Specify the range of GridLines Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Adding custom GridLines to the “automatic” onesPlot with special special gridlinesListPlot: Adding Line with Epilog to automatic GridLines - adjusting StyleHow to create dataset with ordered and hierarchically-grouped rows?GridLines make figure exported to EPS locally bitmap (?!)Labeling GridLines: Callout, Graphics or Labeled?Multiple GridLines of different colorsGridlines in BoxWhisker Chart with Logarithmic Scaling FunctionHow to plot gridlines through centrod of contourplot?Plot vertical gridlines for prime values of `x`
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Specify the range of GridLines
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Adding custom GridLines to the “automatic” onesPlot with special special gridlinesListPlot: Adding Line with Epilog to automatic GridLines - adjusting StyleHow to create dataset with ordered and hierarchically-grouped rows?GridLines make figure exported to EPS locally bitmap (?!)Labeling GridLines: Callout, Graphics or Labeled?Multiple GridLines of different colorsGridlines in BoxWhisker Chart with Logarithmic Scaling FunctionHow to plot gridlines through centrod of contourplot?Plot vertical gridlines for prime values of `x`
$begingroup$
Graphics[Circle[], Frame -> True, GridLines -> Automatic]
This puts grids across a 2D graphic (a circle here).
Is there any way to specify the range of values of GridLines? More specifically I want the part of GridLines that are inside the circle. That is the part of them that is outside the circle should be removed.
Thanks a lot!
grid-layouts grid-lines
asked 6 hours ago
DimitrisDimitris
2,3381332
$endgroup$
add a comment |
$begingroup$
Graphics[Circle[], Frame -> True, GridLines -> Automatic]
This puts grids across a 2D graphic (a circle here).
Is there any way to specify the range of values of GridLines? More specifically I want the part of GridLines that are inside the circle. That is the part of them that is outside the circle should be removed.
Thanks a lot!
grid-layouts grid-lines
asked 6 hours ago
DimitrisDimitris
2,3381332
$endgroup$
add a comment |
$begingroup$
Graphics[Circle[], Frame -> True, GridLines -> Automatic]
This puts grids across a 2D graphic (a circle here).
Is there any way to specify the range of values of GridLines? More specifically I want the part of GridLines that are inside the circle. That is the part of them that is outside the circle should be removed.
Thanks a lot!
grid-layouts grid-lines
asked 6 hours ago
DimitrisDimitris
2,3381332
$endgroup$
Graphics[Circle[], Frame -> True, GridLines -> Automatic]
This puts grids across a 2D graphic (a circle here).
Is there any way to specify the range of values of GridLines? More specifically I want the part of GridLines that are inside the circle. That is the part of them that is outside the circle should be removed.
Thanks a lot!
grid-layouts grid-lines
grid-layouts grid-lines
asked 6 hours ago
DimitrisDimitris
2,3381332
asked 6 hours ago
DimitrisDimitris
2,3381332
asked 6 hours ago
DimitrisDimitris
2,3381332
asked 6 hours ago
DimitrisDimitris
2,3381332
asked 6 hours ago
DimitrisDimitris
2,3381332
2,3381332
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You can use a FilledCurve to create a graphics primitive with a hole in it. For example:
Graphics[
White,
FilledCurve[
Line[Scaled[0,0],Scaled[1,0],Scaled[1,1],Scaled[0,1],Scaled[0,0]],
Line@CirclePoints[.5, 100]
],
Blue,
Circle[0,0,.5]
,
Frame -> True,
GridLines -> Automatic
]
answered 5 hours ago
Carl WollCarl Woll
75.4k3100197
$endgroup$
$begingroup$
This is perfect! Thank you very much. Is is possible also to modify the orientation of these grid lines (for instance by π/4 rad)?
$endgroup$
– Dimitris
4 hours ago
add a comment |
$begingroup$
You may also use parametric plot which can give more flexibility (like the pi/4 rotation you want):
pt = Table[
ParametricPlot[x, x + a, x, 1/2 (-a - Sqrt[2 - a^2]),
1/2 (-a + Sqrt[2 - a^2])], a, -1, 1, .5],
Table[ParametricPlot[x, -x + a, x, 1/2 (a - Sqrt[2 - a^2]),
1/2 (a + Sqrt[2 - a^2])], a, -1, 1, .5] // Flatten;
Show[pt, Graphics[Circle[]], PlotRange -> All, Frame -> True]
where the x range for the gridlines are from
Solve[x + a == Sqrt[1 - x^2], x]
Solve[-x + a == Sqrt[1 - x^2], x]
answered 3 hours ago
egwene sedaiegwene sedai
1,8261021
$endgroup$
$begingroup$
Thanks! How can we modify the code in order to have the center of circle (and the grids translated) to (0.5,0.5)?
