added <set>.replaceIndex(..)...

Signed-off-by: Alex A. Naanou <alex.nanou@gmail.com>

README.md

@@ -39,6 +39,7 @@ Library of JavaScript type extensions, types and utilities.
     - [`<map>.sort(..)`](#mapsort)
   - [`Set`](#set)
     - [`<set>.replace(..)`](#setreplace)
+    - [`<set>.replaceIndex(..)`](#setreplaceindex)
     - [`<set>.unite(..)`](#setunite)
     - [`<set>.intersect(..)`](#setintersect)
     - [`<set>.subtract(..)`](#setsubtract)
@@ -896,13 +897,33 @@ Replace value in set with other value retaining item order
     -> <set>
 ```
 
-This might not be cheap for very large sets, to simply replace 
-the value without sorting use:
+Replace the value without sorting
 ```bnf
 <set>.replace(<old>, <new>, false)
     -> <set>
 ```
 
+Note that when sorting large sets this can get expensive.
+
+
+
+### `<set>.replaceIndex(..)`
+
+Replace item at position in set retaining order
+```bnf
+<set>.replace(<old>, <new>)
+    -> <set>
+```
+
+Replace the value at index without sorting
+```bnf
+<set>.replace(<old>, <new>, false)
+    -> <set>
+```
+
+Note that when sorting large sets this can get expensive.
+
+
 
 ### `<set>.unite(..)`
 

Set.js

@@ -54,10 +54,23 @@ object.Mixin('SetMixin', 'soft', {
 			var order = [...this]
 			// XXX is this fast enough???
 			order[order.lastIndexOf(old)] = value }
-
+		// replace...
+		this.delete(old)
+		this.add(value)
+		ordered
+			&& this.sort(order)
+		return this },
+	replaceIndex: function(index, value, ordered=true){
+		// nothing to do...
+		if(this.size < index || old === value){
+			return this }
+		var order = [...this]
+		var old = order[index]
+		ordered
+			&& (order[index] = value)
+		// replace...
 		this.delete(old)
 		this.add(value)
-
 		ordered
 			&& this.sort(order)
 		return this },

package.json

@@ -1,6 +1,6 @@
 {
   "name": "ig-types",
-  "version": "6.5.1",
+  "version": "6.6.0",
   "description": "Generic JavaScript types and type extensions...",
   "main": "main.js",
   "scripts": {