Define function that behaves almost identically to Mathematica function
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Often I like to define my own functions that are almost exactly the same as Mathematica defined functions, apart from a few tweaks. See this question for example. I want to define them properly so they handle optional arguments correctly. What is a general strategy for accomplishing this? Here's a concrete (esoteric) example. I define myListPlot that is almost identical to ListPlot except that is adds a gridline corresponding to the first data point.
data = Table[RandomReal, {x, 1, 10}]
myListPlot[data_, opts_] := ListPlot[data, GridLines -> {None, {data[[1]]}}, opts]
myListPlot[data, {PlotStyle -> Red, Joined -> True}]
Not too bad. However I have to pass the optional arguments as a list. Instead, I would like to pass the optional arguments in the same way one does with ListPlot. In other words, I would like to modify myListPlot so that I would pass arguments like
myListPlot[data, PlotStyle -> Red, Joined -> True]
Perhaps I'm going about this completely the wrong way. Nevertheless I hope the reader understands what I'm trying to accomplish and can suggest a solution.
functions optional-arguments arguments
asked 2 hours ago
TomTom
1,243919
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add a comment |
$begingroup$
Often I like to define my own functions that are almost exactly the same as Mathematica defined functions, apart from a few tweaks. See this question for example. I want to define them properly so they handle optional arguments correctly. What is a general strategy for accomplishing this? Here's a concrete (esoteric) example. I define myListPlot that is almost identical to ListPlot except that is adds a gridline corresponding to the first data point.
data = Table[RandomReal, {x, 1, 10}]
myListPlot[data_, opts_] := ListPlot[data, GridLines -> {None, {data[[1]]}}, opts]
myListPlot[data, {PlotStyle -> Red, Joined -> True}]
Not too bad. However I have to pass the optional arguments as a list. Instead, I would like to pass the optional arguments in the same way one does with ListPlot. In other words, I would like to modify myListPlot so that I would pass arguments like
myListPlot[data, PlotStyle -> Red, Joined -> True]
Perhaps I'm going about this completely the wrong way. Nevertheless I hope the reader understands what I'm trying to accomplish and can suggest a solution.
functions optional-arguments arguments
asked 2 hours ago
TomTom
1,243919
$endgroup$
$begingroup$
Try changingmyListPlot[data_, opts_]tomyListPlot[data_, opts___].
$endgroup$
– rafalc
1 hour ago
$begingroup$
Works like a charm. Why not go ahead and make it an answer.
$endgroup$
– Tom
1 hour ago
add a comment |
$begingroup$
Often I like to define my own functions that are almost exactly the same as Mathematica defined functions, apart from a few tweaks. See this question for example. I want to define them properly so they handle optional arguments correctly. What is a general strategy for accomplishing this? Here's a concrete (esoteric) example. I define myListPlot that is almost identical to ListPlot except that is adds a gridline corresponding to the first data point.
data = Table[RandomReal, {x, 1, 10}]
myListPlot[data_, opts_] := ListPlot[data, GridLines -> {None, {data[[1]]}}, opts]
myListPlot[data, {PlotStyle -> Red, Joined -> True}]
Not too bad. However I have to pass the optional arguments as a list. Instead, I would like to pass the optional arguments in the same way one does with ListPlot. In other words, I would like to modify myListPlot so that I would pass arguments like
myListPlot[data, PlotStyle -> Red, Joined -> True]
Perhaps I'm going about this completely the wrong way. Nevertheless I hope the reader understands what I'm trying to accomplish and can suggest a solution.
functions optional-arguments arguments
asked 2 hours ago
TomTom
1,243919
$endgroup$
Often I like to define my own functions that are almost exactly the same as Mathematica defined functions, apart from a few tweaks. See this question for example. I want to define them properly so they handle optional arguments correctly. What is a general strategy for accomplishing this? Here's a concrete (esoteric) example. I define myListPlot that is almost identical to ListPlot except that is adds a gridline corresponding to the first data point.
