Version 8 of fold

Updated 2005-09-21 21:37:31 by escargo

NEM: The fold function reduces a list (or other structure) to a single value by "folding" a binary operator between successive elements of the list. To illustrate, if you have a list containing numbers such as:

 1 2 3 4 5

Then to sum that list you can fold a binary + operator between the elements, resulting in something like:

 1 + 2 + 3 + 4 + 5

(assuming infix notation). This is essentially what the "fold" function does. It takes a binary function, together with an initial value (to tack on to the end of the list) and a list:

 proc fold {func init list} {
   if {[llength $list] == 0} {
     return $init
   } else {
     apply $func [lindex $list 0] \
                 [fold $func $init [lrange $list 1 end]]
   }
 }
 proc apply {func args} { uplevel #0 $func $args }

So, we can do our sum over a list by:

 proc + {a b} { expr {$a + $b} }
 fold + 0 {1 2 3 4 5} ;# produces 15 as the answer

You may have noticed that the version of fold above is right-associative. In other words, the result it produces is equivalent to doing:

 1 + (2 + (3 + (4 + (5 + 0))))

With some operators, it might make more sense to have the reduction be done in a left-associative manner. We can define a foldl which does this easily enough. I'll leave that (and converting the definition given above to a more efficient iterative version) as an exercise.

You might wonder if we can perform the reverse operation, i.e., go from a value and a function that takes a single argument and produces a pair of results and from that generate a sequence/list. Well, you can indeed, and this is roughly equivalent to iterators. There is a demonstration of this unfold operation on that page.


LV If the above discussion doesn't seem to match the context of fold that you were seeking, another use of the word fold, in the context of text editing, is the idea of text widget elision, or representing one or more lines of text with a place holder. Some text editors call this text folding.

escargo - This use of fold reminded me of something, and I couldn't remember what. Then I recalled: It reminds me of the APL reduce operator /, where

 +/A

would add up all of the values of the vector A.


RS 2005-09-21: for non-empty lists and expr operators, the following does a very simple fold (but see Elegance vs. performance):

 expr [join $list $op]

The neutral element, if list is empty, would be 0 for "+", 1 for "*"... and the others?

escargo - By "neutral element" do you mean "identity value"? (That is, the assumed element would be whatever is the identify value for the particular operation?)


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