Putting a vertical line in each Histogram using GraphicsGrid
$begingroup$
I'm using GraphicsGrid to show several histograms.
In each histogram, I would like to show 2 vertical lines on the 2.5 and 97.5 percentiles. If I had an isolated histogram I would use Line, and Show. However, I have no idea how to proceed with a GraphicsGrid...
plotting graphics histograms
edited Feb 24 at 10:20
m_goldberg
87k872197
asked Feb 24 at 9:31
An old man in the sea.An old man in the sea.
1,064819
$endgroup$
add a comment |
$begingroup$
I'm using GraphicsGrid to show several histograms.
In each histogram, I would like to show 2 vertical lines on the 2.5 and 97.5 percentiles. If I had an isolated histogram I would use Line, and Show. However, I have no idea how to proceed with a GraphicsGrid...
plotting graphics histograms
edited Feb 24 at 10:20
m_goldberg
87k872197
asked Feb 24 at 9:31
An old man in the sea.An old man in the sea.
1,064819
$endgroup$
$begingroup$
Could you just put your isolated-histogram solution inGraphicsGrid? It'd be easier to diagnose with some code...
$endgroup$
– Chris K
Feb 24 at 10:13
add a comment |
$begingroup$
I'm using GraphicsGrid to show several histograms.
In each histogram, I would like to show 2 vertical lines on the 2.5 and 97.5 percentiles. If I had an isolated histogram I would use Line, and Show. However, I have no idea how to proceed with a GraphicsGrid...
plotting graphics histograms
edited Feb 24 at 10:20
m_goldberg
87k872197
asked Feb 24 at 9:31
An old man in the sea.An old man in the sea.
1,064819
$endgroup$
I'm using GraphicsGrid to show several histograms.
In each histogram, I would like to show 2 vertical lines on the 2.5 and 97.5 percentiles. If I had an isolated histogram I would use Line, and Show. However, I have no idea how to proceed with a GraphicsGrid...
plotting graphics histograms
plotting graphics histograms
edited Feb 24 at 10:20
m_goldberg
87k872197
asked Feb 24 at 9:31
An old man in the sea.An old man in the sea.
1,064819
edited Feb 24 at 10:20
m_goldberg
87k872197
asked Feb 24 at 9:31
An old man in the sea.An old man in the sea.
1,064819
edited Feb 24 at 10:20
m_goldberg
87k872197
edited Feb 24 at 10:20
m_goldberg
87k872197
edited Feb 24 at 10:20
m_goldberg
87k872197
87k872197
asked Feb 24 at 9:31
An old man in the sea.An old man in the sea.
1,064819
asked Feb 24 at 9:31
An old man in the sea.An old man in the sea.
1,064819
asked Feb 24 at 9:31
An old man in the sea.An old man in the sea.
1,064819
1,064819
$begingroup$
Could you just put your isolated-histogram solution inGraphicsGrid? It'd be easier to diagnose with some code...
$endgroup$
– Chris K
Feb 24 at 10:13
add a comment |
$begingroup$
Could you just put your isolated-histogram solution inGraphicsGrid? It'd be easier to diagnose with some code...
$endgroup$
– Chris K
Feb 24 at 10:13
$begingroup$
Could you just put your isolated-histogram solution in
GraphicsGrid? It'd be easier to diagnose with some code...$endgroup$
– Chris K
Feb 24 at 10:13
$begingroup$
Could you just put your isolated-histogram solution in
GraphicsGrid? It'd be easier to diagnose with some code...$endgroup$
– Chris K
Feb 24 at 10:13
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You can use GridLines combined with the option Method ->{"GridLinesInFront" -> True}:
SeedRandom[1]
{data1, data2} = RandomVariate[NormalDistribution[#, 1], 500] & /@ {2, 4};
GraphicsGrid[{Histogram[#, ImageSize -> 300,
GridLines -> {Thread[{Quantile[#, {.025, .975}],
Directive[Opacity[1], Thick, #] & /@ {Red, Blue}}], None},
Method -> {"GridLinesInFront" -> True}] & /@ {data1, data2}}]
answered Feb 24 at 11:02
kglrkglr
186k10203422
$endgroup$
$begingroup$
Do you want to edit this so it's using {0.025, .975} not {0.25, .975}?
$endgroup$
– Eric William Smith
Feb 24 at 13:21
1
$begingroup$
Than you @Eric. Done.
