I have a function in three variables:


I want to reduce this to g[var3] and have it output f[10,20,var3].

Hopefully you get the idea. I want to create a new function g by evaluating f at two of its three inputs, how do I do this?

Any help is appreciated.


closed as off-topic by Anton Antonov, m_goldberg, Roman, Bob Hanlon, Carl Woll May 31 at 22:29

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  • 6
    $\begingroup$ Use f[10,20,var3], that is, drop the blank (_) on the RHS. $\endgroup$ – Carl Woll May 31 at 19:07

B.! @Carl Woll's answer is a great one, it would look like this:


Notice that you don't have to stick to the same naming convention for the expressions which you create another function for. Though, I may suggest a method that looks a bit fancier, and uses pure functions, which I prefer to use in code to save space, though with taking a hit on readability if the user is unfamiliar. Find it below:


Off hand, I don't know if it runs any faster or does much more than make you use pure functions, but hey! It's a different answer :D

Also notice that you don't necessarily need to use a := SetDelayed method of defining the function, and maybe this will help you come up with some neat tricks where you define the initial function one way, use this redefinition/reduction method to define g and then ClearAll and Remove the initial function definition, as the = Set would evaluate the initial function at that very moment, but actually you have to wrap it in an Evaluate if you are to do what I just suggested, but my note about the use of := remains true to form, as that would evaluate f upon execution of your redefined/alternative function g.


Will all accomplish what you desire, and all of these would carry the same required input of g[altv3] to achieve evaluation!

Hope this helps!


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