Revisions to Histogram Bar Line Thickness

deleted 51 characters in body

Source Link

edited Feb 19, 2019 at 21:01

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Update 2: The answers below addressed the question in OP's comment. Based Based on the accepted answer it now seems that removing the bottom of edge of rectangles is not neededessential. In that case, the option ChartBaseStyle gives the desired result (there is no need for custom ChartElementFunctions):

Histogram[data, ChartBaseStyle -> EdgeForm[Thickness[.01]]]

Histogram[data, ChartStyle -> "Pastel", 
 ChartElementFunction -> "GlassRectangle", 
 ChartBaseStyle -> EdgeForm[{Opacity[1, Red], Thickness[.01]}]]

Update: The ChartElementFunction in the original post works only for the default setting for the option BarOrigin. The new function ceF2 works for arbitrary values for the BarOrigin option.

ceF2[cedf_: "GlassRectangle", o : OptionsPattern[]][col_: Black, thickness_: 3] := 
 Module[{or = Charting`ChartStyleInformation["BarOrigin"], ll = Tuples[#][[{1, 2, 4, 3}]]},
   ll = RotateRight[ll, Switch[or, Bottom, 0, Top, 2, Left, 3, Right, 1]] ;
   {ChartElementDataFunction[cedf, o][##], 
    col, AbsoluteThickness[thickness], CapForm["Butt"], Line[ll]}] &

Examples:

Grid[Partition[Histogram[data, ChartStyle -> 1, ImageSize -> 300, 
     PlotLabel -> Style["BarOrigin -> " <> ToString[#], 16, "Panel"], 
     ChartElementFunction -> ceF2["FadingRectangle", "GradientOrigin" -> Top][
       Dynamic[Darker@CurrentValue["Color"]], 5], 
     BarOrigin -> #] & /@ {Bottom, Top, Left, Right}, 2]]

Mathematica graphics

Original post:

A more flexible chart element function that modifies built-in chart element functions to add thick lines:

ClearAll[ceF]
ceF[cedf_: "GlassRectangle", o : OptionsPattern[]][col_: Black, thickness_: 3] := 
  {ChartElementDataFunction[cedf, o][##], 
   col, AbsoluteThickness[thickness], CapForm["Butt"],
   Line[Tuples[#][[{1, 2, 4, 3}]]]} &

Examples:

SeedRandom[42];
data = RandomVariate[NormalDistribution[0, 1], 200];

Histogram[data, ChartStyle -> 1, ChartElementFunction -> ceF[][]]

Mathematica graphics

Histogram[data, ChartStyle -> 1, 
 ChartElementFunction -> 
  ceF["FadingRectangle"][Dynamic[Darker@Darker@CurrentValue["Color"]], 5]]

Mathematica graphics

Histogram[data, ChartStyle -> 1, 
 ChartElementFunction -> 
  ceF["FadingRectangle", "GradientOrigin"->Top][Dynamic[Darker@CurrentValue["Color"]], 5]]

Mathematica graphics

Histogram[data, 5, ChartStyle -> {Red, Green, Blue, Orange, Cyan, Purple}, 
 ChartElementFunction -> 
  ceF["FadingRectangle", "GradientOrigin"->Top][Dynamic[Darker@CurrentValue["Color"]], 5]]

Mathematica graphics

Update 2: The answers below addressed the question in OP's comment. Based the accepted answer it now seems that removing the bottom of edge of rectangles is not needed. In that case, the option ChartBaseStyle gives the desired result (there is no need for custom ChartElementFunctions):

Histogram[data, ChartBaseStyle -> EdgeForm[Thickness[.01]]]

Histogram[data, ChartStyle -> "Pastel", 
 ChartElementFunction -> "GlassRectangle", 
 ChartBaseStyle -> EdgeForm[{Opacity[1, Red], Thickness[.01]}]]

Update: The ChartElementFunction in the original post works only for the default setting for the option BarOrigin. The new function ceF2 works for arbitrary values for the BarOrigin option.

