Say you have the following list of 12 elements

X={"x", "x", "x", "x", "x", "x", "x", "x", "x", "x", "x", "x"}

And I want to turn it into the following sorted list:

{"x1", "x10", "x11", "x12", "x2", "x3", "x4", "x5", "x6", "x7", "x8", "x9"}

How do I achieve this? I've tried numerous methods such as using a "Print[]" within a For loop and StringReplace but I haven't figured it out. I'm new to mathematica. Thanks in advance for the help!

  • $\begingroup$ Hi,brother or sister.If I have help you to solve this problem,click the check that left in my answer. $\endgroup$ – yode Mar 16 '16 at 1:35
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  • $\begingroup$ Sort@Table["x" <> IntegerString[i], {i, 1, 12}] $\endgroup$ – user1066 Mar 16 '16 at 10:00

Basically, the problem is to join each element in the list with its index. The MapIndexed function is handy for this.

The function to apply for each element is a straightforward StringJoin:


Overall, the solution is

Sort@MapIndexed[#1 <> ToString[First[#2]] &, X]

{"x1", "x10", "x11", "x12", "x2", "x3", "x4", "x5", "x6", "x7", "x8", "x9"}

X = {"x", "x", "x", "x", "x", "x", "x", "x", "x", "x", "x", "x"};
n = IntegerString@Range@Length@X;
Sort@MapThread[StringJoin, {X, n}]

{"x1", "x10", "x11", "x12", "x2", "x3", "x4", "x5", "x6", "x7", "x8", "x9"}

Actually,I can tell you another secret to this

Sort@Table[TemplateApply["x<*i*>"], {i, 12}]

{"x1", "x10", "x11", "x12", "x2", "x3", "x4", "x5", "x6", "x7", "x8", "x9"}

  • $\begingroup$ With[ {lst = Table["x", {12}]}, StringJoin[lst[[#]], IntegerString[#, 10, 2]] & /@ Range[Length@lst] ] will get you what you want if sorting is critical. $\endgroup$ – dwa Mar 16 '16 at 2:16
  • $\begingroup$ @dwa Thanks for your good suggestion~ $\endgroup$ – yode Mar 16 '16 at 2:21

Note you could just do:

Sort["x" <> ToString[#] & /@ Range[12]]

If you have list you can also use MapIndexed:

xs = ConstantArray["x", 12];
Sort@MapIndexed[#1 <> ToString[#2[[1]]] &, xs]

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