New answers tagged filling
4
I think what you want is:
Plot[Evaluate[{x^2, x}],
{x, 1, 2},
PlotRange -> All,
Filling -> {1 -> {2}},
PlotStyle->{Blue, Blue}
]
4
The issue caused by differing vertical positions of Axis across plots with filling.
plotswithfilling = Table[
Plot[PDF[NormalDistribution[i, 1], x] Exp[-Sqrt[i]/4] /.
i -> Range[1, 3, 2] // Evaluate, {x, k, k + 1},
PlotRange -> Full, Filling -> Axis, Axes -> None,
PlotStyle -> {ColorData[10][4 + k],
Darker[ColorData[...
2
This is very closely related to this question. Here is a version of kglr's answer adapted to your problem:
ListPlot[{
(*Horizontal line 1*){{0, 5}, {35, 5}},
(*Horizontal line 2*){{0, 10}, {35, 10}},
(*Vertical line 3*){{5, 0}, {5, 35}},
(*Vertical line 4*){{10, 0}, {10, 35}},
(*Horizontal line 5*){{0, 0}, {5, 0}}
},
Joined -> True,
Filling ...
4
threshold = 1.;
Plot[{x^2, x^2/6, ConditionalExpression[0, x <= threshold]}, {x, 0, 2},
PlotStyle -> {Automatic, Orange, None},
Filling -> {2 -> {{3}, {None, LightOrange}}},
PlotRange -> {0, 1}]
and
Plot[{x^2, x^2/6, Boole[x > threshold ]}, {x, 0, 2},
Exclusions -> {{threshold }},
PlotStyle -> {Automatic, Orange, None},
...
4
An alternative approach using a single ParametricPlot3D:
ClearAll[f1, f2, f3, f4, f5]
f1[x_, y_] := {x Sin[x], x Cos[x], -(x/3)}
f2[x_, y_] := {Sqrt[y] x Sin[x], y^2 x Cos[x], -(x/3)}
f3[x_, y_] := {Sqrt[y] x Sin[x], y^2 x Cos[x], -((x*y^2)/3)}
f4[r_: {1, 1.15}][x_, y_] := Module[{t = Rescale[y, r]},
t f2[x, r[[2]]] + (1 - t) f3[x, r[[2]]]]
f5[r_: {...
5
You need to interpolate between the two bounding curves. One way to do this is to multiply one of the bounding curves by $t$, multiply the other by $(1-t)$, and create a ParametricPlot running from $t = 0$ to 1.
originalplot =
ParametricPlot3D[
{
{x*Sin[x], x*Cos[x], -(x/3)},
{Sqrt[y]*x*Sin[x], y^2*x*Cos[x], -(x/3)},
{Sqrt[y]*x*Sin[x], y^...
8
Perhaps you're looking for something like this:
Plot[{
ConditionalExpression[x^2, x < 1],
ConditionalExpression[x^2, x > 1],
ConditionalExpression[2 + 2 x^2, x < 1]
},
{x, 0, 2},
Filling -> {1 -> Bottom}
]
4
If you have version 12.+, you can also use Around to process data1 and use ListLinePlot with the options IntervalMarkers and IntervalMarkersStyle:
data1b = Around[#, {#3, #2} - #] & @@@ data1;
ListLinePlot[data1b,
IntervalMarkers -> "Bands",
IntervalMarkersStyle -> Pink,
PlotRangePadding -> {Automatic, {Automatic, Scaled[.1]}}]
8
ListLinePlot[Transpose[data1],
PlotStyle -> {Directive[Thick, Blue], Directive[Red], Directive[Red]},
Filling -> {2 -> {3}}, FillingStyle -> {Opacity[.25], Pink},
PlotTheme -> "Detailed"]
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