# Handling vector comparisons in “Piecewise” function

Is it possible for the Piecewise function to process a vector comparison? For example, my input vector is

x={0,1,2,3,4}


I would like to define a function that gives me x^2 if x<3.5, and x+3 otherwise. i.e. I would like my output to be {0,1,4,9,7}. Here I define a function:

myfunc[x_] := Piecewise[{{x^2,x<3.5},{x+3,x>=3.5}}]


which obviously does not work. Is there a way to let Mathematica know that I want to compare x with 3.5 elementwise and apply that particular value to the segment as appropriate?

I'm used to R programming where I can use the ifelse function for this job, but I'm relatively new to Mathematica and don't know if this is possible.

Thanks!

• Related question: mathematica.stackexchange.com/questions/2821/… – Leonid Shifrin Apr 1 '12 at 11:15
• Thanks for your solutions; all work perfectly. As a R user I don't feel particularly comfortable with the way mathematica handles vector manipulation. Due to the nature of this problem I couldn't think of any keywords to search for a solution effectively. – David L Apr 1 '12 at 16:53
• Note that solutions using Listable and Map lose the speed advantage offered by vectorized operations in Mathematica. If speed is important, you may wish to use things like UnitStep and Unitize instead. – Leonid Shifrin Apr 1 '12 at 16:57

I'm not in front of a computer now, but I think you should be able to set your function to Listable and use it as-is:

SetAttributes[myfunc, Listable]

• I am in front of a computer. It works. – yohbs Apr 1 '12 at 12:48
• @yohbs: Thanks! – Cassini Apr 1 '12 at 18:41

There is nothing wrong with your function. You can Map it to a list as

 myfunc/@{0,1,2,3,4}


Alternatively, you can use your myfunc to define a new function that accepts lists:

 mynewfunc[y_List] := Map[myfunc, y]


Now

 mynewfunc[{0,1,2,3,4}]


gives

 {0, 1, 4, 9, 7}

• Should it work for a multivariate function that is being evaluated element-wise of two vectors? Because I have a piecewise function of two variables and I want to map two vectors on it but I can't make it work using the suggestion above. – Laura K Feb 5 '18 at 14:54
• @LauraK, I think you can use Apply (@@@) instead of Map (/@) in the case you describe. For example, given ClearAll[myfunc]; myfunc[x_, y_] := Piecewise[{{x^2 + y, x + y < 10}, {x + 3, y > 5}}]; and an input list of pairs, say, xylist = RandomInteger[9, {5, 2}], you can do myfunc @@@ xylist or mynewfunc[y_List] := Apply[myfunc, y, {1}]; mynewfunc[xylist] to get the same result. – kglr Feb 5 '18 at 18:23
• It worked. Thank you very much, @kglr! – Laura K Feb 5 '18 at 19:01