ceF2[cedf_: "GlassRectangle", o : OptionsPattern[]][col_: Black, thickness_: 3] := 
 Module[{or = Charting`ChartStyleInformation["BarOrigin"], ll = Tuples[#][[{1, 2, 4, 3}]]},
   ll = RotateRight[ll, Switch[or, Bottom, 0, Top, 2, Left, 3, Right, 1]] ;
   {ChartElementDataFunction[cedf, o][##], 
    col, AbsoluteThickness[thickness], CapForm["Butt"], Line[ll]}] &

Examples:

Grid[Partition[Histogram[data, ChartStyle -> 1, ImageSize -> 300, 
     PlotLabel -> Style["BarOrigin -> " <> ToString[#], 16, "Panel"], 
     ChartElementFunction -> ceF2["FadingRectangle", "GradientOrigin" -> Top][
       Dynamic[Darker@CurrentValue["Color"]], 5], 
     BarOrigin -> #] & /@ {Bottom, Top, Left, Right}, 2]]

Mathematica graphics

Original post:

A more flexible chart element function that modifies built-in chart element functions to add thick lines:

ClearAll[ceF]
ceF[cedf_: "GlassRectangle", o : OptionsPattern[]][col_: Black, thickness_: 3] := 
  {ChartElementDataFunction[cedf, o][##], 
   col, AbsoluteThickness[thickness], CapForm["Butt"],
   Line[Tuples[#][[{1, 2, 4, 3}]]]} &

Examples:

SeedRandom[42];
data = RandomVariate[NormalDistribution[0, 1], 200];

Histogram[data, ChartStyle -> 1, ChartElementFunction -> ceF[][]]

Mathematica graphics

Histogram[data, ChartStyle -> 1, 
 ChartElementFunction -> 
  ceF["FadingRectangle"][Dynamic[Darker@Darker@CurrentValue["Color"]], 5]]

Mathematica graphics

Histogram[data, ChartStyle -> 1, 
 ChartElementFunction -> 
  ceF["FadingRectangle", "GradientOrigin"->Top][Dynamic[Darker@CurrentValue["Color"]], 5]]

Mathematica graphics

Histogram[data, 5, ChartStyle -> {Red, Green, Blue, Orange, Cyan, Purple}, 
 ChartElementFunction -> 
  ceF["FadingRectangle", "GradientOrigin"->Top][Dynamic[Darker@CurrentValue["Color"]], 5]]

Mathematica graphics

Update 2: Based on the accepted answer it now seems that removing the bottom of edge of rectangles is not essential. In that case, the option ChartBaseStyle gives the desired result (there is no need for custom ChartElementFunctions):

Histogram[data, ChartBaseStyle -> EdgeForm[Thickness[.01]]]

Histogram[data, ChartStyle -> "Pastel", 
 ChartElementFunction -> "GlassRectangle", 
 ChartBaseStyle -> EdgeForm[{Opacity[1, Red], Thickness[.01]}]]

Update: The ChartElementFunction in the original post works only for the default setting for the option BarOrigin. The new function ceF2 works for arbitrary values for the BarOrigin option.

ceF2[cedf_: "GlassRectangle", o : OptionsPattern[]][col_: Black, thickness_: 3] := 
 Module[{or = Charting`ChartStyleInformation["BarOrigin"], ll = Tuples[#][[{1, 2, 4, 3}]]},
   ll = RotateRight[ll, Switch[or, Bottom, 0, Top, 2, Left, 3, Right, 1]] ;
   {ChartElementDataFunction[cedf, o][##], 
    col, AbsoluteThickness[thickness], CapForm["Butt"], Line[ll]}] &

Examples:

Grid[Partition[Histogram[data, ChartStyle -> 1, ImageSize -> 300, 
     PlotLabel -> Style["BarOrigin -> " <> ToString[#], 16, "Panel"], 
     ChartElementFunction -> ceF2["FadingRectangle", "GradientOrigin" -> Top][
       Dynamic[Darker@CurrentValue["Color"]], 5], 
     BarOrigin -> #] & /@ {Bottom, Top, Left, Right}, 2]]

