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ELI5 - "Map-filter-reduce"

Namc on December 04, 2017

Many developers that I have lately interacted with associate map-reduce-filter with functional programming, and they are actually dreaded by the co...

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Harry • Dec 5 '17 • Edited

You could simplify your JavaScript solution!

const nums = [0,1,2,3,4,5,6,7,8,9];

const ans = nums
  .map(num => num*num)
  .filter(num => num % 2 === 0)
  .reduce((total, num) => total + num, 0);

IsaacLeimgruber • Dec 5 '17

I definitely prefer chained expressions c:

Kasey Speakman • Dec 4 '17 • Edited

Many devs are probably already familiar with map/filter/reduce, but they know it in a different form: SQL.

SELECT is map

  SELECT sales_volume > 10000 AS is_preferred_member ...

WHERE is filter

  SELECT ...
   WHERE order_status = 'open'

GROUP BY or aggregate functions like SUM are forms of reduce

  SELECT SUM(order_total) ...

Once I realized that, it seemed only natural that the same capabilities be available in my programming language for in-memory collections too. You could already do them in for loops of course, but adding these explicit operations to the language saves the code and mental weight from the overhead of the loop. And more importantly, it makes the commonly-performed operation obvious, whereas a for loop has to be read line-by-line to make sure of what it is doing.

Idan Arye • Dec 5 '17

When working with map and friends, you usually want to use iterators/generators/enumerators/streams/whatever-they-are-called-in-your-language-of-choice instead of lists. You are only going to iterate them once - it's wasteful to allocate memory for it when these combinators can usually read it one value at a time and "pass" the value they emit to the next combinator.

If we take your Python example:

map(lambda x : x*x, [i for i in range(10)])

This allocates a list to hold your 10 numbers - which is redundant, because map can just read them one at a time. This is better:

map(lambda x : x*x, (i for i in range(10)))

(i for i in range(10)) will not allocate a list for holding all 10 numbers. Instead, it will create an iterator that generates the next number on demand. With 10 numbers the difference is neglectable - but when you need to handle larger lists it can be quite serious, and the syntax for doing it the right way is easy.

This holds for other languages too - each with it's own syntax/api.

Peter Holloway • Dec 5 '17 • Edited

Even

map(lambda x : x*x, (i for i in range(10)))

is excessive if you're using a comprehension anyway.

(i*i for i in range(10))

Idan Arye • Dec 5 '17

Your version doesn't do the job though...

Peter Holloway • Dec 6 '17

It doesn't? What's the difference?

Idan Arye • Dec 6 '17

The purpose of the original snippet was to demonstrate the map combinator. Your version doesn't do it.

Idan Arye • Dec 5 '17

It seems, though, like it doesn't improve things in all languages: stackoverflow.com/q/47653131/794380

Dian Fay • Dec 4 '17

Also important to note that both map and filter can be redundant if you're already using reduce. Your JavaScript example could be written like this:

const numbers = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9];

const ans = numbers.reduce(function(total, num) {
  if (num % 2 === 0) { // replaces filter
    return total += num * num; // replaces map
  }

  return total;
});

Obviously this doesn't demonstrate what map and filter do, which is less useful from that perspective! The big deal about it is that it only makes one pass. This doesn't matter much with ten elements but with bigger collections it can save a lot of time.

João Bueno • Dec 12 '17

Totally agree Dian :)

Just to register another way to write the same, I would code like this:

const ans = numbers.reduce((total, num) => (
  num % 2 === 0 // replaces filter
    ? total + num * num // replaces map
    : total
))

Regards!

Dimitar Venkov • Dec 5 '17 • Edited

for the sake of completeness, if we were interested in the total sum and are not concerned about making our code follow functional programming methods, we could rewrite the python one liner as:

sum(i*i for i in range(10) if not i % 2)

Idan Arye • Dec 5 '17

You don't have to ignore language constructs to "follow functional programming methods". In Haskell, for example, you can use the combinators:

Prelude> sum (map (\x -> x * x) (filter (\x -> x `mod` 2 == 0) [0..9]))
120

Or you can use list comprehension(like in Python):

Prelude> sum [x * x | x <- [0..9], x `mod` 2 == 0]
120

The second approach is also functional programming.

Max Cerrina • Dec 6 '17

That single line explained more of map-filter-reduce than the articles I've been reading has! Not directed at the author, to be clear, it was a genuinely great article--I just have been puzzling over it because something has been tripping around in my brain and BAM, this comment did it. My bachelor's is in math and so the map-filter-reduce descriptions sounded so close and yet so far to what I kept almost thinking and that particular one-line tripped the right memory circuits!

Namc • Dec 6 '17

Is there something I missed that I could have done better?

I tried to make it as simple as possible, and I'm sorry if it did not simplify things for you.

Max Cerrina • Dec 8 '17

No, not at all! I tried to make that clear, and failed. Your article was wonderfully written--it's just a topic that I've been chewing on for awhile and has eluded me, and that random one-liner reminded me of past math classes and was the final link, as it were. I really loved your piece!

Evan Oman • Dec 4 '17 • Edited

Java 8+

-> IntStream.range(0,10).                         
>> map(xs -> xs*xs).                              
>> filter(xs -> xs % 2 == 0).                     
>> reduce(0, (xs_1, xs_2) -> xs_1 + xs_2)         
|  Expression value is: 120                       
|    assigned to temporary variable $7 of type int

Scala

scala> (0 to 9).          
     | map(xs => xs*xs).  
     | filter(_ % 2 == 0).
     | reduce(_+_)        
res0: Int = 120

Note: Both Java and Scala have sum methods on these objects which would be equivalent the reduce I used above.

IsaacLeimgruber • Dec 5 '17

Is scala popular now?

de-russification • Dec 4 '17 • Edited

Ruby version can be written as:

def main
  nums = [*0..9]
  p nums 
  p nums.map { |x| x * x }
        .select(&:even?)
        .reduce(:+)
end
main

IsaacLeimgruber • Dec 5 '17

I'm more dreaded by fold_left/ fold_right than map, flatmap, reduce, reduce_right, reduce_left, filter and groupby