Give a element rank list of a list

Assuming your lists are positive integers as in OP, this should be quite fast, particularly if duplication is high. Fiddle with using Union vs Sort@DeleteDuplicates - each will have an advantage depending on data.

Normal[SparseArray[
   With[{us = Sort@DeleteDuplicates@#}, us -> Range@Length@us]][[#]]]&@yourlist

For any kind of data, this s/b quick also:

Replace[#, PositionIndex[Sort@DeleteDuplicates@#][[All, 1]], 1] &@yourlist

or probably a bit quicker yet for general data:

Replace[#, 
   With[{s = Sort@DeleteDuplicates@#}, 
    AssociationThread[s, Range@Length@s]], 1] &@yourlist

A possible approach:

position[list_List] := Module[{n = 1, f}, 
   f[x_] := f[x] = n++; 
   f /@ Union[list]; 
   f /@ list
];

Examples:

SeedRandom[417];
list1 = RandomChoice[Range[22, 28], 4];

position[list1]
(* {1, 3, 4, 2} *)

SeedRandom[416];
list2 = RandomChoice[Range[22, 28], 4];

position[list2]
(* {3, 1, 2, 1} *)

A few timings:

SeedRandom[416];
list3 = RandomChoice[Range[22, 28], 10^6];

position[list3]; // AbsoluteTiming
(* {0.672818, Null} *)

SeedRandom[417];
list4 = RandomChoice[Range[10^5], 10^5];

position[list4]; // AbsoluteTiming
(* {0.379642, Null} *)

A replacement-Rule-based approach:

rankList[lst_List] := lst /. Dispatch[Thread[# -> Range@Length@#] &@ Union@lst]

It's a little bit of speed-up over Xavier's solution (position):

SeedRandom[416];
list3 = RandomChoice[Range[22, 28], 10^6];
rankList[list3]; // AbsoluteTiming // First
position[list3]; // AbsoluteTiming // First
rankList[list3] === position[list3]
(* 0.287793 *)
(* 0.584785 *)
(* True *)

SeedRandom[417];
list4 = RandomChoice[Range[10^5], 10^5];
rankList[list4]; // AbsoluteTiming // First
position[list4]; // AbsoluteTiming // First
rankList[list4] === position[list4]
(* 0.108716 *)
(* 0.281810 *)
(* True *)

---

Note: Dispatch is absolutely crucial here. Without it, the second example runs for a long time. Alternatively, use Associations (suggested by Xavier in a comment), which serve the same purpose:

rankList1[lst_List] := lst /. (AssociationThread[# -> Range@Length@#] &@Union@lst)