$endgroup$
– Dimitris
1 hour ago
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 a FilledCurve to create a graphics primitive with a hole in it. For example:
Graphics[
White,
FilledCurve[
Line[Scaled[0,0],Scaled[1,0],Scaled[1,1],Scaled[0,1],Scaled[0,0]],
Line@CirclePoints[.5, 100]
],
Blue,
Circle[0,0,.5]
,
Frame -> True,
GridLines -> Automatic
]
answered 5 hours ago
Carl WollCarl Woll
75.4k3100197
$endgroup$
$begingroup$
This is perfect! Thank you very much. Is is possible also to modify the orientation of these grid lines (for instance by π/4 rad)?
$endgroup$
– Dimitris
4 hours ago
add a comment |
$begingroup$
You can use a FilledCurve to create a graphics primitive with a hole in it. For example:
Graphics[
White,
FilledCurve[
Line[Scaled[0,0],Scaled[1,0],Scaled[1,1],Scaled[0,1],Scaled[0,0]],
Line@CirclePoints[.5, 100]
],
Blue,
Circle[0,0,.5]
,
Frame -> True,
GridLines -> Automatic
]
answered 5 hours ago
Carl WollCarl Woll
75.4k3100197
$endgroup$
$begingroup$
This is perfect! Thank you very much. Is is possible also to modify the orientation of these grid lines (for instance by π/4 rad)?
$endgroup$
– Dimitris
4 hours ago
add a comment |
$begingroup$
You can use a FilledCurve to create a graphics primitive with a hole in it. For example:
Graphics[
White,
FilledCurve[
Line[Scaled[0,0],Scaled[1,0],Scaled[1,1],Scaled[0,1],Scaled[0,0]],
Line@CirclePoints[.5, 100]
],
Blue,
Circle[0,0,.5]
,
Frame -> True,
GridLines -> Automatic
]
answered 5 hours ago
Carl WollCarl Woll
75.4k3100197
$endgroup$
You can use a FilledCurve to create a graphics primitive with a hole in it. For example:
Graphics[
White,
FilledCurve[
Line[Scaled[0,0],Scaled[1,0],Scaled[1,1],Scaled[0,1],Scaled[0,0]],
Line@CirclePoints[.5, 100]
],
Blue,
Circle[0,0,.5]
,
Frame -> True,
GridLines -> Automatic
]
answered 5 hours ago
Carl WollCarl Woll
75.4k3100197
answered 5 hours ago
Carl WollCarl Woll
75.4k3100197
answered 5 hours ago
Carl WollCarl Woll
75.4k3100197
answered 5 hours ago
Carl WollCarl Woll
75.4k3100197
75.4k3100197
$begingroup$
This is perfect! Thank you very much. Is is possible also to modify the orientation of these grid lines (for instance by π/4 rad)?
$endgroup$
– Dimitris
4 hours ago
add a comment |
$begingroup$
This is perfect! Thank you very much. Is is possible also to modify the orientation of these grid lines (for instance by π/4 rad)?
$endgroup$
– Dimitris
4 hours ago
$begingroup$
This is perfect! Thank you very much. Is is possible also to modify the orientation of these grid lines (for instance by π/4 rad)?
$endgroup$
– Dimitris
4 hours ago
$begingroup$
This is perfect! Thank you very much. Is is possible also to modify the orientation of these grid lines (for instance by π/4 rad)?
