Javascript - Execute an array of functions against a single argument

Question

I have a class that has two functions. Add, which adds a function to an array, and Execute which executes the array of function against its argument. My code is:

class LazyEvaluation {
  constructor(){
    this.functionQueue = []
  }

  add(fn){
    this.functionQueue.push(fn)
    return this
  }

  evaluate(target){
    for (let i = 0; i < this.functionQueue.length; i++){
      let newArray = target.map(this.functionQueue[i])
      return newArray
    }
  }
}

As it stands, this code works when i only have one function in the array. My problem is as soon as there is more than one function in the array the execute function creates a new array for each function.

For example, when the functionQueue array has the following two functions:

(function timesTwo(a){ return a * 2 })
(function addOne(a) { return a + 1 })

And the execute function is given [1, 2, 3]

The output I need is [3, 5, 7] however I am getting two separate outputs of [2, 4, 6] and [2, 3, 4]

How do I ensure that the execute function doesn't create a new array for each function in the functionQueue?

user3297291 · Accepted Answer · 2018-06-06 09:34:19Z

2

You're returning too early in your for loop. You actually have to reduce your functions passing along the values from target:

class LazyEvaluation {
  constructor(){
    this.functionQueue = []
  }

  add(fn){
    this.functionQueue.push(fn)
    return this
  }

  evaluate(target){
    return target.map(x => 
      this.functionQueue.reduce((y, f) => f(y), x)
    );
  }
}

const sums = new LazyEvaluation();
sums.add(x => x * 2);
sums.add(x => x + 1);

console.log(
  sums.evaluate([1,2,3])
);

edited Jun 6, 2018 at 9:34

answered Jun 6, 2018 at 9:29

user3297291

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7 Comments

T.J. Crowder Over a year ago

That won't produce the desired output.

user3297291 Over a year ago

@T.J.Crowder Ah, read a bit too fast there and jumped on the first "mistake" I saw in the code. Fixed it now! Thanks for the heads up

T.J. Crowder Over a year ago

That's still not the output the OP said they want.

user3297291 Over a year ago

Because I chose different example data... It's not about reproducing the exact example outcome, it's about figuring out the logic, right? I expect the OP to read my code snippet and see it meets the requirement, not to only check the outcome of the console.log. Do you agree that's a valid approach?

LewMoore Over a year ago

Thanks, this worked! - just for the sake of understanding, would it be possible to explain why this worked? So, the results from the map of target is passed in to the reduce?

|

T.J. Crowder · Accepted Answer · 2018-06-06 09:33:00Z

1

For example, when the functionQueue array has the following two functions:
(function timesTwo(a){ return a * 2 })
(function addOne(a) { return a + 1 })
And the execute function is given [1, 2, 3]

The output I need is [3, 5, 7] however I am getting two separate outputs of [2, 4, 6] and [2, 3, 4]

You need to feed the result of the first function into the second, etc.:

evaluate(target){
  return target.map(value => {
      for (const f of this.functionQueue) {
          value = f(value);
      }
      return value;
  });
}

Live Example:

class LazyEvaluation {
  constructor(){
    this.functionQueue = []
  }

  add(fn){
    this.functionQueue.push(fn)
    return this
  }

  evaluate(target){
    return target.map(value => {
        for (const f of this.functionQueue) {
            value = f(value);
        }
        return value;
    });
  }
}

const l = new LazyEvaluation();
l.add(function timesTwo(a){ return a * 2 });
l.add(function addOne(a) { return a + 1 });
console.log(l.evaluate([1, 2, 3]));

answered Jun 6, 2018 at 9:33

T.J. Crowder

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5 Comments

Michael Over a year ago

Is const f of this.functionQueue valid? const shouldn't be a variable that never going to be assigned again?

T.J. Crowder Over a year ago

@Michael - Yes, it's valid -- did you notice the runnable example? :-) The semantics of const and let "variables" declared in loops are different from the behavior of var variables in loops: A different variable/constant is created for each loop iteration. for-of and for-in never try to modify the value of that variable, so you can use a constant. for does modify it, so you have to use let in that case (even though, again, it's a different variable each time).

T.J. Crowder Over a year ago

@Michael - The new behavior is hugely useful in addressing the common closures in loops problem.

Michael Over a year ago

I noticed the runnable example, and even run it in my chrome console to be sure :) I know about let f of this.functionQueue but const j... sounds illogical for me.

T.J. Crowder Over a year ago

@Michael - :-) It's logical if you dig deep enough, it's really a different variable/constant each time. for-of is syntactic sugar for acquiring an iterator, calling its next method to get a result object, and then assigning a variable/constant to the result's value property; then repeating the next call the next time. Here's a long-winded example: jsfiddle.net/x098rj7L

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