Mathematica graphics

Original post:

A more flexible chart element function that modifies built-in chart element functions to add thick lines:

ClearAll[ceF]
ceF[cedf_: "GlassRectangle", o : OptionsPattern[]][col_: Black, thickness_: 3] := 
  {ChartElementDataFunction[cedf, o][##], 
   col, AbsoluteThickness[thickness], CapForm["Butt"],
   Line[Tuples[#][[{1, 2, 4, 3}]]]} &

Examples:

SeedRandom[42];
data = RandomVariate[NormalDistribution[0, 1], 200];

Histogram[data, ChartStyle -> 1, ChartElementFunction -> ceF[][]]

Mathematica graphics

Histogram[data, ChartStyle -> 1, 
 ChartElementFunction -> 
  ceF["FadingRectangle"][Dynamic[Darker@Darker@CurrentValue["Color"]], 5]]

Mathematica graphics

Histogram[data, ChartStyle -> 1, 
 ChartElementFunction -> 
  ceF["FadingRectangle", "GradientOrigin"->Top][Dynamic[Darker@CurrentValue["Color"]], 5]]

Mathematica graphics

Histogram[data, 5, ChartStyle -> {Red, Green, Blue, Orange, Cyan, Purple}, 
 ChartElementFunction -> 
  ceF["FadingRectangle", "GradientOrigin"->Top][Dynamic[Darker@CurrentValue["Color"]], 5]]

Mathematica graphics

deleted 7 characters in body

Source Link

edited Feb 19, 2019 at 20:34

kglr

400.5k
18
488
929

Update 2: The answers below addressed the question in OP's comment. FromBased the accepted answer, it is now clearseems that removing the bottom of edge of rectangles is not needed.

In that case, the option ChartBaseStyle gives the resireddesired result (there is no need for custom ChartElementFunctions):

Histogram[data, ChartBaseStyle -> EdgeForm[Thickness[.01]]]

Histogram[data, ChartStyle -> "Pastel", 
 ChartElementFunction -> "GlassRectangle", 
 ChartBaseStyle -> EdgeForm[{Opacity[1, Red], Thickness[.01]}]]

Update: The ChartElementFunction in the original post works only for the default setting for the option BarOrigin. The new function ceF2 works for arbitrary values for the BarOrigin option.

ceF2[cedf_: "GlassRectangle", o : OptionsPattern[]][col_: Black, thickness_: 3] := 
 Module[{or = Charting`ChartStyleInformation["BarOrigin"], ll = Tuples[#][[{1, 2, 4, 3}]]},
   ll = RotateRight[ll, Switch[or, Bottom, 0, Top, 2, Left, 3, Right, 1]] ;
   {ChartElementDataFunction[cedf, o][##], 
    col, AbsoluteThickness[thickness], CapForm["Butt"], Line[ll]}] &

Examples:

Grid[Partition[Histogram[data, ChartStyle -> 1, ImageSize -> 300, 
     PlotLabel -> Style["BarOrigin -> " <> ToString[#], 16, "Panel"], 
     ChartElementFunction -> ceF2["FadingRectangle", "GradientOrigin" -> Top][
       Dynamic[Darker@CurrentValue["Color"]], 5], 
     BarOrigin -> #] & /@ {Bottom, Top, Left, Right}, 2]]

Mathematica graphics

Original post:

A more flexible chart element function that modifies built-in chart element functions to add thick lines:

ClearAll[ceF]
ceF[cedf_: "GlassRectangle", o : OptionsPattern[]][col_: Black, thickness_: 3] := 
  {ChartElementDataFunction[cedf, o][##], 
   col, AbsoluteThickness[thickness], CapForm["Butt"],
   Line[Tuples[#][[{1, 2, 4, 3}]]]} &

Examples:

SeedRandom[42];
data = RandomVariate[NormalDistribution[0, 1], 200];

Histogram[data, ChartStyle -> 1, ChartElementFunction -> ceF[][]]