$endgroup$
– kglr
Feb 24 at 13:42
$begingroup$
Many thanks for the answer. ;)
$endgroup$
– An old man in the sea.
Feb 24 at 16:01
$begingroup$
@Anoldmaninthesea., myt pleasure. Thank you for the accept.
$endgroup$
– kglr
Feb 24 at 17:20
add a comment |
$begingroup$
Combining two sets of graphics objects with Show in a graphics grid is not difficult as long as the sets are compatible. That means, at least, all the objects in both lists should be plotted in the same coordinate system and have the same image size.
Here is an example using some graphics I contrived.
Draw random group of $n$ circles
circles[n_] :=
Module[{r, cntr},
r := RandomReal[.25];
cntr := RandomReal[1, {2}];
Graphics[
Table[{EdgeForm[Black], Hue[RandomReal], Disk[cntr, r]}, n],
PlotRange -> {{0, 1}, {0, 1}},
PlotRangeClipping -> True,
Frame -> True]]
Draw two random vertical lines with the left one red and the right one blue.
lines :=
Module[{lf, rt},
lf := With[{x = RandomReal[.48]}, {Red, Line[{{x, 0}, {x, 1}}]}];
rt := With[{x = RandomReal[{.52, 1}]}, {Blue, Line[{{x, 0}, {x, 1}}]}];
Graphics[{lf, rt},
PlotRange -> {{0, 1}, {0, 1}},
PlotRangeClipping -> True,
Frame -> True]]
Now the following simple function will combined any two lists of graphics that are compatible in sense mentioned in the preamble to this answer. The rather elaborate argument patterns on the lefthand side of the SetDelayed expression represent my attempt to enforce the compatibility of the arguments.
makeGrid[g1 : {_Graphics ..}, g2 : {_Graphics ..}, rows_Integer /; rows > 0] /;
Length[g1] == Length[g2] && Mod[Length[g1], rows] == 0 :=
GraphicsGrid @ Apply[Show, Partition[Transpose[{g1, g2}], rows], {2}]
So let's make a 4 x 4 graphics grid from a list of four circles groups and a list of four pairs of vertical lines.
SeedRandom[4];
makeGrid[Table[circles[8], 4], Table[lines, 4], 2]
answered Feb 24 at 12:30
m_goldbergm_goldberg
87k872197
$endgroup$
$begingroup$
many thanks for the answer ;)
$endgroup$
– An old man in the sea.
Feb 24 at 16:02
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can use GridLines combined with the option Method ->{"GridLinesInFront" -> True}:
SeedRandom[1]
{data1, data2} = RandomVariate[NormalDistribution[#, 1], 500] & /@ {2, 4};
GraphicsGrid[{Histogram[#, ImageSize -> 300,
GridLines -> {Thread[{Quantile[#, {.025, .975}],
Directive[Opacity[1], Thick, #] & /@ {Red, Blue}}], None},
Method -> {"GridLinesInFront" -> True}] & /@ {data1, data2}}]
answered Feb 24 at 11:02
kglrkglr
186k10203422
$endgroup$
$begingroup$
Do you want to edit this so it's using {0.025, .975} not {0.25, .975}?
$endgroup$
– Eric William Smith
Feb 24 at 13:21
1
$begingroup$
Than you @Eric. Done.
$endgroup$
– kglr
Feb 24 at 13:42
$begingroup$
Many thanks for the answer. ;)
$endgroup$
– An old man in the sea.
Feb 24 at 16:01
$begingroup$
@Anoldmaninthesea., myt pleasure. Thank you for the accept.
$endgroup$
– kglr
Feb 24 at 17:20
add a comment |
$begingroup$
You can use GridLines combined with the option Method ->{"GridLinesInFront" -> True}:
SeedRandom[1]
{data1, data2} = RandomVariate[NormalDistribution[#, 1], 500] & /@ {2, 4};
GraphicsGrid[{Histogram[#, ImageSize -> 300,
GridLines -> {Thread[{Quantile[#, {.025, .975}],
Directive[Opacity[1], Thick, #] & /@ {Red, Blue}}], None},
Method -> {"GridLinesInFront" -> True}] & /@ {data1, data2}}]
answered Feb 24 at 11:02
kglrkglr
186k10203422
$endgroup$
$begingroup$
Do you want to edit this so it's using {0.025, .975} not {0.25, .975}?
$endgroup$
– Eric William Smith
Feb 24 at 13:21
1
$begingroup$
Than you @Eric. Done.
