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
87.9k872198
asked Feb 24 at 9:31
An old man in the sea.An old man in the sea.
1,094919
$endgroup$
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
87.9k872198
asked Feb 24 at 9:31
An old man in the sea.An old man in the sea.
1,094919
$endgroup$
$begingroup$
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
87.9k872198
asked Feb 24 at 9:31
An old man in the sea.An old man in the sea.
1,094919
$endgroup$
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
87.9k872198
asked Feb 24 at 9:31
An old man in the sea.An old man in the sea.
1,094919
edited Feb 24 at 10:20
m_goldberg
87.9k872198
asked Feb 24 at 9:31
An old man in the sea.An old man in the sea.
1,094919
edited Feb 24 at 10:20
m_goldberg
87.9k872198
edited Feb 24 at 10:20
m_goldberg
87.9k872198
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
asked Feb 24 at 9:31
An old man in the sea.An old man in the sea.
1,094919
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 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
190k10206424
$endgroup$
$begingroup$
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
87.9k872198
$endgroup$
$begingroup$
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 thinkmakeGridis 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
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "387"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f192106%2fputting-a-vertical-line-in-each-histogram-using-graphicsgrid%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
190k10206424
$endgroup$
$begingroup$
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
190k10206424
$endgroup$
$begingroup$
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
190k10206424
$endgroup$
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
190k10206424
edited Feb 24 at 13:42
answered Feb 24 at 11:02
kglrkglr
190k10206424
answered Feb 24 at 11:02
kglrkglr
190k10206424
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
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
87.9k872198
$endgroup$
$begingroup$
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 thinkmakeGridis 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
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
87.9k872198
$endgroup$
$begingroup$
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 thinkmakeGridis 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
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
87.9k872198
$endgroup$
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
87.9k872198
edited Mar 24 at 0:47
answered Feb 24 at 12:30
m_goldbergm_goldberg
87.9k872198
answered Feb 24 at 12:30
m_goldbergm_goldberg
87.9k872198
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 thinkmakeGridis 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
add a comment |
$begingroup$
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 thinkmakeGridis 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
add a comment |
Thanks for contributing an answer to Mathematica Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f192106%2fputting-a-vertical-line-in-each-histogram-using-graphicsgrid%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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