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Putting a vertical line in each Histogram using GraphicsGrid

Weird vertical line when trying to plot a histogramPlot and Histogram together in Mathematica 7Issue with combining weighted data log histogramsHow to compare Histogram Data?What happened with good old Histogram?Removing a histogram's vertical edgesHorizontal histogram axis and ticks positioningHow to do a parametric histogram?Histogram Bar Line ThicknessDrawing a vertical line at the mean of a bell curve

1

$begingroup$

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

share|improve this question

edited 2 hours ago

m_goldberg

86.7k872196

asked 3 hours ago

An old man in the sea.An old man in the sea.

1,052819

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Could you just put your isolated-histogram solution in GraphicsGrid? It'd be easier to diagnose with some code...
  $endgroup$
  – Chris K
  3 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

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

share|improve this question

edited 2 hours ago

m_goldberg

86.7k872196

asked 3 hours ago

An old man in the sea.An old man in the sea.

1,052819

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Could you just put your isolated-histogram solution in GraphicsGrid? It'd be easier to diagnose with some code...
  $endgroup$
  – Chris K
  3 hours ago
  

  
  
  
  

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

share|improve this question

edited 2 hours ago

m_goldberg

86.7k872196

asked 3 hours ago

An old man in the sea.An old man in the sea.

1,052819

$endgroup$

share|improve this question

edited 2 hours ago

m_goldberg

86.7k872196

asked 3 hours ago

An old man in the sea.An old man in the sea.

1,052819

share|improve this question

edited 2 hours ago

m_goldberg

86.7k872196

asked 3 hours ago

An old man in the sea.An old man in the sea.

1,052819

edited 2 hours ago

m_goldberg

86.7k872196

edited 2 hours ago

m_goldberg

86.7k872196

asked 3 hours ago

An old man in the sea.An old man in the sea.

1,052819

asked 3 hours ago

An old man in the sea.An old man in the sea.

1,052819

2 Answers
2

active

oldest

votes

2

$begingroup$

You can use GridLines combined with the option Method ->{"GridLinesInFront" -> True}:

{data1, data2} = RandomVariate[NormalDistribution[#, 1], 100] & /@ {2, 4};

GraphicsGrid[{Histogram[#, ImageSize -> 300, 

     GridLines -> {Thread[{Quantile[#, {.25, .975}], 

         Directive[Opacity[1], Thick, #] & /@ {Red, Blue}}], None}, 

     Method -> {"GridLinesInFront" -> True}] & /@ {data1, data2}}]

share|improve this answer

answered 2 hours ago

kglrkglr

185k10202421

$endgroup$

add a comment | 

0

$begingroup$

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]

share|improve this answer

edited 37 mins ago

answered 47 mins ago

m_goldbergm_goldberg

86.7k872196

$endgroup$

add a comment | 

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2

$begingroup$

You can use GridLines combined with the option Method ->{"GridLinesInFront" -> True}:

{data1, data2} = RandomVariate[NormalDistribution[#, 1], 100] & /@ {2, 4};

GraphicsGrid[{Histogram[#, ImageSize -> 300, 

     GridLines -> {Thread[{Quantile[#, {.25, .975}], 

         Directive[Opacity[1], Thick, #] & /@ {Red, Blue}}], None}, 

     Method -> {"GridLinesInFront" -> True}] & /@ {data1, data2}}]

share|improve this answer

answered 2 hours ago

kglrkglr

185k10202421

$endgroup$

add a comment | 

2

$begingroup$

You can use GridLines combined with the option Method ->{"GridLinesInFront" -> True}:

{data1, data2} = RandomVariate[NormalDistribution[#, 1], 100] & /@ {2, 4};

GraphicsGrid[{Histogram[#, ImageSize -> 300, 

     GridLines -> {Thread[{Quantile[#, {.25, .975}], 

         Directive[Opacity[1], Thick, #] & /@ {Red, Blue}}], None}, 

     Method -> {"GridLinesInFront" -> True}] & /@ {data1, data2}}]

share|improve this answer

answered 2 hours ago

kglrkglr

185k10202421

$endgroup$

add a comment | 

$begingroup$

You can use GridLines combined with the option Method ->{"GridLinesInFront" -> True}:

{data1, data2} = RandomVariate[NormalDistribution[#, 1], 100] & /@ {2, 4};

GraphicsGrid[{Histogram[#, ImageSize -> 300, 

     GridLines -> {Thread[{Quantile[#, {.25, .975}], 

         Directive[Opacity[1], Thick, #] & /@ {Red, Blue}}], None}, 

     Method -> {"GridLinesInFront" -> True}] & /@ {data1, data2}}]

share|improve this answer

answered 2 hours ago

kglrkglr

185k10202421

$endgroup$

{data1, data2} = RandomVariate[NormalDistribution[#, 1], 100] & /@ {2, 4};

GraphicsGrid[{Histogram[#, ImageSize -> 300, 

     GridLines -> {Thread[{Quantile[#, {.25, .975}], 

         Directive[Opacity[1], Thick, #] & /@ {Red, Blue}}], None}, 

     Method -> {"GridLinesInFront" -> True}] & /@ {data1, data2}}]

share|improve this answer

answered 2 hours ago

kglrkglr

185k10202421

answered 2 hours ago

kglrkglr

185k10202421

answered 2 hours ago

kglrkglr

185k10202421

0

$begingroup$

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]

share|improve this answer

edited 37 mins ago

answered 47 mins ago

m_goldbergm_goldberg

86.7k872196

$endgroup$

add a comment | 

0

$begingroup$

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]

share|improve this answer

edited 37 mins ago

answered 47 mins ago

m_goldbergm_goldberg

86.7k872196

$endgroup$

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]

share|improve this answer

edited 37 mins ago

answered 47 mins ago

m_goldbergm_goldberg

86.7k872196

$endgroup$

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]]

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]]

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}]

SeedRandom[4];

makeGrid[Table[circles[8], 4], Table[lines[], 4], 2]

share|improve this answer

edited 37 mins ago

answered 47 mins ago

m_goldbergm_goldberg

86.7k872196

answered 47 mins ago

m_goldbergm_goldberg

86.7k872196

answered 47 mins ago

m_goldbergm_goldberg

86.7k872196