First of all: I'm a beginner with Mathematica.

  1. I'd like to define a function called Rounding which does the following:

    • if for $x = n+a, n \in \mathbb{N}, a\in [0,1)$ I have $a < 0.5$ then the function should round down to $n$. Otherwise it should round up to $n+1$.

    I did this:

     Rounding[x_] := If[x - Floor[x] < 0.5, Floor[x], Ceiling[x]]

    It works.

  2. I'd like to create a list LIST of random integers and THEN create a LIST2 which rounds the elements of LIST with my Rounding function.

    LIST = Table[RandomInteger[{-10,10}]/2, {i,1,10}]
    LIST2 = Table[Round[...]???

I tried choosing each element of LIST by using Part[LIST,{k,1,10}] but I don't get it.

  • $\begingroup$ It is a general question. Do we think that we are here to make somebody's homework? $\endgroup$ Apr 9, 2014 at 8:08

1 Answer 1


In this instance, I would make your Rounding function work just the way the built in Round function does, where it will thread over lists:

Round[{1, 1.3, 0.5, 1.7, 2}] === {1, 1, 0, 2, 2}

By adding the Listable attribute to Rounding you can accomplish the same thing:

Attributes[Rounding] = {Listable};
Rounding[x_] := If[x - Floor[x] < 0.5, Floor[x], Ceiling[x]];


Rounding[{1, 1.3, 0.5, 1.7, 2}] === {1, 1, 1, 2, 2}

Then to round LIST using this function, you can use

LIST2 = Rounding[LIST]

EDIT to add: If you need to do it the boring way because it's a homework problem (;)), you can use Table to accomplish the same thing:

LIST2 = Table[Rounding[e], {e, LIST}]

In Table (and Do and even Sum) you can have the "iterator" (i.e., the last argument) range over the elements of a list directly, instead of having to index into the list, like so:

LIST2 = Table[Rounding[List[[i]]], {i, Length[LIST]}]

I think the former is easier to read, it's definitely easier to type, and best of all, it can be considerably faster to boot.

  • $\begingroup$ Hi - can I do it without Listable? I'm not allowed to use it yet =/ $\endgroup$
    – K. L.
    Apr 9, 2014 at 15:53
  • $\begingroup$ Great help, I thank you so much. $\endgroup$
    – K. L.
    Apr 9, 2014 at 16:39

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