[This post needs better tags than I could come up with. Edits to the tags would be particularly welcome.]

I realize that it is trivial to define a function that takes an interval (i.e. two endpoints, $a < b$), and an integer $n > 1$, and returns a list of $n$ evenly spaced points $x_1 = a, x_2, \dots, x_{n-1}, x_n = b$,1 but still, given that this functionality is frequently needed, and is commonplace in other scientific computing environments, I am surprised not to be able to find a Mathematica built-in for it in the docs.

Did I miss it?

(BTW: I'd love to lay my hand on some stash of such "useful functions that should be included-by-default in Mathematica but aren't".)

1 For example,

linearmesh[a_, b_, n_Integer /; n > 1] := Range[a, b, (b - a)/(n - 1)]
linearmesh[10, 20, 300] // N // Short

Some may prefer this instead:

linearmesh2[a_, b_, n_Integer /; n > 0] := Range[a, b, (b - a)/n]
linearmesh2[10, 20, 300] // N // Short


I'm sure there are more clueful ways to implement this...

  • $\begingroup$ 1) No, you didn't miss it -- it's called Range. If you frequently need certain convenience wrappers for Range, then put them in your initialization package. 2) As to a list of "useful functions that should be included-by-default in Mathematica but aren't", I suggest that what you would want to see on such a list would be very different from what I would want to see, and this would be true for any pair of users, making a universal version of such a list essentially impossible. The range of the interests of the user base is simply too broad. $\endgroup$
    – m_goldberg
    Sep 20, 2013 at 17:12
  • 2
    $\begingroup$ Maybe FindDivisions? Depends on how exact you need to be. $\endgroup$
    – chuy
    Sep 20, 2013 at 18:28
  • $\begingroup$ @m_goldberg, instead of pairwise comparison b/w users' desired functions, what about pooling all users and rank ordering preferences. Think of FAQs or even SE which relies on tagging and synonyms to relate similar questions. $\endgroup$ Dec 16, 2014 at 14:44

4 Answers 4


The three argument form of Array is convenient and I've used this a few times in different answers (example). The definition is quite simple:

linearmesh[a_, b_, n_Integer] := Array[# &, n, {a, b}]

and it gives you the same answers as your first example. Note that in your second example (with spacing $(b-a)/n$), you have 301 samples. This definition is used when you want 300 partitions, not 300 points.

  • $\begingroup$ P.S. I thought this must have been asked before, but the only post I could find was a question on MATLAB's logspace. $\endgroup$
    – rm -rf
    Sep 20, 2013 at 18:05
  • 2
    $\begingroup$ You can also use Identity instead of # &. $\endgroup$
    – klimenkov
    Oct 9, 2016 at 11:40

In version 10.1 the function Subdivide was introduced which does precisely that.

(* {0, 1/10, 1/5, 3/10, 2/5, 1/2, 3/5, 7/10, 4/5, 9/10, 1} *)

Subdivide[10, 5]
(* {0, 2, 4, 6, 8, 10} *)

Note that the number 5 equals the number of intervals not the amount of entries in the list (which is higher by one).

  • $\begingroup$ Unfortunately, Subdivide is not very convenient when it comes to extremities which you have to handle manually (in contrast to Array). Array[# &, 1, {2, -2}] returns {0} while Subdivide[2, -2, 0] returns error. I could've settle with min or max or mean, but not with an error... $\endgroup$ Sep 20, 2016 at 10:04

This function works exactly like MATLAB's linspace as it gives you n points (rather than n+1):

linspace[x0_, x1_, n_] := Range[x0, x1, (x1 - x0)/(n - 1)];
  • 2
    $\begingroup$ How is this different from the OP's linearmesh? $\endgroup$
    – Michael E2
    Dec 16, 2014 at 12:14

From the Manual https://reference.wolfram.com/language/ref/FindDivisions.html this can be easily solved:


will produce {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} which are 10 in number though...

  • 4
    $\begingroup$ The manual also says: "finds a list of about n "nice" numbers that divide the interval around to into equally spaced parts." The keyword here is "about". $\endgroup$ Jul 1, 2015 at 21:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.