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Gantt Chart like rectangles with log scale
Poets of the 19th centuryIs there a way to output histogram x and y data without using the Histogram chart function?BarChart and Log Scale results in different bar originsIssue with ListDensityPlot when using logarithmic scaleHow to plot BarChart with error bars on logarithmic scale in Mathematica 9?How to plot two y axis? or combine(merge) two plots? Should handle faceted column as wellJumping from one dimension (1D) to two dimensions (2D)Jumping from one dimension (1D) to two dimensions (2D) Part IIWavelet Data Visualization (modus and phase) ListPlotBar chart scale in multi-chart visualisationUpdate: Combining DistributionChart and BoxWhiskerChart
$begingroup$
I need to plot something that looks like a gant chart, but are actual data ranges. These ranges need to be on a log scale. Graphics does not offer a Scalingfunction. A LogPlot with ErrorBars does not work either and would not look ideal. Is it possible to draw rectangles with one axis being log scale?
E.g.:
Min and Max values of the rectangles are the known data.
plotting charts
$endgroup$
add a comment |
$begingroup$
I need to plot something that looks like a gant chart, but are actual data ranges. These ranges need to be on a log scale. Graphics does not offer a Scalingfunction. A LogPlot with ErrorBars does not work either and would not look ideal. Is it possible to draw rectangles with one axis being log scale?
E.g.:
Min and Max values of the rectangles are the known data.
plotting charts
$endgroup$
1
$begingroup$
Can you point to an example (even a non-Mathematica example)?
$endgroup$
– David G. Stork
5 hours ago
add a comment |
$begingroup$
I need to plot something that looks like a gant chart, but are actual data ranges. These ranges need to be on a log scale. Graphics does not offer a Scalingfunction. A LogPlot with ErrorBars does not work either and would not look ideal. Is it possible to draw rectangles with one axis being log scale?
E.g.:
Min and Max values of the rectangles are the known data.
plotting charts
$endgroup$
I need to plot something that looks like a gant chart, but are actual data ranges. These ranges need to be on a log scale. Graphics does not offer a Scalingfunction. A LogPlot with ErrorBars does not work either and would not look ideal. Is it possible to draw rectangles with one axis being log scale?
E.g.:
Min and Max values of the rectangles are the known data.
plotting charts
plotting charts
edited 5 hours ago
Mockup Dungeon
asked 5 hours ago
Mockup DungeonMockup Dungeon
888613
888613
1
$begingroup$
Can you point to an example (even a non-Mathematica example)?
$endgroup$
– David G. Stork
5 hours ago
add a comment |
1
$begingroup$
Can you point to an example (even a non-Mathematica example)?
$endgroup$
– David G. Stork
5 hours ago
1
1
$begingroup$
Can you point to an example (even a non-Mathematica example)?
$endgroup$
– David G. Stork
5 hours ago
$begingroup$
Can you point to an example (even a non-Mathematica example)?
$endgroup$
– David G. Stork
5 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
SeedRandom[1]
dates = Sort /@ RandomInteger[10, 1000, 5, 2];
data = MapIndexed[Thread@#, #2[[1]] &, dates];
labels = CharacterRange["A", "E"];
data2 = MapIndexed[Labeled[#, #2[[1]], Style[labels[[#2[[1]]]], White,
FontSize -> Scaled[.03]], Center] &, N[GeometricMean /@ dates]];
ListLogLinearPlot
Show[ListLogLinearPlot[data,
BaseStyle -> Directive[AbsoluteThickness[36], CapForm["Butt"]],
Joined -> True, AspectRatio -> 1/2, ImageSize -> Large,
PlotRange -> 0, 6, Frame -> True,
FrameTicks -> None, None, 50, 100, 200, 500, 1000, Automatic],
ListLogLinearPlot[data2, PlotMarkers -> ""]]
TimeLinePlot
TimelinePlot[Labeled[#, Style[#2, White, FontSize -> Scaled[.03]], Center] & @@@
Transpose[Interval /@ (Map[N@*Log, dates, -1]), labels],
PlotStyle -> Directive[AbsoluteThickness[40], CapForm["Butt"]],
Spacings -> .1, PlotMarkers -> "", Frame -> True,
AspectRatio -> 1/2, PerformanceGoal -> "Speed",
FrameTicks -> Automatic, Automatic,
Transpose[N@Log@#, # &@50, 100, 200, 500], None]
Thanks: @David G. Stork for the GeometricMean
idea to center the labels.
