Need help defining my own rounding function

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

I’d like to define a function called Rounding which does the following:

if for x=n+a,n∈N,a∈[0,1)x = n+a, n \in \mathbb{N}, a\in [0,1) I have a<0.5a < 0.5 then the function should round down to nn. Otherwise it should round up to n+1n+1. I did this: Rounding[x_] := If[x - Floor[x] < 0.5, Floor[x], Ceiling[x]] It works. 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. =================      It is a general question. Do we think that we are here to make somebody's homework? – Alexei Boulbitch Apr 9 '14 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]]; Now 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.      Hi - can I do it without Listable? I'm not allowed to use it yet =/ – K. L. Apr 9 '14 at 15:53      Great help, I thank you so much. – K. L. Apr 9 '14 at 16:39