data = Table[RandomReal, {x, 1, 10}]
myListPlot[data_, opts_] := ListPlot[data, GridLines -> {None, {data[[1]]}}, opts]
myListPlot[data, {PlotStyle -> Red, Joined -> True}]
Not too bad. However I have to pass the optional arguments as a list. Instead, I would like to pass the optional arguments in the same way one does with ListPlot. In other words, I would like to modify myListPlot so that I would pass arguments like
myListPlot[data, PlotStyle -> Red, Joined -> True]
Perhaps I'm going about this completely the wrong way. Nevertheless I hope the reader understands what I'm trying to accomplish and can suggest a solution.
functions optional-arguments arguments
functions optional-arguments arguments
asked 2 hours ago
TomTom
1,243919
asked 2 hours ago
TomTom
1,243919
asked 2 hours ago
TomTom
1,243919
asked 2 hours ago
TomTom
1,243919
asked 2 hours ago
TomTom
1,243919
1,243919
$begingroup$
Try changingmyListPlot[data_, opts_]tomyListPlot[data_, opts___].
$endgroup$
– rafalc
1 hour ago
$begingroup$
Works like a charm. Why not go ahead and make it an answer.
$endgroup$
– Tom
1 hour ago
add a comment |
$begingroup$
Try changingmyListPlot[data_, opts_]tomyListPlot[data_, opts___].
$endgroup$
– rafalc
1 hour ago
$begingroup$
Works like a charm. Why not go ahead and make it an answer.
$endgroup$
– Tom
1 hour ago
$begingroup$
Try changing
myListPlot[data_, opts_] to myListPlot[data_, opts___].$endgroup$
– rafalc
1 hour ago
$begingroup$
Try changing
myListPlot[data_, opts_] to myListPlot[data_, opts___].$endgroup$
– rafalc
1 hour ago
$begingroup$
Works like a charm. Why not go ahead and make it an answer.
$endgroup$
– Tom
1 hour ago
$begingroup$
Works like a charm. Why not go ahead and make it an answer.
$endgroup$
– Tom
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If you want to constrain it to only options from ListPlot, you could use OptionsPattern in combination with FilterRulesand Options.
myListPlot[data_, opts : OptionsPattern] :=
ListPlot[data, GridLines -> {None, {data[[1]]}},
FilterRules[{opts}, Options[ListPlot]]]
which results in:
myListPlot[data, PlotStyle -> Red, Joined -> True]
answered 1 hour ago
chuychuy
9,2231740
$endgroup$
add a comment |
$begingroup$
The usual way to define a Wolfram Language function that takes n arguments and an arbitrary number of options is like this:
f[arg1_, ..., argn_, opts___] := ...
A little bit of pattern matching background (taken from the WL reference):
_any single expression
x_any single expression, to be namedx
__any sequence of one or more expressions
x__sequence namedx
x__hsequence of expressions, all of whose heads areh
___any sequence of zero or more expressions
x___sequence of zero or more expressions namedx
x___hsequence of zero or more expressions, all of whose heads areh
answered 1 hour ago
rafalcrafalc
617212
$endgroup$
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
$begingroup$
If you want to constrain it to only options from ListPlot, you could use OptionsPattern in combination with FilterRulesand Options.
myListPlot[data_, opts : OptionsPattern] :=
ListPlot[data, GridLines -> {None, {data[[1]]}},
FilterRules[{opts}, Options[ListPlot]]]
which results in:
myListPlot[data, PlotStyle -> Red, Joined -> True]
answered 1 hour ago
chuychuy
9,2231740
$endgroup$
add a comment |
$begingroup$
If you want to constrain it to only options from ListPlot, you could use OptionsPattern in combination with FilterRulesand Options.
myListPlot[data_, opts : OptionsPattern] :=
ListPlot[data, GridLines -> {None, {data[[1]]}},
FilterRules[{opts}, Options[ListPlot]]]
which results in:
myListPlot[data, PlotStyle -> Red, Joined -> True]
answered 1 hour ago
chuychuy
9,2231740
$endgroup$
add a comment |
$begingroup$
If you want to constrain it to only options from ListPlot, you could use OptionsPattern in combination with FilterRulesand Options.