$endgroup$
– kglr
Feb 24 at 13:42
$begingroup$
Many thanks for the answer. ;)
$endgroup$
– An old man in the sea.
Feb 24 at 16:01
$begingroup$
@Anoldmaninthesea., myt pleasure. Thank you for the accept.
$endgroup$
– kglr
Feb 24 at 17:20
add a comment |
$begingroup$
You can use GridLines combined with the option Method ->{"GridLinesInFront" -> True}:
SeedRandom[1]
{data1, data2} = RandomVariate[NormalDistribution[#, 1], 500] & /@ {2, 4};
GraphicsGrid[{Histogram[#, ImageSize -> 300,
GridLines -> {Thread[{Quantile[#, {.025, .975}],
Directive[Opacity[1], Thick, #] & /@ {Red, Blue}}], None},
Method -> {"GridLinesInFront" -> True}] & /@ {data1, data2}}]
answered Feb 24 at 11:02
kglrkglr
186k10203422
$endgroup$
You can use GridLines combined with the option Method ->{"GridLinesInFront" -> True}:
SeedRandom[1]
{data1, data2} = RandomVariate[NormalDistribution[#, 1], 500] & /@ {2, 4};
GraphicsGrid[{Histogram[#, ImageSize -> 300,
GridLines -> {Thread[{Quantile[#, {.025, .975}],
Directive[Opacity[1], Thick, #] & /@ {Red, Blue}}], None},
Method -> {"GridLinesInFront" -> True}] & /@ {data1, data2}}]
answered Feb 24 at 11:02
kglrkglr
186k10203422
edited Feb 24 at 13:42
answered Feb 24 at 11:02
kglrkglr
186k10203422
answered Feb 24 at 11:02
kglrkglr
186k10203422
answered Feb 24 at 11:02
kglrkglr
186k10203422
186k10203422
$begingroup$
Do you want to edit this so it's using {0.025, .975} not {0.25, .975}?
$endgroup$
– Eric William Smith
Feb 24 at 13:21
1
$begingroup$
Than you @Eric. Done.
$endgroup$
– kglr
Feb 24 at 13:42
$begingroup$
Many thanks for the answer. ;)
$endgroup$
– An old man in the sea.
Feb 24 at 16:01
$begingroup$
@Anoldmaninthesea., myt pleasure. Thank you for the accept.
$endgroup$
– kglr
Feb 24 at 17:20
add a comment |
$begingroup$
Do you want to edit this so it's using {0.025, .975} not {0.25, .975}?
$endgroup$
– Eric William Smith
Feb 24 at 13:21
1
$begingroup$
Than you @Eric. Done.
$endgroup$
– kglr
Feb 24 at 13:42
$begingroup$
Many thanks for the answer. ;)
$endgroup$
– An old man in the sea.
Feb 24 at 16:01
$begingroup$
@Anoldmaninthesea., myt pleasure. Thank you for the accept.
$endgroup$
– kglr
Feb 24 at 17:20
$begingroup$
Do you want to edit this so it's using {0.025, .975} not {0.25, .975}?
$endgroup$
– Eric William Smith
Feb 24 at 13:21
$begingroup$
Do you want to edit this so it's using {0.025, .975} not {0.25, .975}?
$endgroup$
– Eric William Smith
Feb 24 at 13:21
1
1
$begingroup$
Than you @Eric. Done.
$endgroup$
– kglr
Feb 24 at 13:42
$begingroup$
Than you @Eric. Done.
$endgroup$
– kglr
Feb 24 at 13:42
$begingroup$
Many thanks for the answer. ;)
$endgroup$
– An old man in the sea.
Feb 24 at 16:01
$begingroup$
Many thanks for the answer. ;)
$endgroup$
– An old man in the sea.
Feb 24 at 16:01
$begingroup$
@Anoldmaninthesea., myt pleasure. Thank you for the accept.
$endgroup$
– kglr
Feb 24 at 17:20
$begingroup$
@Anoldmaninthesea., myt pleasure. Thank you for the accept.
$endgroup$
– kglr
Feb 24 at 17:20
add a comment |
$begingroup$
Combining two sets of graphics objects with Show in a graphics grid is not difficult as long as the sets are compatible. That means, at least, all the objects in both lists should be plotted in the same coordinate system and have the same image size.
Here is an example using some graphics I contrived.