Mathematica graphics

Histogram[data, ChartStyle -> 1, 
 ChartElementFunction -> 
  ceF["FadingRectangle"][Dynamic[Darker@Darker@CurrentValue["Color"]], 5]]

Mathematica graphics

Histogram[data, ChartStyle -> 1, 
 ChartElementFunction -> 
  ceF["FadingRectangle", "GradientOrigin"->Top][Dynamic[Darker@CurrentValue["Color"]], 5]]

Mathematica graphics

Histogram[data, 5, ChartStyle -> {Red, Green, Blue, Orange, Cyan, Purple}, 
 ChartElementFunction -> 
  ceF["FadingRectangle", "GradientOrigin"->Top][Dynamic[Darker@CurrentValue["Color"]], 5]]

Mathematica graphics

Update 2: The answers below addressed the question in OP's comment. From the accepted answer, it is now clear that removing the bottom of edge of rectangles is not needed.

In that case, the option ChartBaseStyle gives the resired result (there is no need for custom ChartElementFunctions):

Histogram[data, ChartBaseStyle -> EdgeForm[Thickness[.01]]]

Histogram[data, ChartStyle -> "Pastel", 
 ChartElementFunction -> "GlassRectangle", 
 ChartBaseStyle -> EdgeForm[{Opacity[1, Red], Thickness[.01]}]]

Update: The ChartElementFunction in the original post works only for the default setting for the option BarOrigin. The new function ceF2 works for arbitrary values for the BarOrigin option.

ceF2[cedf_: "GlassRectangle", o : OptionsPattern[]][col_: Black, thickness_: 3] := 
 Module[{or = Charting`ChartStyleInformation["BarOrigin"], ll = Tuples[#][[{1, 2, 4, 3}]]},
   ll = RotateRight[ll, Switch[or, Bottom, 0, Top, 2, Left, 3, Right, 1]] ;
   {ChartElementDataFunction[cedf, o][##], 
    col, AbsoluteThickness[thickness], CapForm["Butt"], Line[ll]}] &

Examples:

Grid[Partition[Histogram[data, ChartStyle -> 1, ImageSize -> 300, 
     PlotLabel -> Style["BarOrigin -> " <> ToString[#], 16, "Panel"], 
     ChartElementFunction -> ceF2["FadingRectangle", "GradientOrigin" -> Top][
       Dynamic[Darker@CurrentValue["Color"]], 5], 
     BarOrigin -> #] & /@ {Bottom, Top, Left, Right}, 2]]

Mathematica graphics

Original post:

A more flexible chart element function that modifies built-in chart element functions to add thick lines:

ClearAll[ceF]
ceF[cedf_: "GlassRectangle", o : OptionsPattern[]][col_: Black, thickness_: 3] := 
  {ChartElementDataFunction[cedf, o][##], 
   col, AbsoluteThickness[thickness], CapForm["Butt"],
   Line[Tuples[#][[{1, 2, 4, 3}]]]} &

Examples:

SeedRandom[42];
data = RandomVariate[NormalDistribution[0, 1], 200];

Histogram[data, ChartStyle -> 1, ChartElementFunction -> ceF[][]]

Mathematica graphics

Histogram[data, ChartStyle -> 1, 
 ChartElementFunction -> 
  ceF["FadingRectangle"][Dynamic[Darker@Darker@CurrentValue["Color"]], 5]]

Mathematica graphics

Histogram[data, ChartStyle -> 1, 
 ChartElementFunction -> 
  ceF["FadingRectangle", "GradientOrigin"->Top][Dynamic[Darker@CurrentValue["Color"]], 5]]

Mathematica graphics

Histogram[data, 5, ChartStyle -> {Red, Green, Blue, Orange, Cyan, Purple}, 
 ChartElementFunction -> 
  ceF["FadingRectangle", "GradientOrigin"->Top][Dynamic[Darker@CurrentValue["Color"]], 5]]

Mathematica graphics

Update 2: The answers below addressed the question in OP's comment. Based the accepted answer it now seems that removing the bottom of edge of rectangles is not needed. In that case, the option ChartBaseStyle gives the desired result (there is no need for custom ChartElementFunctions):

