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Giving Plot options defined outside of the Plot expression
All curves in plot have the same style. Cannot be fixed with Evaluate[]Force Plot Area size to be equal excluding axesWhen is Evaluate needed within function arguments?How can I change the position of my plot legends?How do I show the combined plot from the do-loop?Change all Options of Plot DynamicallyPlot with plot markers without using ListPlothow can I change the length/size ticks in a framed plot?ListPlot with many variablesHow can one control the style of the cut off markers on a plot?
$begingroup$
When I use the following code with ListPlot, it produces a plot without any errors:
graphs = {ImageSize -> Full, Frame -> True};
ListPlot[Table[x, {x, 1, 2, .01}], graphs]
However, the same thing doesn't work for Plot:
graphs = {ImageSize -> Full, Frame -> True};
Plot[x, {x, 1, 2}, graphs]
Why? What is the simple notation change that I need to make it work?
plotting options
edited 7 mins ago
J. M. is slightly pensive♦
97.9k10304464
asked 4 hours ago
axsvl77axsvl77
383212
$endgroup$
add a comment |
$begingroup$
When I use the following code with ListPlot, it produces a plot without any errors:
graphs = {ImageSize -> Full, Frame -> True};
ListPlot[Table[x, {x, 1, 2, .01}], graphs]
However, the same thing doesn't work for Plot:
graphs = {ImageSize -> Full, Frame -> True};
Plot[x, {x, 1, 2}, graphs]
Why? What is the simple notation change that I need to make it work?
plotting options
edited 7 mins ago
J. M. is slightly pensive♦
97.9k10304464
asked 4 hours ago
axsvl77axsvl77
383212
$endgroup$
add a comment |
$begingroup$
When I use the following code with ListPlot, it produces a plot without any errors:
graphs = {ImageSize -> Full, Frame -> True};
ListPlot[Table[x, {x, 1, 2, .01}], graphs]
However, the same thing doesn't work for Plot:
graphs = {ImageSize -> Full, Frame -> True};
Plot[x, {x, 1, 2}, graphs]
Why? What is the simple notation change that I need to make it work?
plotting options
edited 7 mins ago
J. M. is slightly pensive♦
97.9k10304464
asked 4 hours ago
axsvl77axsvl77
383212
$endgroup$
When I use the following code with ListPlot, it produces a plot without any errors:
graphs = {ImageSize -> Full, Frame -> True};
ListPlot[Table[x, {x, 1, 2, .01}], graphs]
However, the same thing doesn't work for Plot:
graphs = {ImageSize -> Full, Frame -> True};
Plot[x, {x, 1, 2}, graphs]
Why? What is the simple notation change that I need to make it work?
plotting options
plotting options
edited 7 mins ago
J. M. is slightly pensive♦
97.9k10304464
asked 4 hours ago
axsvl77axsvl77
383212
edited 7 mins ago
J. M. is slightly pensive♦
97.9k10304464
asked 4 hours ago
axsvl77axsvl77
383212
edited 7 mins ago
J. M. is slightly pensive♦
97.9k10304464
edited 7 mins ago
J. M. is slightly pensive♦
97.9k10304464
edited 7 mins ago
J. M. is slightly pensive♦
97.9k10304464
97.9k10304464
asked 4 hours ago
axsvl77axsvl77
383212
asked 4 hours ago
axsvl77axsvl77
383212
asked 4 hours ago
axsvl77axsvl77
383212
383212
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You can use
Plot[x, {x, 1, 2}, Evaluate@graphs]
Why?
The reason Plot[x, {x, 1, 2}, graphs] doesn't work and ListPlot[Table[x, {x, 1, 2, .01}], graphs]does is that Plot has attribute HoldAll ("all arguments (..) maintained in an unevaluated form")
Attributes[Plot]
{HoldAll, Protected, ReadProtected}
whereas ListPlot doesn't:
Attributes[ListPlot]
{Protected, ReadProtected}
answered 3 hours ago
kglrkglr
188k10204422
$endgroup$
add a comment |
$begingroup$
You can also use With because it makes the needed substitution before Plot sees any of its arguments.
options = {ImageSize -> Full, Frame -> True};
With[{opts = options}, Plot[x, {x, 1, 2}, opts]
answered 11 mins ago
m_goldbergm_goldberg
87.4k872198
$endgroup$
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
Plot[x, {x, 1, 2}, Evaluate@graphs]
Why?
