I would like to diagonally hatch the bars in BarChart and to have three different background colors for the different bars, like this:
Any ideas how to reproduce this image? The closest post I found is here.
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4$\begingroup$ Related: How do I plot a histogram with hatched shading? $\endgroup$– rm -rf ♦Oct 15, 2013 at 4:35
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$\begingroup$ @rm-rf The hatching part of this question is a duplicate indeed. The background part, however, is unique to this question. Therefore, I hesitate to vote to close. $\endgroup$– Sjoerd C. de VriesOct 15, 2013 at 6:29
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$\begingroup$ @SjoerdC.deVries Fair enough. Changed comment to "Related" $\endgroup$– rm -rf ♦Oct 15, 2013 at 7:05
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$\begingroup$ Since v12.1 HatchFilling can be used in BarCharts $\endgroup$– Sjoerd C. de VriesMar 18, 2020 at 21:55
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2 Answers
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With a bit of manual parameters:
barFilled[gap_, h_, seg_][{{xmin_, xmax_}, {ymin_, ymax_}}, ___] :=
Module[{width, line, yt, yb, lend},
{yb, yt} = Sort[{ymin, ymax}];
width = xmax - xmin;
line = Table[{{xmin, i}, {xmax, i + width}}, {i, yb, yt - width, h/seg}];
lend = line[[-1, 1, 2]];
line = {Line[line],
Line[Table[{{xmin + i, yb}, {xmax, yb + width - i}}, {i, h/seg, width, h/seg}]],
Line[Table[{{xmin, lend + i}, {xmax - (lend + width - yt) - i,yt}}, {i, h/seg, width + h/seg, h/seg}]]};
{{Opacity[.2], EdgeForm[], Rectangle[{xmin, -h}, {xmax + gap, h}]},
{CapForm["Butt"], line}, {FaceForm[], Rectangle[{xmin, ymin}, {xmax, ymax}]}}]
BarChart[{2, -1, 1.5, -3, 3, 2.5}, BarSpacing -> 2,
ChartElementFunction -> barFilled[.65, 3, 35], ChartStyle -> 61,
GridLines -> {None, Automatic}]
Like AimForClarity mentioned, to avoid empty strip, we could replace 0 with some dummy value + meta and define barFilled for that value. For example:
barFilled[gap_, h_, seg_][{{xmin_, xmax_}, {ymin_, ymax_}}, _, {None}] :=
{{Opacity[.2], EdgeForm[], Rectangle[{xmin, -h}, {xmax + gap, h}]}}
BarChart[{2, -1, 0, -2, 1, 2} /. {0 -> (1 -> None)}, BarSpacing -> 2, ChartElementFunction -> barFilled[.65, 2, 35],
ChartStyle -> 61, GridLines -> {None, Automatic}]
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$\begingroup$ Does the grid lines have to be on top of the charts and axes ticks labels? $\endgroup$– mmalOct 15, 2013 at 8:45
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$\begingroup$ @mmal grid lines are behind bars. It's a side effect of setting FaceForm[]. If you don't want that, you can insert one more rectangle before drawing lines. For example, .., {FaceForm[White], Rectangle[{xmin, ymin}, {xmax, ymax}]}, {CapForm["Butt"], line}, ... $\endgroup$– halmirOct 15, 2013 at 12:14
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2$\begingroup$ @mmal To have grid lines on top of the charts and axes ticks labels one can use
Method -> {"GridLinesInFront" -> True}option. $\endgroup$ Oct 15, 2013 at 21:12 -
$\begingroup$ Thanks, the halmir's method does exactly what I was thinking about. Using
Frame->Trueclips the grid lines to frame region. $\endgroup$– mmalOct 16, 2013 at 5:52 -
$\begingroup$ I hav not been able to get around this silly problem, that if one of the elements is 0 in the array, the barFilled function does not seem to be called and so there is an empty strip, any ideas how to fix this? $\endgroup$ Oct 19, 2013 at 4:31
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We can use ChartBaseStyle -> HatchFilling[] (HatchFilling is introduced in version 12.1) to get rectangles with hatched filling. Using the (undocumented) option "BarStartingValue" -> 0 the horizontal coordinates of bars become successive integers starting with 0 (independent of the BarSpacing option value). Then, it becomes an easy task to get the coordinates of the background rectangles using Scaled to construct epilog primitives:
epilog[col_] := MapIndexed[{Opacity[.2], #,
Rectangle[Scaled[{0, -1}, {#2[[1]] - 1, 0}], Scaled[{0, 1}, {#2[[1]], 0}]]} &,
col]
Examples:
data = {2, -1, 1.5, -3, 3, 2.5};
colors = ColorData[61] /@ Range[Length @ data];
BarChart[data,
ChartBaseStyle -> HatchFilling[],
BarSpacing -> 2,
ChartStyle -> 61, Axes -> {False, True},
GridLines -> {None, Automatic},
PlotRange -> {{0, Length @ data}, All},
"BarStartingValue" -> 0,
Epilog -> {Line[{{0, 0}, {Length @ data, 0}}],
epilog[PadRight[colors, Length @ data,colors]]}]
Use BarSpacing -> 1 to get
With data = {2, -1, 0, -3, 3, 2.5}; we get