Histogram[data, ChartBaseStyle -> EdgeForm[Thickness[.01]]]

Histogram[data, ChartStyle -> "Pastel", 
 ChartElementFunction -> "GlassRectangle", 
 ChartBaseStyle -> EdgeForm[{Opacity[1, Red], Thickness[.01]}]]

Update: The ChartElementFunction in the original post works only for the default setting for the option BarOrigin. The new function ceF2 works for arbitrary values for the BarOrigin option.

ceF2[cedf_: "GlassRectangle", o : OptionsPattern[]][col_: Black, thickness_: 3] := 
 Module[{or = Charting`ChartStyleInformation["BarOrigin"], ll = Tuples[#][[{1, 2, 4, 3}]]},
   ll = RotateRight[ll, Switch[or, Bottom, 0, Top, 2, Left, 3, Right, 1]] ;
   {ChartElementDataFunction[cedf, o][##], 
    col, AbsoluteThickness[thickness], CapForm["Butt"], Line[ll]}] &

Examples:

Grid[Partition[Histogram[data, ChartStyle -> 1, ImageSize -> 300, 
     PlotLabel -> Style["BarOrigin -> " <> ToString[#], 16, "Panel"], 
     ChartElementFunction -> ceF2["FadingRectangle", "GradientOrigin" -> Top][
       Dynamic[Darker@CurrentValue["Color"]], 5], 
     BarOrigin -> #] & /@ {Bottom, Top, Left, Right}, 2]]

Mathematica graphics

Original post:

A more flexible chart element function that modifies built-in chart element functions to add thick lines:

ClearAll[ceF]
ceF[cedf_: "GlassRectangle", o : OptionsPattern[]][col_: Black, thickness_: 3] := 
  {ChartElementDataFunction[cedf, o][##], 
   col, AbsoluteThickness[thickness], CapForm["Butt"],
   Line[Tuples[#][[{1, 2, 4, 3}]]]} &

Examples:

SeedRandom[42];
data = RandomVariate[NormalDistribution[0, 1], 200];

Histogram[data, ChartStyle -> 1, ChartElementFunction -> ceF[][]]

Mathematica graphics

Histogram[data, ChartStyle -> 1, 
 ChartElementFunction -> 
  ceF["FadingRectangle"][Dynamic[Darker@Darker@CurrentValue["Color"]], 5]]

Mathematica graphics

Histogram[data, ChartStyle -> 1, 
 ChartElementFunction -> 
  ceF["FadingRectangle", "GradientOrigin"->Top][Dynamic[Darker@CurrentValue["Color"]], 5]]

Mathematica graphics

Histogram[data, 5, ChartStyle -> {Red, Green, Blue, Orange, Cyan, Purple}, 
 ChartElementFunction -> 
  ceF["FadingRectangle", "GradientOrigin"->Top][Dynamic[Darker@CurrentValue["Color"]], 5]]

Mathematica graphics

added 455 characters in body

Source Link

edited Feb 19, 2019 at 20:22

kglr

400.5k
18
488
929

Update 2: The answers below addressed the question in OP's comment. From the accepted answer, it is now clear that removing the bottom of edge of rectangles is not needed.

In that case, the option ChartBaseStyle gives the resired result (there is no need for custom ChartElementFunctions):

Histogram[data, ChartBaseStyle -> EdgeForm[Thickness[.01]]]

Histogram[data, ChartStyle -> "Pastel", 
 ChartElementFunction -> "GlassRectangle", 
 ChartBaseStyle -> EdgeForm[{Opacity[1, Red], Thickness[.01]}]]

Update: The ChartElementFunction in the original post works only for the default setting for the option BarOrigin. The new function ceF2 works for arbitrary values for the BarOrigin option.

ceF2[cedf_: "GlassRectangle", o : OptionsPattern[]][col_: Black, thickness_: 3] := 
 Module[{or = Charting`ChartStyleInformation["BarOrigin"], ll = Tuples[#][[{1, 2, 4, 3}]]},
   ll = RotateRight[ll, Switch[or, Bottom, 0, Top, 2, Left, 3, Right, 1]] ;
   {ChartElementDataFunction[cedf, o][##], 
    col, AbsoluteThickness[thickness], CapForm["Butt"], Line[ll]}] &