The reason Plot[x, {x, 1, 2}, graphs] doesn't work and ListPlot[Table[x, {x, 1, 2, .01}], graphs]does is that Plot has attribute HoldAll ("all arguments (..) maintained in an unevaluated form")
Attributes[Plot]
{HoldAll, Protected, ReadProtected}
whereas ListPlot doesn't:
Attributes[ListPlot]
{Protected, ReadProtected}
answered 3 hours ago
kglrkglr
188k10204422
$endgroup$
add a comment |
$begingroup$
You can use
Plot[x, {x, 1, 2}, Evaluate@graphs]
Why?
The reason Plot[x, {x, 1, 2}, graphs] doesn't work and ListPlot[Table[x, {x, 1, 2, .01}], graphs]does is that Plot has attribute HoldAll ("all arguments (..) maintained in an unevaluated form")
Attributes[Plot]
{HoldAll, Protected, ReadProtected}
whereas ListPlot doesn't:
Attributes[ListPlot]
{Protected, ReadProtected}
answered 3 hours ago
kglrkglr
188k10204422
$endgroup$
add a comment |
$begingroup$
You can use
Plot[x, {x, 1, 2}, Evaluate@graphs]
Why?
The reason Plot[x, {x, 1, 2}, graphs] doesn't work and ListPlot[Table[x, {x, 1, 2, .01}], graphs]does is that Plot has attribute HoldAll ("all arguments (..) maintained in an unevaluated form")
Attributes[Plot]
{HoldAll, Protected, ReadProtected}
whereas ListPlot doesn't:
Attributes[ListPlot]
{Protected, ReadProtected}
answered 3 hours ago
kglrkglr
188k10204422
$endgroup$
You can use
Plot[x, {x, 1, 2}, Evaluate@graphs]
Why?
The reason Plot[x, {x, 1, 2}, graphs] doesn't work and ListPlot[Table[x, {x, 1, 2, .01}], graphs]does is that Plot has attribute HoldAll ("all arguments (..) maintained in an unevaluated form")
Attributes[Plot]
{HoldAll, Protected, ReadProtected}
whereas ListPlot doesn't:
Attributes[ListPlot]
{Protected, ReadProtected}
answered 3 hours ago
kglrkglr
188k10204422
edited 3 hours ago
answered 3 hours ago
kglrkglr
188k10204422
answered 3 hours ago
kglrkglr
188k10204422
answered 3 hours ago
kglrkglr
188k10204422
188k10204422
add a comment |
add a comment |
$begingroup$
You can also use With because it makes the needed substitution before Plot sees any of its arguments.
options = {ImageSize -> Full, Frame -> True};
With[{opts = options}, Plot[x, {x, 1, 2}, opts]
answered 11 mins ago
m_goldbergm_goldberg
87.4k872198
$endgroup$
add a comment |
$begingroup$
You can also use With because it makes the needed substitution before Plot sees any of its arguments.
options = {ImageSize -> Full, Frame -> True};
With[{opts = options}, Plot[x, {x, 1, 2}, opts]
answered 11 mins ago
m_goldbergm_goldberg
87.4k872198
$endgroup$
add a comment |
$begingroup$
You can also use With because it makes the needed substitution before Plot sees any of its arguments.
options = {ImageSize -> Full, Frame -> True};
With[{opts = options}, Plot[x, {x, 1, 2}, opts]
answered 11 mins ago
m_goldbergm_goldberg
87.4k872198
$endgroup$
You can also use With because it makes the needed substitution before Plot sees any of its arguments.
options = {ImageSize -> Full, Frame -> True};
With[{opts = options}, Plot[x, {x, 1, 2}, opts]
answered 11 mins ago
m_goldbergm_goldberg
87.4k872198
answered 11 mins ago
m_goldbergm_goldberg
87.4k872198
answered 11 mins ago
m_goldbergm_goldberg
87.4k872198
answered 11 mins ago
m_goldbergm_goldberg
87.4k872198
87.4k872198
add a comment |
add a comment |
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