See also: Poets of the 19th century
$endgroup$
1
$begingroup$
The only needed improvement: take the geometric mean of each bar's endpoints to get the "middle" placement of the letter.Mean
is simply incorrect. TryGeometricMean
.
$endgroup$
– David G. Stork
4 hours ago
$begingroup$
Thank you @David! I was struggling with exactly that issue.
$endgroup$
– kglr
4 hours ago
add a comment |
$begingroup$
LogLinearPlot[1,
t, .1, 100,
PlotStyle -> White,
Epilog -> Rectangle[Log[1], .5, Log[50], 1],
Red, Rectangle[Log[30], 1.5, Log[80], 2],
Text[Style["a", White, 18], Log[7], .75],
Text[Style["b", White, 18], Log[53], 1.75]]
$endgroup$
$begingroup$
This is nice, was wondering how to useRectangle[]
. Thank you!
$endgroup$
– mjw
4 hours ago
add a comment |
$begingroup$
Here is a start ...
A = 10^5; B = 10;
rectangle[start_, length_, number_] := RegionPlot[Log10[start] <= x <= Log10[start + length] && number <= y <= number + .75, x, 0, Log10[A], y, 0, B];
Then we place a few rectangle[]
's in Show
[].
Show[rectangle[100, 300, 2], rectangle[200, 2000, 3], rectangle[1300, 3000, 4], rectangle[1800, 9000, 5]]
We now need to change the tick-marks and x-axis labels. I wanted to use ScalingFunction->"Log10",Automatic
but this is not a valid argument for RegionPlot[]
.
Here is how it looks right now:
With your edits, we can get the x-axis looking better! Currently the x-axis label is $log_10 x $ and not $x$ as we would like.
$endgroup$
$begingroup$
Axis-Ticks should not be a problem. I'll check tomorrow. It's now night here.
$endgroup$
– Mockup Dungeon
4 hours ago
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
SeedRandom[1]
dates = Sort /@ RandomInteger[10, 1000, 5, 2];
data = MapIndexed[Thread@#, #2[[1]] &, dates];
labels = CharacterRange["A", "E"];
data2 = MapIndexed[Labeled[#, #2[[1]], Style[labels[[#2[[1]]]], White,
FontSize -> Scaled[.03]], Center] &, N[GeometricMean /@ dates]];
ListLogLinearPlot
Show[ListLogLinearPlot[data,
BaseStyle -> Directive[AbsoluteThickness[36], CapForm["Butt"]],
Joined -> True, AspectRatio -> 1/2, ImageSize -> Large,
PlotRange -> 0, 6, Frame -> True,
FrameTicks -> None, None, 50, 100, 200, 500, 1000, Automatic],
ListLogLinearPlot[data2, PlotMarkers -> ""]]
TimeLinePlot
TimelinePlot[Labeled[#, Style[#2, White, FontSize -> Scaled[.03]], Center] & @@@
Transpose[Interval /@ (Map[N@*Log, dates, -1]), labels],
PlotStyle -> Directive[AbsoluteThickness[40], CapForm["Butt"]],
Spacings -> .1, PlotMarkers -> "", Frame -> True,
AspectRatio -> 1/2, PerformanceGoal -> "Speed",
FrameTicks -> Automatic, Automatic,
Transpose[N@Log@#, # &@50, 100, 200, 500], None]
Thanks: @David G. Stork for the GeometricMean
idea to center the labels.
See also: Poets of the 19th century
$endgroup$
1
$begingroup$
The only needed improvement: take the geometric mean of each bar's endpoints to get the "middle" placement of the letter.Mean
is simply incorrect. TryGeometricMean
.
$endgroup$
– David G. Stork
4 hours ago
$begingroup$
Thank you @David! I was struggling with exactly that issue.