Draw random group of $n$ circles
circles[n_] :=
Module[{r, cntr},
r := RandomReal[.25];
cntr := RandomReal[1, {2}];
Graphics[
Table[{EdgeForm[Black], Hue[RandomReal], Disk[cntr, r]}, n],
PlotRange -> {{0, 1}, {0, 1}},
PlotRangeClipping -> True,
Frame -> True]]
Draw two random vertical lines with the left one red and the right one blue.
lines :=
Module[{lf, rt},
lf := With[{x = RandomReal[.48]}, {Red, Line[{{x, 0}, {x, 1}}]}];
rt := With[{x = RandomReal[{.52, 1}]}, {Blue, Line[{{x, 0}, {x, 1}}]}];
Graphics[{lf, rt},
PlotRange -> {{0, 1}, {0, 1}},
PlotRangeClipping -> True,
Frame -> True]]
Now the following simple function will combined any two lists of graphics that are compatible in sense mentioned in the preamble to this answer. The rather elaborate argument patterns on the lefthand side of the SetDelayed expression represent my attempt to enforce the compatibility of the arguments.
makeGrid[g1 : {_Graphics ..}, g2 : {_Graphics ..}, rows_Integer /; rows > 0] /;
Length[g1] == Length[g2] && Mod[Length[g1], rows] == 0 :=
GraphicsGrid @ Apply[Show, Partition[Transpose[{g1, g2}], rows], {2}]
So let's make a 4 x 4 graphics grid from a list of four circles groups and a list of four pairs of vertical lines.
SeedRandom[4];
makeGrid[Table[circles[8], 4], Table[lines, 4], 2]
answered Feb 24 at 12:30
m_goldbergm_goldberg
87k872197
$endgroup$
$begingroup$
many thanks for the answer ;)
$endgroup$
– An old man in the sea.
Feb 24 at 16:02
add a comment |
$begingroup$
Combining two sets of graphics objects with Show in a graphics grid is not difficult as long as the sets are compatible. That means, at least, all the objects in both lists should be plotted in the same coordinate system and have the same image size.
Here is an example using some graphics I contrived.
Draw random group of $n$ circles
circles[n_] :=
Module[{r, cntr},
r := RandomReal[.25];
cntr := RandomReal[1, {2}];
Graphics[
Table[{EdgeForm[Black], Hue[RandomReal], Disk[cntr, r]}, n],
PlotRange -> {{0, 1}, {0, 1}},
PlotRangeClipping -> True,
Frame -> True]]
Draw two random vertical lines with the left one red and the right one blue.
lines :=
Module[{lf, rt},
lf := With[{x = RandomReal[.48]}, {Red, Line[{{x, 0}, {x, 1}}]}];
rt := With[{x = RandomReal[{.52, 1}]}, {Blue, Line[{{x, 0}, {x, 1}}]}];
Graphics[{lf, rt},
PlotRange -> {{0, 1}, {0, 1}},
PlotRangeClipping -> True,
Frame -> True]]
Now the following simple function will combined any two lists of graphics that are compatible in sense mentioned in the preamble to this answer. The rather elaborate argument patterns on the lefthand side of the SetDelayed expression represent my attempt to enforce the compatibility of the arguments.
makeGrid[g1 : {_Graphics ..}, g2 : {_Graphics ..}, rows_Integer /; rows > 0] /;
Length[g1] == Length[g2] && Mod[Length[g1], rows] == 0 :=
GraphicsGrid @ Apply[Show, Partition[Transpose[{g1, g2}], rows], {2}]
So let's make a 4 x 4 graphics grid from a list of four circles groups and a list of four pairs of vertical lines.
SeedRandom[4];
makeGrid[Table[circles[8], 4], Table[lines, 4], 2]
answered Feb 24 at 12:30
m_goldbergm_goldberg
87k872197
$endgroup$
$begingroup$
many thanks for the answer ;)
$endgroup$
– An old man in the sea.
Feb 24 at 16:02
add a comment |
$begingroup$
Combining two sets of graphics objects with Show in a graphics grid is not difficult as long as the sets are compatible. That means, at least, all the objects in both lists should be plotted in the same coordinate system and have the same image size.
Here is an example using some graphics I contrived.
