Menu
I want to write a
Module Arg[f_,n_]
that takes a function f (having <=n arguments) and a natural number n and outputs the n-th argument of the function f.
As an example, suppose that f is defined by
f[a_,b_]=a^2+b^2.
Then,
Arg[f[s,t],1]
should be s;
while
Arg[f[u,v],2]
should be v.
My question is whether this is possible. If so, what should I write in the place of "???" below?
Arg[f_,n_] := Module[{}, ??? ]
Note that I don't want to specify a_ and b_ in the definition of Arg like
Arg[f_,a_,b_,n_]
EDIT: "Arg" is just my name for the module not the internal function Arg of Mathematica.
965
to answer your question concerning calling functions without their parameters. it's possible for functions where default values for all parameters are set e.g.: def foo(bar = "default string"): print(bar) print(foo()) # prints: default string print(foo("hello world!")) # prints: hello world!
You can use a default argument in Python if you wish to call your function without passing parameters. The function parameter takes the default value if the parameter is not supplied during the function call.
Defining a Function Without Parameters We can call the function by typing its name followed by parentheses () . When we call this function, it will print the current date. Note that, this output can be different for you. The output will be the date in which you call the function.
How to call a function without using the function name to send parameters? Explanation: None.
Perhaps
SetAttributes[arg, HoldFirst];
arg[f_[x___], n_] := {x}[[n]]
f[a_, b_] := a^2 + b^2.
arg[f[arg[f[s, t], 1], t], 1]
arg[f[s, t], 2]
(*
-> s
-> t
*)
arg[ArcTan[f[Cos@Sin@x, x], t], 1]
(*
-> x^2. + Cos[Sin[x]]^2
*)
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With