$endgroup$
– kglr
4 hours ago
add a comment |
$begingroup$
SeedRandom[1]
dates = Sort /@ RandomInteger[10, 1000, 5, 2];
data = MapIndexed[Thread@#, #2[[1]] &, dates];
labels = CharacterRange["A", "E"];
data2 = MapIndexed[Labeled[#, #2[[1]], Style[labels[[#2[[1]]]], White,
FontSize -> Scaled[.03]], Center] &, N[GeometricMean /@ dates]];
ListLogLinearPlot
Show[ListLogLinearPlot[data,
BaseStyle -> Directive[AbsoluteThickness[36], CapForm["Butt"]],
Joined -> True, AspectRatio -> 1/2, ImageSize -> Large,
PlotRange -> 0, 6, Frame -> True,
FrameTicks -> None, None, 50, 100, 200, 500, 1000, Automatic],
ListLogLinearPlot[data2, PlotMarkers -> ""]]
TimeLinePlot
TimelinePlot[Labeled[#, Style[#2, White, FontSize -> Scaled[.03]], Center] & @@@
Transpose[Interval /@ (Map[N@*Log, dates, -1]), labels],
PlotStyle -> Directive[AbsoluteThickness[40], CapForm["Butt"]],
Spacings -> .1, PlotMarkers -> "", Frame -> True,
AspectRatio -> 1/2, PerformanceGoal -> "Speed",
FrameTicks -> Automatic, Automatic,
Transpose[N@Log@#, # &@50, 100, 200, 500], None]
Thanks: @David G. Stork for the GeometricMean
idea to center the labels.
See also: Poets of the 19th century
$endgroup$
1
$begingroup$
The only needed improvement: take the geometric mean of each bar's endpoints to get the "middle" placement of the letter.Mean
is simply incorrect. TryGeometricMean
.
$endgroup$
– David G. Stork
4 hours ago
$begingroup$
Thank you @David! I was struggling with exactly that issue.
$endgroup$
– kglr
4 hours ago
add a comment |
$begingroup$
SeedRandom[1]
dates = Sort /@ RandomInteger[10, 1000, 5, 2];
data = MapIndexed[Thread@#, #2[[1]] &, dates];
labels = CharacterRange["A", "E"];
data2 = MapIndexed[Labeled[#, #2[[1]], Style[labels[[#2[[1]]]], White,
FontSize -> Scaled[.03]], Center] &, N[GeometricMean /@ dates]];
ListLogLinearPlot
Show[ListLogLinearPlot[data,
BaseStyle -> Directive[AbsoluteThickness[36], CapForm["Butt"]],
Joined -> True, AspectRatio -> 1/2, ImageSize -> Large,
PlotRange -> 0, 6, Frame -> True,
FrameTicks -> None, None, 50, 100, 200, 500, 1000, Automatic],
ListLogLinearPlot[data2, PlotMarkers -> ""]]
TimeLinePlot
TimelinePlot[Labeled[#, Style[#2, White, FontSize -> Scaled[.03]], Center] & @@@
Transpose[Interval /@ (Map[N@*Log, dates, -1]), labels],
PlotStyle -> Directive[AbsoluteThickness[40], CapForm["Butt"]],
Spacings -> .1, PlotMarkers -> "", Frame -> True,
AspectRatio -> 1/2, PerformanceGoal -> "Speed",
FrameTicks -> Automatic, Automatic,
Transpose[N@Log@#, # &@50, 100, 200, 500], None]
Thanks: @David G. Stork for the GeometricMean
idea to center the labels.
See also: Poets of the 19th century
$endgroup$
SeedRandom[1]
dates = Sort /@ RandomInteger[10, 1000, 5, 2];
data = MapIndexed[Thread@#, #2[[1]] &, dates];
labels = CharacterRange["A", "E"];
data2 = MapIndexed[Labeled[#, #2[[1]], Style[labels[[#2[[1]]]], White,
FontSize -> Scaled[.03]], Center] &, N[GeometricMean /@ dates]];
ListLogLinearPlot
Show[ListLogLinearPlot[data,
BaseStyle -> Directive[AbsoluteThickness[36], CapForm["Butt"]],
Joined -> True, AspectRatio -> 1/2, ImageSize -> Large,
PlotRange -> 0, 6, Frame -> True,
FrameTicks -> None, None, 50, 100, 200, 500, 1000, Automatic],
ListLogLinearPlot[data2, PlotMarkers -> ""]]
TimeLinePlot
TimelinePlot[Labeled[#, Style[#2, White, FontSize -> Scaled[.03]], Center] & @@@
Transpose[Interval /@ (Map[N@*Log, dates, -1]), labels],
PlotStyle -> Directive[AbsoluteThickness[40], CapForm["Butt"]],
Spacings -> .1, PlotMarkers -> "", Frame -> True,
AspectRatio -> 1/2, PerformanceGoal -> "Speed",
FrameTicks -> Automatic, Automatic,
Transpose[N@Log@#, # &@50, 100, 200, 500], None]
Thanks: @David G. Stork for the GeometricMean
idea to center the labels.