Draw random group of $n$ circles
circles[n_] :=
Module[{r, cntr},
r := RandomReal[.25];
cntr := RandomReal[1, {2}];
Graphics[
Table[{EdgeForm[Black], Hue[RandomReal], Disk[cntr, r]}, n],
PlotRange -> {{0, 1}, {0, 1}},
PlotRangeClipping -> True,
Frame -> True]]
Draw two random vertical lines with the left one red and the right one blue.
lines :=
Module[{lf, rt},
lf := With[{x = RandomReal[.48]}, {Red, Line[{{x, 0}, {x, 1}}]}];
rt := With[{x = RandomReal[{.52, 1}]}, {Blue, Line[{{x, 0}, {x, 1}}]}];
Graphics[{lf, rt},
PlotRange -> {{0, 1}, {0, 1}},
PlotRangeClipping -> True,
Frame -> True]]
Now the following simple function will combined any two lists of graphics that are compatible in sense mentioned in the preamble to this answer. The rather elaborate argument patterns on the lefthand side of the SetDelayed expression represent my attempt to enforce the compatibility of the arguments.
makeGrid[g1 : {_Graphics ..}, g2 : {_Graphics ..}, rows_Integer /; rows > 0] /;
Length[g1] == Length[g2] && Mod[Length[g1], rows] == 0 :=
GraphicsGrid @ Apply[Show, Partition[Transpose[{g1, g2}], rows], {2}]
So let's make a 4 x 4 graphics grid from a list of four circles groups and a list of four pairs of vertical lines.
SeedRandom[4];
makeGrid[Table[circles[8], 4], Table[lines, 4], 2]
answered Feb 24 at 12:30
m_goldbergm_goldberg
87k872197
$endgroup$
Combining two sets of graphics objects with Show in a graphics grid is not difficult as long as the sets are compatible. That means, at least, all the objects in both lists should be plotted in the same coordinate system and have the same image size.
Here is an example using some graphics I contrived.
Draw random group of $n$ circles
circles[n_] :=
Module[{r, cntr},
r := RandomReal[.25];
cntr := RandomReal[1, {2}];
Graphics[
Table[{EdgeForm[Black], Hue[RandomReal], Disk[cntr, r]}, n],
PlotRange -> {{0, 1}, {0, 1}},
PlotRangeClipping -> True,
Frame -> True]]
Draw two random vertical lines with the left one red and the right one blue.
lines :=
Module[{lf, rt},
lf := With[{x = RandomReal[.48]}, {Red, Line[{{x, 0}, {x, 1}}]}];
rt := With[{x = RandomReal[{.52, 1}]}, {Blue, Line[{{x, 0}, {x, 1}}]}];
Graphics[{lf, rt},
PlotRange -> {{0, 1}, {0, 1}},
PlotRangeClipping -> True,
Frame -> True]]
Now the following simple function will combined any two lists of graphics that are compatible in sense mentioned in the preamble to this answer. The rather elaborate argument patterns on the lefthand side of the SetDelayed expression represent my attempt to enforce the compatibility of the arguments.
makeGrid[g1 : {_Graphics ..}, g2 : {_Graphics ..}, rows_Integer /; rows > 0] /;
Length[g1] == Length[g2] && Mod[Length[g1], rows] == 0 :=
GraphicsGrid @ Apply[Show, Partition[Transpose[{g1, g2}], rows], {2}]
So let's make a 4 x 4 graphics grid from a list of four circles groups and a list of four pairs of vertical lines.
SeedRandom[4];
makeGrid[Table[circles[8], 4], Table[lines, 4], 2]
answered Feb 24 at 12:30
m_goldbergm_goldberg
87k872197
edited Feb 24 at 12:40
answered Feb 24 at 12:30
m_goldbergm_goldberg
87k872197
answered Feb 24 at 12:30
m_goldbergm_goldberg
87k872197
answered Feb 24 at 12:30
m_goldbergm_goldberg
87k872197
87k872197
$begingroup$
many thanks for the answer ;)
$endgroup$
– An old man in the sea.
Feb 24 at 16:02
add a comment |
$begingroup$
many thanks for the answer ;)
$endgroup$
– An old man in the sea.
Feb 24 at 16:02
$begingroup$
many thanks for the answer ;)
$endgroup$
– An old man in the sea.
Feb 24 at 16:02
$begingroup$
many thanks for the answer ;)
$endgroup$
– An old man in the sea.
Feb 24 at 16:02
add a comment |
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$begingroup$
Could you just put your isolated-histogram solution in
GraphicsGrid? It'd be easier to diagnose with some code...$endgroup$
– Chris K
Feb 24 at 10:13