Rotate a list. (Creates and returns a new list.)

    proc lrotate {xs {n 1}} {
      if {!$n} {return $xs}
      if {$n<0} {return [lrotate $xs [expr {[llength $xs]+$n}]]}
      
      foreach x [lassign $xs y] {
        lappend ys $x
      }
      lappend ys $y
      lrotate $ys [incr n -1]
    }

By default, rotates a list ''left'' by one element (moving the head to the tail).

Examples:

    % lrotate {1 2 3 4 5}
    2 3 4 5 1

    % lrotate {1 2 3 4 5} -1
    5 1 2 3 4

    % lrotate {1 2 3 4 5} 2
    3 4 5 1 2

    % lrotate {1 2 3 4 5} 0
    1 2 3 4 5

I'm sure someone can come up with a better implementation...

[AM] I could not resist this challenge :). Here is mine (note that allows any number for the shift):

======
proc lrotate2 {xs {n 1}} {
    set length [llength $xs]
    if {$n == 0 || $length == 0 } {return $xs}
    set n [expr {$n % $length}]
    return [concat [lrange $xs $n end] [lrange $xs 0 $n-1]]
}    
======
[KPV] This is a fairly well-know technical interview question. One quick solution is to concatenate the list with itself and extract out $length items starting at $n. 
======
proc lrotate3 {xs {n 1}} {
    set length [llength $xs]
    if {$n == 0 || $length == 0 } {return $xs}
    set n [expr {($n % $length) - 1}]
    return [lrange [concat $xs $xs] $n+1 $length+$n]
}
======