Re: repeating elements in a list

• To: mathgroup at smc.vnet.net
• Subject: [mg54978] Re: [mg54878] repeating elements in a list
• From: János <janos.lobb at yale.edu>
• Date: Tue, 8 Mar 2005 05:04:37 -0500 (EST)
• References: <200503050634.BAA00963@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```On Mar 5, 2005, at 1:34 AM, Torsten Coym wrote:

> Hello,
> suppose I have list
> lst={A, B, C}
> and I want to create a new list, where all elements are repeated n
> times
> so that (for n=3):
> newlst={A, A, A, B, B, B, C, C, C}
> I have the following code to do this:
> n=3;
> newlst=Flatten[Table[Table[lst[[i]], {n}], {i, Length[lst]}]];
> but I'm pretty sure, that there must be a more elegant way to solve the
> problem.
> thank you

I was thinking to generalize a less efficient solution like:

In[197]:=
Flatten[Cases[Tuples[lst, n],
{x1_, x2_, x3_} /;
x1 == x2 == x3]]
Out[197]=
{a, a, a, b, b, b, c, c, c}

If I construct the pattern {x1_,x2_,x3_ } this way:
In[254]:=
Table[ToExpression[
StringJoin["x", ToString[
i], "_"]], {i, 1, n}]
Out[254]=
{x1_, x2_, x3_}

then it still works and gives:
In[249]:=
Cases[Tuples[lst, n],
Table[ToExpression[
StringJoin["x",
ToString[i], "_"]],
{i, 1, n}] /; x1 == x2 ==
x3]
Out[249]=
{{a, a, a}, {b, b, b},
{c, c, c}}

If I construct the condition this way:
In[263]:=
Flatten[Fold[Equal,
First[Table[ToExpression[
StringJoin["x",
ToString[i]]],
{i, 1, n}]],
Rest[Table[ToExpression[
StringJoin["x",
ToString[i]]],
{i, 1, n}]]]]
Out[263]=
x1 == x2 == x3

and insert into the Case statement, then it does  not work.  Any good
explanation for it ?  What is the right way to build up a condition
programatically ?