myListPlot[data_, opts : OptionsPattern] :=
ListPlot[data, GridLines -> {None, {data[[1]]}},
FilterRules[{opts}, Options[ListPlot]]]
which results in:
myListPlot[data, PlotStyle -> Red, Joined -> True]
answered 1 hour ago
chuychuy
9,2231740
$endgroup$
If you want to constrain it to only options from ListPlot, you could use OptionsPattern in combination with FilterRulesand Options.
myListPlot[data_, opts : OptionsPattern] :=
ListPlot[data, GridLines -> {None, {data[[1]]}},
FilterRules[{opts}, Options[ListPlot]]]
which results in:
myListPlot[data, PlotStyle -> Red, Joined -> True]
answered 1 hour ago
chuychuy
9,2231740
answered 1 hour ago
chuychuy
9,2231740
answered 1 hour ago
chuychuy
9,2231740
answered 1 hour ago
chuychuy
9,2231740
9,2231740
add a comment |
add a comment |
$begingroup$
The usual way to define a Wolfram Language function that takes n arguments and an arbitrary number of options is like this:
f[arg1_, ..., argn_, opts___] := ...
A little bit of pattern matching background (taken from the WL reference):
_any single expression
x_any single expression, to be namedx
__any sequence of one or more expressions
x__sequence namedx
x__hsequence of expressions, all of whose heads areh
___any sequence of zero or more expressions
x___sequence of zero or more expressions namedx
x___hsequence of zero or more expressions, all of whose heads areh
answered 1 hour ago
rafalcrafalc
617212
$endgroup$
add a comment |
$begingroup$
The usual way to define a Wolfram Language function that takes n arguments and an arbitrary number of options is like this:
f[arg1_, ..., argn_, opts___] := ...
A little bit of pattern matching background (taken from the WL reference):
_any single expression
x_any single expression, to be namedx
__any sequence of one or more expressions
x__sequence namedx
x__hsequence of expressions, all of whose heads areh
___any sequence of zero or more expressions
x___sequence of zero or more expressions namedx
x___hsequence of zero or more expressions, all of whose heads areh
answered 1 hour ago
rafalcrafalc
617212
$endgroup$
add a comment |
$begingroup$
The usual way to define a Wolfram Language function that takes n arguments and an arbitrary number of options is like this:
f[arg1_, ..., argn_, opts___] := ...
A little bit of pattern matching background (taken from the WL reference):
_any single expression
x_any single expression, to be namedx
__any sequence of one or more expressions
x__sequence namedx
x__hsequence of expressions, all of whose heads areh
___any sequence of zero or more expressions
x___sequence of zero or more expressions namedx
x___hsequence of zero or more expressions, all of whose heads areh
answered 1 hour ago
rafalcrafalc
617212
$endgroup$
The usual way to define a Wolfram Language function that takes n arguments and an arbitrary number of options is like this:
f[arg1_, ..., argn_, opts___] := ...
A little bit of pattern matching background (taken from the WL reference):
_any single expression
x_any single expression, to be namedx
__any sequence of one or more expressions
x__sequence namedx
x__hsequence of expressions, all of whose heads areh
___any sequence of zero or more expressions
x___sequence of zero or more expressions namedx
x___hsequence of zero or more expressions, all of whose heads areh
answered 1 hour ago
rafalcrafalc
617212
answered 1 hour ago
rafalcrafalc
617212
answered 1 hour ago
rafalcrafalc
617212
answered 1 hour ago
rafalcrafalc
617212
617212
add a comment |
add a comment |
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$begingroup$
Try changing
myListPlot[data_, opts_]tomyListPlot[data_, opts___].$endgroup$
– rafalc
1 hour ago
$begingroup$
Works like a charm. Why not go ahead and make it an answer.
$endgroup$
– Tom
1 hour ago