Examples:

Grid[Partition[Histogram[data, ChartStyle -> 1, ImageSize -> 300, 
     PlotLabel -> Style["BarOrigin -> " <> ToString[#], 16, "Panel"], 
     ChartElementFunction -> ceF2["FadingRectangle", "GradientOrigin" -> Top][
       Dynamic[Darker@CurrentValue["Color"]], 5], 
     BarOrigin -> #] & /@ {Bottom, Top, Left, Right}, 2]]

Mathematica graphics

Original post:

A more flexible chart element function that modifies built-in chart element functions to add thick lines:

ClearAll[ceF]
ceF[cedf_: "GlassRectangle", o : OptionsPattern[]][col_: Black, thickness_: 3] := 
  {ChartElementDataFunction[cedf, o][##], 
   col, AbsoluteThickness[thickness], CapForm["Butt"],
   Line[Tuples[#][[{1, 2, 4, 3}]]]} &

Examples:

SeedRandom[42];
data = RandomVariate[NormalDistribution[0, 1], 200];

Histogram[data, ChartStyle -> 1, ChartElementFunction -> ceF[][]]

Mathematica graphics

Histogram[data, ChartStyle -> 1, 
 ChartElementFunction -> 
  ceF["FadingRectangle"][Dynamic[Darker@Darker@CurrentValue["Color"]], 5]]

Mathematica graphics

Histogram[data, ChartStyle -> 1, 
 ChartElementFunction -> 
  ceF["FadingRectangle", "GradientOrigin"->Top][Dynamic[Darker@CurrentValue["Color"]], 5]]

Mathematica graphics

Histogram[data, 5, ChartStyle -> {Red, Green, Blue, Orange, Cyan, Purple}, 
 ChartElementFunction -> 
  ceF["FadingRectangle", "GradientOrigin"->Top][Dynamic[Darker@CurrentValue["Color"]], 5]]

Mathematica graphics

Update: The ChartElementFunction in the original post works only for the default setting for the option BarOrigin. The new function ceF2 works for arbitrary values for the BarOrigin option.

ceF2[cedf_: "GlassRectangle", o : OptionsPattern[]][col_: Black, thickness_: 3] := 
 Module[{or = Charting`ChartStyleInformation["BarOrigin"], ll = Tuples[#][[{1, 2, 4, 3}]]},
   ll = RotateRight[ll, Switch[or, Bottom, 0, Top, 2, Left, 3, Right, 1]] ;
   {ChartElementDataFunction[cedf, o][##], 
    col, AbsoluteThickness[thickness], CapForm["Butt"], Line[ll]}] &

Examples:

Grid[Partition[Histogram[data, ChartStyle -> 1, ImageSize -> 300, 
     PlotLabel -> Style["BarOrigin -> " <> ToString[#], 16, "Panel"], 
     ChartElementFunction -> ceF2["FadingRectangle", "GradientOrigin" -> Top][
       Dynamic[Darker@CurrentValue["Color"]], 5], 
     BarOrigin -> #] & /@ {Bottom, Top, Left, Right}, 2]]

Mathematica graphics

Original post:

A more flexible chart element function that modifies built-in chart element functions to add thick lines:

ClearAll[ceF]
ceF[cedf_: "GlassRectangle", o : OptionsPattern[]][col_: Black, thickness_: 3] := 
  {ChartElementDataFunction[cedf, o][##], 
   col, AbsoluteThickness[thickness], CapForm["Butt"],
   Line[Tuples[#][[{1, 2, 4, 3}]]]} &

Examples:

SeedRandom[42];
data = RandomVariate[NormalDistribution[0, 1], 200];

Histogram[data, ChartStyle -> 1, ChartElementFunction -> ceF[][]]