See also: Poets of the 19th century
edited 12 mins ago
answered 4 hours ago
kglrkglr
189k10205422
189k10205422
1
$begingroup$
The only needed improvement: take the geometric mean of each bar's endpoints to get the "middle" placement of the letter.Mean
is simply incorrect. TryGeometricMean
.
$endgroup$
– David G. Stork
4 hours ago
$begingroup$
Thank you @David! I was struggling with exactly that issue.
$endgroup$
– kglr
4 hours ago
add a comment |
1
$begingroup$
The only needed improvement: take the geometric mean of each bar's endpoints to get the "middle" placement of the letter.Mean
is simply incorrect. TryGeometricMean
.
$endgroup$
– David G. Stork
4 hours ago
$begingroup$
Thank you @David! I was struggling with exactly that issue.
$endgroup$
– kglr
4 hours ago
1
1
$begingroup$
The only needed improvement: take the geometric mean of each bar's endpoints to get the "middle" placement of the letter.
Mean
is simply incorrect. Try GeometricMean
.$endgroup$
– David G. Stork
4 hours ago
$begingroup$
The only needed improvement: take the geometric mean of each bar's endpoints to get the "middle" placement of the letter.
Mean
is simply incorrect. Try GeometricMean
.$endgroup$
– David G. Stork
4 hours ago
$begingroup$
Thank you @David! I was struggling with exactly that issue.
$endgroup$
– kglr
4 hours ago
$begingroup$
Thank you @David! I was struggling with exactly that issue.
$endgroup$
– kglr
4 hours ago
add a comment |
$begingroup$
LogLinearPlot[1,
t, .1, 100,
PlotStyle -> White,
Epilog -> Rectangle[Log[1], .5, Log[50], 1],
Red, Rectangle[Log[30], 1.5, Log[80], 2],
Text[Style["a", White, 18], Log[7], .75],
Text[Style["b", White, 18], Log[53], 1.75]]
$endgroup$
$begingroup$
This is nice, was wondering how to useRectangle[]
. Thank you!
$endgroup$
– mjw
4 hours ago
add a comment |
$begingroup$
LogLinearPlot[1,
t, .1, 100,
PlotStyle -> White,
Epilog -> Rectangle[Log[1], .5, Log[50], 1],
Red, Rectangle[Log[30], 1.5, Log[80], 2],
Text[Style["a", White, 18], Log[7], .75],
Text[Style["b", White, 18], Log[53], 1.75]]
$endgroup$
$begingroup$
This is nice, was wondering how to useRectangle[]
. Thank you!
$endgroup$
– mjw
4 hours ago
add a comment |
$begingroup$
LogLinearPlot[1,
t, .1, 100,
PlotStyle -> White,
Epilog -> Rectangle[Log[1], .5, Log[50], 1],
Red, Rectangle[Log[30], 1.5, Log[80], 2],
Text[Style["a", White, 18], Log[7], .75],
Text[Style["b", White, 18], Log[53], 1.75]]
$endgroup$
LogLinearPlot[1,
t, .1, 100,
PlotStyle -> White,
Epilog -> Rectangle[Log[1], .5, Log[50], 1],
Red, Rectangle[Log[30], 1.5, Log[80], 2],
Text[Style["a", White, 18], Log[7], .75],
Text[Style["b", White, 18], Log[53], 1.75]]
edited 4 hours ago
answered 4 hours ago
David G. StorkDavid G. Stork
24.6k22153
24.6k22153
$begingroup$
This is nice, was wondering how to useRectangle[]
. Thank you!
$endgroup$
– mjw
4 hours ago
add a comment |
$begingroup$
This is nice, was wondering how to useRectangle[]
. Thank you!
$endgroup$
– mjw
4 hours ago
$begingroup$
This is nice, was wondering how to use
Rectangle[]
. Thank you!$endgroup$
– mjw
4 hours ago
$begingroup$
This is nice, was wondering how to use
Rectangle[]
. Thank you!$endgroup$
– mjw
4 hours ago
add a comment |
$begingroup$
Here is a start ...
