How can I force certain strings to be sorted last in javascript?

Question

I'm trying to sort an array of strings alphabetically, except for certain strings (e.g. "NA" and "Wild") that should always be placed last. The sorting priority should be sorted_values_alphabetically < "NA" < "Wild".

If we have the following array:

["Wild", "sit", "ipsum", "dolor", "NA", "amet", "lorem"];

I would like it to be sorted as:

["amet", "dolor", "ipsum", "lorem", "sit", "NA", "Wild"];

I was thinking something like

arr.sort(function(a,b) {
  var aVal = a, bVal = b;

  // Hack to make values < "NA" < "Wild"
  if (aVal == "NA") aVal = "zzz" + aVal;
  if (bVal == "NA") bVal = "zzz" + bVal;
  if (aVal == "Wild") aVal = "zzzz" + aVal;
  if (bVal == "Wild") bVal = "zzzz" + bVal;

  return aVal.toLowerCase().localeCompare(bVal.toLowerCase());
});

But this probably wouldn't work for all Unicode characters.

I'm also interested in performant algorithms!

Performance

Just FYI, T. J. Crowder's algorithm is slightly more performant via jsPerf. Altohugh I prefer Halcyon's more concise approach!

Answer 1

So basically, all strings are less than "Wild", and all strings except "Wild" are less than "NA". The function you pass into sort should return a negative number if a < b, 0 if a == b, or a positive number if a > b. So you handle the special cases by returning appropriate values:

arr.sort(function(a,b) {

  // Everything is less than "Wild"
  if (a === "Wild") {
    return 1;  // a is greater than b
  }
  if (b === "Wild") {
    return -1; // b is greater than a
  }

  // Everything else is less than "NA"
  if (a === "NA") {
    return 1;  // a is greater than b
  }
  if (b === "NA") {
    return -1; // b is greater than a
  }

  // Normal result
  return a.toLowerCase().localeCompare(a.toLowerCase());
});

Live Example (source)

(Obviously, the detailed comments make that look longer than it actually is...)