticks" and ignore the red highlighting ...


4

Specifying FrameTicks adds a "legend" to MatrixPlot's row: legend = {"Hello", "How are you?", "Good Bye"}; MatrixPlot[{{1, 2, 1}, {3, 0, 1}, {0, 0, -1}}, FrameTicks -> { {True (* Left *), Thread[{Range@Length@legend, legend}](* Right *)}, {True (* Bottom *), False (* Top *)}}]



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