If I start out with a list A with $n$ elements in it, how do I append elements to it until it has a length $N$?

For example if A={1,2,3,4}, and suppose I want to append the element 3.14 to A until the length of A is 10. I tried doing this with a While loop as follows:


However Mathematica gets stuck computing it. How can one do this?


3 Answers 3


If you introduce a length bound already, there is absolutely no reason for Append/AppendTo. Such a loop with Append/AppendTo has runtime complexity $O(n^2)$ because Append/AppendTo has to perform a copy of the whole list (which grows longer and longer). So a better practice is to initialize A as a list with the maximal length, to introduce an iteration counter i that actually stores the current length of the list, and just to write into A[[i]]. This way, only one entry of A is updated and the loop has complexity $O(n)$.

n = 10;
A = ConstantArray[0., n];
A[[1;;4]] = Range[1.,4.];
i = 4;
While[i < n,
 A[[i]] = 3.14


While[Length[A] < 10,
A = Append[A, 3.14]]

Which gives

{1, 2, 3, 4, 3.14, 3.14, 3.14, 3.14, 3.14, 3.14}

When you call Length[A] inside the loop using Append[] it only remembers the A that you originally defined, so you need to explicitly 'update' its value. As mentioned in the comment below, AppendTo[A, 3.14] should do the trick:

While[Length[A] < 10,
AppendTo[A, 3.14]]

Worth reading the documentation of AppendTo[].

  • 3
    $\begingroup$ it might be worth mentioning that AppendTo[A, 3.14] also works since AppendTo automatically updates A while Append does not. $\endgroup$
    – Nasser
    Apr 6, 2020 at 5:44

PadRight can be used.

In a general setting you will need to do PadRight[A, N, element].

As in this case, you can use PadRight[A, 10, 3.14] and assign that list to A itself for appending. This is much faster than While loop approach for larger sets.

A = PadRight[A,10,3.14]
(*{1, 2, 3, 4, 3.14, 3.14, 3.14, 3.14, 3.14, 3.14}*)

Another approach:

A = Join[A, ConstantArray[elem, N-n]]

Here, we can use

A = Join[A, ConstantArray[3.14, 6]]

Performance: (as on Intel i5, 8 GB RAM)

A = {1,2,3,4}
(A = PadRight[A, 3.14, 10^6])//RepeatedTiming

B = {1,2,3,4}
(B = Join[B, ConstantArray[3.14, 10^6-4]])//RepeatedTiming

A == B

{0.014, (a big list)}

{0.072, (a big list)}



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