This is more of a conceptual problem.

Suppose I have three lists.

x1 = {1, 2, 9, 10};
x2 = {5, 6, 7};
x3 = {9, 10, 13};

Now, I wish to keep only the first two elements of each list. More precisely, I want

x1= {1, 2};
x2 = {5, 6};
x3 = {9, 10};

One easy way to do is,

x1=x1[[{1,2}; x2=x2[[{1,2}]]; x3=x3[[{1,2}]];

So far, it is okay. But it won't work if I try to do the same using the Do[] loop. For example,


do not assign new values to x1,x2,x3.

I also tried,


But it didn't work.

I don't understand the reason behind this. What will be the correct way to assign variables/lists in a Do[] loop?

  • $\begingroup$ You can use #[[{1, 2}]] & /@ {x1, x2, x3} $\endgroup$
    – fred85
    Sep 1, 2021 at 17:48
  • $\begingroup$ Yes, but I want to assign the new list of variables into x1, x2, x3. Your code extracts the first two elements but won't store them in x1, x2, x3. $\endgroup$
    – Epsilon
    Sep 1, 2021 at 17:53
  • 1
    $\begingroup$ How about you create a function say f[x_List]:=x[[{1,2}]]; then use x1 = f[x1] $\endgroup$
    – fred85
    Sep 1, 2021 at 18:25

1 Answer 1


Mathematica is a term rewriting system. It will replace x1, x2, x3 by their values as soon as possible. One solution is:

{x1,x2,x3} = #[[{1,2}]]&/@{x1,x2,x3}

We can make it shorter as:

{x1, x2, x3} //= #[[{1, 2}]] & /@ # & 

Here //= is the "apply to" assignment operator. Basically x //= f is shorthand for x=f[x]. Because {x1,x2,x3} appear on the left side of the assignment they won't be replaced by their values (due to the HoldFirst attribute of assignment), and the proper assignment will take place.

  • $\begingroup$ I didn't know this //=. Thanks for pointing this out. $\endgroup$
    – Epsilon
    Sep 1, 2021 at 18:33
  • $\begingroup$ You can also obtain the assignments with precise evaluation control (eg. using the Hold, Release :> operators) for example as Hold[z = z[[{1, 2}]]] /. {z :> x1} , but that seems rather tedious and to go against "the grain" of the language, so to speak. $\endgroup$ Sep 1, 2021 at 18:40

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