Skip to main content

code review - How to deal with the condition that a function own many options?


Assming that I have a function myFunc which has some options.



Options[myFunc]={ firstOpt->1, secondOpt->"A", thirdOpt->True };

where, I set the values of firstOpt to 1 or 2,and set the value secondOpt to "A" or "B".


myFunc[arg1_,arg2_,OptionsPattern[]]:=
Module[{method},
method= OptionValue/@{firstOpt,secondOpt,thirdOpt};
Switch[
method,
{1,"A",True},subFunc1[...],
{2,"A",True},subFunc2[...],

{1,"B",True},subFunc3[...],
{2,"B",True},subFunc4[...],
{1,"A",False},subFunc4[...],
{2,"A",False},subFunc5[...],
{1,"B",False},subFunc6[...],
{2,"B",False},subFunc7[...],
]
]

Obviously, this is a fussy and awkward solution.So I would like to know how to deal with condition when myFunc has many options.



Or is it possible to know Mathematica how to deal with many options? For instance,


 Length@Options@ArrayPlot
(*48*)



An example(Implementation of Runge-Kutta Algorithm)


 (*MiddlePoint formula*)
middlePointOrderTwo[{xn_, yn_}, step_, func_] :=
Module[{K1, K2},
K1 = func[xn, yn];

K2 = func[xn + 1/2 step, yn + 1/2 step K1];
{xn + step, yn + step K2}
]
(*Henu formula of order 2*)
henuOrderTwo[{xn_, yn_}, step_, func_] :=
Module[{K1, K2},
K1 = func[xn, yn];
K2 = func[xn + 2/3 step, yn + 2/3 step K1];
{xn + step, yn + 1/4 step (K1 + 3 K2)}
]

(*Henu formula of order 2*)
henuOrderThree[{xn_, yn_}, step_, func_] :=
Module[{K1, K2, K3},
K1 = func[xn, yn];
K2 = func[xn + 1/3 step, yn + 1/3 step K1];
K3 = func[xn + 2/3 step, yn + 2/3 step K2];
{xn + step, yn + 1/4 step (K1 + 3 K3)}
]
(*Kutta formula of order 3*)
kuttaOrderThree[{xn_, yn_}, step_, func_] :=

Module[{K1, K2, K3},
K1 = func[xn, yn];
K2 = func[xn + 1/2 step, yn + 1/2 step K1];
K3 = func[xn + step, yn - step K1 + 2 step K2];
{xn + step, yn + 1/6 step (K1 + 4 K2 + K3)}
]
(*Runge-Kutta formula of order 4*)
rungeKuttaOrderFour[{xn_, yn_}, step_, func_] :=
Module[{K1, K2, K3, K4},
K1 = func[xn, yn];

K2 = func[xn + 1/2 step, yn + 1/2 step K1];
K3 = func[xn + 1/2 step, yn + 1/2 step K2];
K4 = func[xn + step, yn + step K3];
{xn + step, yn + 1/6 step (K1 + 2 K2 + 2 K3 + K4)}
]



  rungeKuttaFormula[{a_, b_}, ya_, step_, func_, OptionsPattern[]] :=
Module[{OrderMethod, num},
OrderMethod = OptionValue[SolvingOrderMethod];

num = IntegerPart[(b - a)/step];
Switch[
OrderMethod,
{2, "Henu"},
NestList[
henuOrderTwo[#, step, func] &, {a, ya}, num],
{2, "MiddlePoint"},
NestList[
middlePointOrderTwo[#, step, func] &, {a, ya}, num],
{3, "Henu"},

NestList[
henuOrderThree[#, step, func] &, {a, ya}, num],
{3, "Kutta"},
NestList[
kuttaOrderThree[#, step, func] &, {a, ya}, num],
{4, "RungeKutta"},
NestList[
rungeKuttaOrderFour[#, step, func] &, {a, ya}, num]]
]


Comments

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...

dynamic - How can I make a clickable ArrayPlot that returns input?

I would like to create a dynamic ArrayPlot so that the rectangles, when clicked, provide the input. Can I use ArrayPlot for this? Or is there something else I should have to use? Answer ArrayPlot is much more than just a simple array like Grid : it represents a ranged 2D dataset, and its visualization can be finetuned by options like DataReversed and DataRange . These features make it quite complicated to reproduce the same layout and order with Grid . Here I offer AnnotatedArrayPlot which comes in handy when your dataset is more than just a flat 2D array. The dynamic interface allows highlighting individual cells and possibly interacting with them. AnnotatedArrayPlot works the same way as ArrayPlot and accepts the same options plus Enabled , HighlightCoordinates , HighlightStyle and HighlightElementFunction . data = {{Missing["HasSomeMoreData"], GrayLevel[ 1], {RGBColor[0, 1, 1], RGBColor[0, 0, 1], GrayLevel[1]}, RGBColor[0, 1, 0]}, {GrayLevel[0], GrayLevel...

list manipulation - Selecting multiple columns from a matrix?

Sample data: data = { {{2013, 1, 1}, 24.13, 167.67, 231.82}, {{2013, 1, 2}, 32.15, 170.92, 225.99}, {{2013, 1, 3}, 35.43, 172.68, 221.67}, {{2013, 1, 4}, 36.73, 173.05, 218.32}, {{2013, 1, 5}, 58.19, 165.96, 197.05}, {{2013, 1, 6}, 69.99, 163.50, 187.52}, {{2013, 1, 7}, 71.37, 154.21, 175.58}, {{2013, 1, 8}, 72.51, 149.66, 163.25}}; I want a DateListPlot with three graphs, so for a matrix formed by columns 1 and 2, one for columns 1 and 3, and 1 for columns 1 and 4. At the moment I'm using this code: data2 = Transpose[{data[[All, 1]], data[[All, 2]]}]; data3 = Transpose[{data[[All, 1]], data[[All, 3]]}]; data4 = Transpose[{data[[All, 1]], data[[All, 4]]}]; DateListPlot[{data2, data3, data4}, Joined -> True, Filling -> {3 -> {1}}] but I have a hunch that this can be done more efficiently. I don't like the Transpose s in particular. Any ideas? edit (for extra credit) What if I need to multiply the second column by 2, which in my solution is simp...