$endgroup$
– Dimitris
4 hours ago
add a comment |
$begingroup$
You may also use parametric plot which can give more flexibility (like the pi/4 rotation you want):
pt = Table[
ParametricPlot[x, x + a, x, 1/2 (-a - Sqrt[2 - a^2]),
1/2 (-a + Sqrt[2 - a^2])], a, -1, 1, .5],
Table[ParametricPlot[x, -x + a, x, 1/2 (a - Sqrt[2 - a^2]),
1/2 (a + Sqrt[2 - a^2])], a, -1, 1, .5] // Flatten;
Show[pt, Graphics[Circle[]], PlotRange -> All, Frame -> True]
where the x range for the gridlines are from
Solve[x + a == Sqrt[1 - x^2], x]
Solve[-x + a == Sqrt[1 - x^2], x]
answered 3 hours ago
egwene sedaiegwene sedai
1,8261021
$endgroup$
$begingroup$
Thanks! How can we modify the code in order to have the center of circle (and the grids translated) to (0.5,0.5)?
$endgroup$
– Dimitris
1 hour ago
add a comment |
$begingroup$
You may also use parametric plot which can give more flexibility (like the pi/4 rotation you want):
pt = Table[
ParametricPlot[x, x + a, x, 1/2 (-a - Sqrt[2 - a^2]),
1/2 (-a + Sqrt[2 - a^2])], a, -1, 1, .5],
Table[ParametricPlot[x, -x + a, x, 1/2 (a - Sqrt[2 - a^2]),
1/2 (a + Sqrt[2 - a^2])], a, -1, 1, .5] // Flatten;
Show[pt, Graphics[Circle[]], PlotRange -> All, Frame -> True]
where the x range for the gridlines are from
Solve[x + a == Sqrt[1 - x^2], x]
Solve[-x + a == Sqrt[1 - x^2], x]
answered 3 hours ago
egwene sedaiegwene sedai
1,8261021
$endgroup$
$begingroup$
Thanks! How can we modify the code in order to have the center of circle (and the grids translated) to (0.5,0.5)?
$endgroup$
– Dimitris
1 hour ago
add a comment |
$begingroup$
You may also use parametric plot which can give more flexibility (like the pi/4 rotation you want):
pt = Table[
ParametricPlot[x, x + a, x, 1/2 (-a - Sqrt[2 - a^2]),
1/2 (-a + Sqrt[2 - a^2])], a, -1, 1, .5],
Table[ParametricPlot[x, -x + a, x, 1/2 (a - Sqrt[2 - a^2]),
1/2 (a + Sqrt[2 - a^2])], a, -1, 1, .5] // Flatten;
Show[pt, Graphics[Circle[]], PlotRange -> All, Frame -> True]
where the x range for the gridlines are from
Solve[x + a == Sqrt[1 - x^2], x]
Solve[-x + a == Sqrt[1 - x^2], x]
answered 3 hours ago
egwene sedaiegwene sedai
1,8261021
$endgroup$
You may also use parametric plot which can give more flexibility (like the pi/4 rotation you want):
pt = Table[
ParametricPlot[x, x + a, x, 1/2 (-a - Sqrt[2 - a^2]),
1/2 (-a + Sqrt[2 - a^2])], a, -1, 1, .5],
Table[ParametricPlot[x, -x + a, x, 1/2 (a - Sqrt[2 - a^2]),
1/2 (a + Sqrt[2 - a^2])], a, -1, 1, .5] // Flatten;
Show[pt, Graphics[Circle[]], PlotRange -> All, Frame -> True]
where the x range for the gridlines are from
Solve[x + a == Sqrt[1 - x^2], x]
Solve[-x + a == Sqrt[1 - x^2], x]
answered 3 hours ago
egwene sedaiegwene sedai
1,8261021
answered 3 hours ago
egwene sedaiegwene sedai
1,8261021
answered 3 hours ago
egwene sedaiegwene sedai
1,8261021
answered 3 hours ago
egwene sedaiegwene sedai
1,8261021
1,8261021
$begingroup$
Thanks! How can we modify the code in order to have the center of circle (and the grids translated) to (0.5,0.5)?
$endgroup$
– Dimitris
1 hour ago
add a comment |
$begingroup$
Thanks! How can we modify the code in order to have the center of circle (and the grids translated) to (0.5,0.5)?
$endgroup$
– Dimitris
1 hour ago
$begingroup$
Thanks! How can we modify the code in order to have the center of circle (and the grids translated) to (0.5,0.5)?
$endgroup$
– Dimitris
1 hour ago
$begingroup$
Thanks! How can we modify the code in order to have the center of circle (and the grids translated) to (0.5,0.5)?
$endgroup$
– Dimitris
1 hour ago
add a comment |
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