Mathematica graphics

Histogram[data, ChartStyle -> 1, 
 ChartElementFunction -> 
  ceF["FadingRectangle"][Dynamic[Darker@Darker@CurrentValue["Color"]], 5]]

Mathematica graphics

Histogram[data, ChartStyle -> 1, 
 ChartElementFunction -> 
  ceF["FadingRectangle", "GradientOrigin"->Top][Dynamic[Darker@CurrentValue["Color"]], 5]]

Mathematica graphics

Histogram[data, 5, ChartStyle -> {Red, Green, Blue, Orange, Cyan, Purple}, 
 ChartElementFunction -> 
  ceF["FadingRectangle", "GradientOrigin"->Top][Dynamic[Darker@CurrentValue["Color"]], 5]]

Mathematica graphics

Update 2: The answers below addressed the question in OP's comment. From the accepted answer, it is now clear that removing the bottom of edge of rectangles is not needed.

In that case, the option ChartBaseStyle gives the resired result (there is no need for custom ChartElementFunctions):

Histogram[data, ChartBaseStyle -> EdgeForm[Thickness[.01]]]

Histogram[data, ChartStyle -> "Pastel", 
 ChartElementFunction -> "GlassRectangle", 
 ChartBaseStyle -> EdgeForm[{Opacity[1, Red], Thickness[.01]}]]

Update: The ChartElementFunction in the original post works only for the default setting for the option BarOrigin. The new function ceF2 works for arbitrary values for the BarOrigin option.

ceF2[cedf_: "GlassRectangle", o : OptionsPattern[]][col_: Black, thickness_: 3] := 
 Module[{or = Charting`ChartStyleInformation["BarOrigin"], ll = Tuples[#][[{1, 2, 4, 3}]]},
   ll = RotateRight[ll, Switch[or, Bottom, 0, Top, 2, Left, 3, Right, 1]] ;
   {ChartElementDataFunction[cedf, o][##], 
    col, AbsoluteThickness[thickness], CapForm["Butt"], Line[ll]}] &

Examples:

Grid[Partition[Histogram[data, ChartStyle -> 1, ImageSize -> 300, 
     PlotLabel -> Style["BarOrigin -> " <> ToString[#], 16, "Panel"], 
     ChartElementFunction -> ceF2["FadingRectangle", "GradientOrigin" -> Top][
       Dynamic[Darker@CurrentValue["Color"]], 5], 
     BarOrigin -> #] & /@ {Bottom, Top, Left, Right}, 2]]

Mathematica graphics

Original post:

A more flexible chart element function that modifies built-in chart element functions to add thick lines:

ClearAll[ceF]
ceF[cedf_: "GlassRectangle", o : OptionsPattern[]][col_: Black, thickness_: 3] := 
  {ChartElementDataFunction[cedf, o][##], 
   col, AbsoluteThickness[thickness], CapForm["Butt"],
   Line[Tuples[#][[{1, 2, 4, 3}]]]} &

Examples:

SeedRandom[42];
data = RandomVariate[NormalDistribution[0, 1], 200];

Histogram[data, ChartStyle -> 1, ChartElementFunction -> ceF[][]]

Mathematica graphics

Histogram[data, ChartStyle -> 1, 
 ChartElementFunction -> 
  ceF["FadingRectangle"][Dynamic[Darker@Darker@CurrentValue["Color"]], 5]]

Mathematica graphics

Histogram[data, ChartStyle -> 1, 
 ChartElementFunction -> 
  ceF["FadingRectangle", "GradientOrigin"->Top][Dynamic[Darker@CurrentValue["Color"]], 5]]

Mathematica graphics

Histogram[data, 5, ChartStyle -> {Red, Green, Blue, Orange, Cyan, Purple}, 
 ChartElementFunction -> 
  ceF["FadingRectangle", "GradientOrigin"->Top][Dynamic[Darker@CurrentValue["Color"]], 5]]

Mathematica graphics

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edited Mar 26, 2017 at 6:05

kglr

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edited Mar 26, 2017 at 6:00

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edited Mar 26, 2017 at 5:03

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created Feb 21, 2017 at 1:56

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