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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 do I get a plot with a certain size?how can I change the length/size ticks in a framed plot?DateListPlot in webMathematica doesn't show FrameTicks properlyHow can one control the style of the cut off markers on a plot?How to set labelled frame ticks on a plotAbout the OptionsPattern[] approach of inheriting OptionsI don't understand the Epilog function
$begingroup$
How can I give Plot formating expressions on a separate line just like ListPlot?
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
asked 6 hours ago
axsvl77axsvl77
383212
$endgroup$
add a comment |
$begingroup$
How can I give Plot formating expressions on a separate line just like ListPlot?
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
asked 6 hours ago
axsvl77axsvl77
383212
$endgroup$
add a comment |
$begingroup$
How can I give Plot formating expressions on a separate line just like ListPlot?
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
asked 6 hours ago
axsvl77axsvl77
383212
$endgroup$
How can I give Plot formating expressions on a separate line just like ListPlot?
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
asked 6 hours ago
axsvl77axsvl77
383212
asked 6 hours ago
axsvl77axsvl77
383212
edited 2 hours ago
axsvl77
asked 6 hours ago
axsvl77axsvl77
383212
asked 6 hours ago
axsvl77axsvl77
383212
asked 6 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 6 hours ago
kglrkglr
188k10204422
$endgroup$
$begingroup$
Alternatively, you can injectgraphusingWithas in m_goldberg's answer or usingPlot[x, x, 1, 2, #] &@graphs.
$endgroup$
– kglr
1 hour ago
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 2 hours 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 6 hours ago
kglrkglr
188k10204422
$endgroup$
$begingroup$
Alternatively, you can injectgraphusingWithas in m_goldberg's answer or usingPlot[x, x, 1, 2, #] &@graphs.
$endgroup$
– kglr
1 hour ago
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 6 hours ago
kglrkglr
188k10204422
$endgroup$
$begingroup$
Alternatively, you can injectgraphusingWithas in m_goldberg's answer or usingPlot[x, x, 1, 2, #] &@graphs.
$endgroup$
– kglr
1 hour ago
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 6 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 6 hours ago
kglrkglr
188k10204422
edited 5 hours ago
answered 6 hours ago
kglrkglr
188k10204422
answered 6 hours ago
kglrkglr
188k10204422
answered 6 hours ago
kglrkglr
188k10204422
188k10204422
$begingroup$
Alternatively, you can injectgraphusingWithas in m_goldberg's answer or usingPlot[x, x, 1, 2, #] &@graphs.
$endgroup$
– kglr
1 hour ago
add a comment |
$begingroup$
Alternatively, you can injectgraphusingWithas in m_goldberg's answer or usingPlot[x, x, 1, 2, #] &@graphs.
$endgroup$
– kglr
1 hour ago
$begingroup$
Alternatively, you can inject
graph using With as in m_goldberg's answer or using Plot[x, x, 1, 2, #] &@graphs.$endgroup$
– kglr
1 hour ago
$begingroup$
Alternatively, you can inject
graph using With as in m_goldberg's answer or using Plot[x, x, 1, 2, #] &@graphs.$endgroup$
– kglr
1 hour ago
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 2 hours 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 2 hours 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 2 hours 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 2 hours ago
m_goldbergm_goldberg
87.4k872198
answered 2 hours ago
m_goldbergm_goldberg
87.4k872198
answered 2 hours ago
m_goldbergm_goldberg
87.4k872198
answered 2 hours ago
m_goldbergm_goldberg
87.4k872198
87.4k872198
add a comment |
add a comment |
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