Putting a vertical line in each Histogram using GraphicsGrid












3












$begingroup$


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











$endgroup$












  • $begingroup$
    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
















3












$begingroup$


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











$endgroup$












  • $begingroup$
    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














3












3








3





$begingroup$


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











$endgroup$




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






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Feb 24 at 10:20









m_goldberg

87.9k872198




87.9k872198










asked Feb 24 at 9:31









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

1,094919




1,094919












  • $begingroup$
    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
















$begingroup$
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










2 Answers
2






active

oldest

votes


















4












$begingroup$

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}}]


enter image description here






share|improve this answer











$endgroup$













  • $begingroup$
    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



















3












$begingroup$

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]


grid






share|improve this answer











$endgroup$













  • $begingroup$
    many thanks for the answer ;)
    $endgroup$
    – An old man in the sea.
    Feb 24 at 16:02










  • $begingroup$
    I've been trying to use your solution for another problem. However, I can't understand the MakeGrid function... Could you write it in a simpler way? Thanks ;)
    $endgroup$
    – An old man in the sea.
    Mar 23 at 9:23












  • $begingroup$
    @Anoldmaninthesea. I'm sorry you are having difficulties. I think makeGrid is about as simple as possible, but I might have an idea of simple different from yours. The only part that I think might be considered complicated is the constraints imposed on the arguments to insure that garbage doesn't get passed into the function, You can remove them at some risk, but if you are careful of what you pass to the function, it should still work. If there is something in the body of the function you don't understand, open a chat room and we could discuss it there.
    $endgroup$
    – m_goldberg
    Mar 24 at 0:12












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

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}}]


enter image description here






share|improve this answer











$endgroup$













  • $begingroup$
    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
















4












$begingroup$

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}}]


enter image description here






share|improve this answer











$endgroup$













  • $begingroup$
    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














4












4








4





$begingroup$

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}}]


enter image description here






share|improve this answer











$endgroup$



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}}]


enter image description here







share|improve this answer














share|improve this answer



share|improve this answer








edited Feb 24 at 13:42

























answered Feb 24 at 11:02









kglrkglr

190k10206424




190k10206424












  • $begingroup$
    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






  • 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











3












$begingroup$

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]


grid






share|improve this answer











$endgroup$













  • $begingroup$
    many thanks for the answer ;)
    $endgroup$
    – An old man in the sea.
    Feb 24 at 16:02










  • $begingroup$
    I've been trying to use your solution for another problem. However, I can't understand the MakeGrid function... Could you write it in a simpler way? Thanks ;)
    $endgroup$
    – An old man in the sea.
    Mar 23 at 9:23












  • $begingroup$
    @Anoldmaninthesea. I'm sorry you are having difficulties. I think makeGrid is about as simple as possible, but I might have an idea of simple different from yours. The only part that I think might be considered complicated is the constraints imposed on the arguments to insure that garbage doesn't get passed into the function, You can remove them at some risk, but if you are careful of what you pass to the function, it should still work. If there is something in the body of the function you don't understand, open a chat room and we could discuss it there.
    $endgroup$
    – m_goldberg
    Mar 24 at 0:12
















3












$begingroup$

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]


grid






share|improve this answer











$endgroup$













  • $begingroup$
    many thanks for the answer ;)
    $endgroup$
    – An old man in the sea.
    Feb 24 at 16:02










  • $begingroup$
    I've been trying to use your solution for another problem. However, I can't understand the MakeGrid function... Could you write it in a simpler way? Thanks ;)
    $endgroup$
    – An old man in the sea.
    Mar 23 at 9:23












  • $begingroup$
    @Anoldmaninthesea. I'm sorry you are having difficulties. I think makeGrid is about as simple as possible, but I might have an idea of simple different from yours. The only part that I think might be considered complicated is the constraints imposed on the arguments to insure that garbage doesn't get passed into the function, You can remove them at some risk, but if you are careful of what you pass to the function, it should still work. If there is something in the body of the function you don't understand, open a chat room and we could discuss it there.
    $endgroup$
    – m_goldberg
    Mar 24 at 0:12














3












3








3





$begingroup$

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]


grid






share|improve this answer











$endgroup$



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]


grid







share|improve this answer














share|improve this answer



share|improve this answer








edited Mar 24 at 0:47

























answered Feb 24 at 12:30









m_goldbergm_goldberg

87.9k872198




87.9k872198












  • $begingroup$
    many thanks for the answer ;)
    $endgroup$
    – An old man in the sea.
    Feb 24 at 16:02










  • $begingroup$
    I've been trying to use your solution for another problem. However, I can't understand the MakeGrid function... Could you write it in a simpler way? Thanks ;)
    $endgroup$
    – An old man in the sea.
    Mar 23 at 9:23












  • $begingroup$
    @Anoldmaninthesea. I'm sorry you are having difficulties. I think makeGrid is about as simple as possible, but I might have an idea of simple different from yours. The only part that I think might be considered complicated is the constraints imposed on the arguments to insure that garbage doesn't get passed into the function, You can remove them at some risk, but if you are careful of what you pass to the function, it should still work. If there is something in the body of the function you don't understand, open a chat room and we could discuss it there.
    $endgroup$
    – m_goldberg
    Mar 24 at 0:12


















  • $begingroup$
    many thanks for the answer ;)
    $endgroup$
    – An old man in the sea.
    Feb 24 at 16:02










  • $begingroup$
    I've been trying to use your solution for another problem. However, I can't understand the MakeGrid function... Could you write it in a simpler way? Thanks ;)
    $endgroup$
    – An old man in the sea.
    Mar 23 at 9:23












  • $begingroup$
    @Anoldmaninthesea. I'm sorry you are having difficulties. I think makeGrid is about as simple as possible, but I might have an idea of simple different from yours. The only part that I think might be considered complicated is the constraints imposed on the arguments to insure that garbage doesn't get passed into the function, You can remove them at some risk, but if you are careful of what you pass to the function, it should still work. If there is something in the body of the function you don't understand, open a chat room and we could discuss it there.
    $endgroup$
    – m_goldberg
    Mar 24 at 0:12
















$begingroup$
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












$begingroup$
I've been trying to use your solution for another problem. However, I can't understand the MakeGrid function... Could you write it in a simpler way? Thanks ;)
$endgroup$
– An old man in the sea.
Mar 23 at 9:23






$begingroup$
I've been trying to use your solution for another problem. However, I can't understand the MakeGrid function... Could you write it in a simpler way? Thanks ;)
$endgroup$
– An old man in the sea.
Mar 23 at 9:23














$begingroup$
@Anoldmaninthesea. I'm sorry you are having difficulties. I think makeGrid is about as simple as possible, but I might have an idea of simple different from yours. The only part that I think might be considered complicated is the constraints imposed on the arguments to insure that garbage doesn't get passed into the function, You can remove them at some risk, but if you are careful of what you pass to the function, it should still work. If there is something in the body of the function you don't understand, open a chat room and we could discuss it there.
$endgroup$
– m_goldberg
Mar 24 at 0:12




$begingroup$
@Anoldmaninthesea. I'm sorry you are having difficulties. I think makeGrid is about as simple as possible, but I might have an idea of simple different from yours. The only part that I think might be considered complicated is the constraints imposed on the arguments to insure that garbage doesn't get passed into the function, You can remove them at some risk, but if you are careful of what you pass to the function, it should still work. If there is something in the body of the function you don't understand, open a chat room and we could discuss it there.
$endgroup$
– m_goldberg
Mar 24 at 0:12


















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