A = 10^5; B = 10;
rectangle[start_, length_, number_] := RegionPlot[Log10[start] <= x <= Log10[start + length] && number <= y <= number + .75, x, 0, Log10[A], y, 0, B];
Then we place a few rectangle[]
's in Show
[].
Show[rectangle[100, 300, 2], rectangle[200, 2000, 3], rectangle[1300, 3000, 4], rectangle[1800, 9000, 5]]
We now need to change the tick-marks and x-axis labels. I wanted to use ScalingFunction->"Log10",Automatic
but this is not a valid argument for RegionPlot[]
.
Here is how it looks right now:
With your edits, we can get the x-axis looking better! Currently the x-axis label is $log_10 x $ and not $x$ as we would like.
$endgroup$
$begingroup$
Axis-Ticks should not be a problem. I'll check tomorrow. It's now night here.
$endgroup$
– Mockup Dungeon
4 hours ago
add a comment |
$begingroup$
Here is a start ...
A = 10^5; B = 10;
rectangle[start_, length_, number_] := RegionPlot[Log10[start] <= x <= Log10[start + length] && number <= y <= number + .75, x, 0, Log10[A], y, 0, B];
Then we place a few rectangle[]
's in Show
[].
Show[rectangle[100, 300, 2], rectangle[200, 2000, 3], rectangle[1300, 3000, 4], rectangle[1800, 9000, 5]]
We now need to change the tick-marks and x-axis labels. I wanted to use ScalingFunction->"Log10",Automatic
but this is not a valid argument for RegionPlot[]
.
Here is how it looks right now:
With your edits, we can get the x-axis looking better! Currently the x-axis label is $log_10 x $ and not $x$ as we would like.
$endgroup$
$begingroup$
Axis-Ticks should not be a problem. I'll check tomorrow. It's now night here.
$endgroup$
– Mockup Dungeon
4 hours ago
add a comment |
$begingroup$
Here is a start ...
A = 10^5; B = 10;
rectangle[start_, length_, number_] := RegionPlot[Log10[start] <= x <= Log10[start + length] && number <= y <= number + .75, x, 0, Log10[A], y, 0, B];
Then we place a few rectangle[]
's in Show
[].
Show[rectangle[100, 300, 2], rectangle[200, 2000, 3], rectangle[1300, 3000, 4], rectangle[1800, 9000, 5]]
We now need to change the tick-marks and x-axis labels. I wanted to use ScalingFunction->"Log10",Automatic
but this is not a valid argument for RegionPlot[]
.
Here is how it looks right now:
With your edits, we can get the x-axis looking better! Currently the x-axis label is $log_10 x $ and not $x$ as we would like.
$endgroup$
Here is a start ...
A = 10^5; B = 10;
rectangle[start_, length_, number_] := RegionPlot[Log10[start] <= x <= Log10[start + length] && number <= y <= number + .75, x, 0, Log10[A], y, 0, B];
Then we place a few rectangle[]
's in Show
[].
Show[rectangle[100, 300, 2], rectangle[200, 2000, 3], rectangle[1300, 3000, 4], rectangle[1800, 9000, 5]]
We now need to change the tick-marks and x-axis labels. I wanted to use ScalingFunction->"Log10",Automatic
but this is not a valid argument for RegionPlot[]
.
Here is how it looks right now:
With your edits, we can get the x-axis looking better! Currently the x-axis label is $log_10 x $ and not $x$ as we would like.
edited 4 hours ago
answered 5 hours ago
mjwmjw
5879
5879
$begingroup$
Axis-Ticks should not be a problem. I'll check tomorrow. It's now night here.
$endgroup$
– Mockup Dungeon
4 hours ago
add a comment |
$begingroup$
Axis-Ticks should not be a problem. I'll check tomorrow. It's now night here.
$endgroup$
– Mockup Dungeon
4 hours ago
$begingroup$
Axis-Ticks should not be a problem. I'll check tomorrow. It's now night here.
$endgroup$
– Mockup Dungeon
4 hours ago
$begingroup$
Axis-Ticks should not be a problem. I'll check tomorrow. It's now night here.
$endgroup$
– Mockup Dungeon
4 hours ago
add a comment |
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Can you point to an example (even a non-Mathematica example)?
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– David G. Stork
5 hours ago