How to group array with multiple properties

You've clarified that every item can only have one of the three properites is_sweet/is_spicy/is_bitter equal to true.

Given this assumption, one approach is that we can group the items based on the value of the quantity x.is_sweet*4 + x.is_spicy*2 + x.is_bitter*1 for the item x. This quantity is equal to 1 for those who have only is_bitter: true, 2 for item having only is_spicy: true, and 4 for item having only is_sweet: true.

Once you group according to that quantity, you can use _.values to get rid of the keys created by _.groupBy:

_.values(_.groupBy(foodsData, x => x.is_sweet*4 + x.is_spicy*2 + x.is_bitter*1))

Obviously you can give a descriptive name to the predicate:

const flavour = food => food.is_sweet*4 + food.is_spicy*2 + food.is_bitter*1;
let groupedFoods = _.values(_.groupBy(foodsData, flavour))

Notice that in case items are allowed to have more than one of those three keys true, the code above will still give sensible results, in the sense that you will have the eight groups relative to [is_sweet, is_spicy, is_bitter] equal to [0,0,0], [0,0,1], [0,1,0], [0,1,1], [1,0,0], [1,0,1], [1,1,0], and [1,1,1].

In hindsight, we don't need that the predicate flavour pulls out a number for each entry food; based on last paragraph, we are just happy if it pulls the three values food.is_sweet, food.is_spicy, and food.is_bitter, and puts them in an array. Lodash has _.over for this, so we could define

const flavour = _.over([is_sweet, is_spicy, is_bitter]);

A few words about the readability of the solution above.

• _.groupBy is a higher level abstraction than reduce-like (member or free) functions; in other words the latter functions are much more powerful than you need for the usecase in the question, and therefore they are less expressive than the former. Indeed, _.groupBy could well be implemented in terms of reduce. Here's a brutal implementation to show that it is possible:

  const groupBy = (things, f) => {
    return things.reduce((acc, item) => {
      if (acc[f(item)] == undefined)
        acc[f(item)] = [item];
      else
        acc[f(item)].push(item);
      return acc;
    }, {});
  }
  // both these give the same result:
    groupBy([1,2,3,4,4,4], x => x % 2 == 0);  // mine
  _.groupBy([1,2,3,4,4,4], x => x % 2 == 0);  // lodash's
  

• The solution is textually extremely short: it doesn't matter how "complicate" or "fancy" it looks; once you get used to this way of doing things (aka the right way), you end up reading the code. Look at it again:

  _.values(_.groupBy(foodsData, flavour))
  

  That's already fairly readable, but if you use lodash/fp module, you can even improve it:

  _.values(_.groupBy(flavour, foodsData))
  

  How far is that from the following English sentence?

  "get the values of the result of grouping by flavor the foods"

  Saying that's not readable just means denying the reality of things.

• Longer solutions using lower lever constructs like for loops or (a bit better) reduce-like utilities force you to parse the code, instead. There's no way you understand what those solutions do without carefully inspecting their bodies. Even if you decided to pull the function (acc, item) => { … } out of the call to reduce and give it a name, ending up with input.reduce(fun, acc) how would you even name it? It is a function which pushes elements on top of the arrays in an intermediate result. Except in trivial use cases, you're not gonna find a good name for such a mouthful. You're just moving the problem somewhere else. People reading this code will still be puzzled.

• _.over is maybe a bit scary too:

  const flavour = _.over([is_sweet, is_spicy, is_bitter]);
  

  But really, what does it cost tha you just learn and accept what _.over means? I mean, it is just like learning a new language, nothing more. And the explanation of _.over is fairly simple:

  _.over([iteratees=[_.identity]])

  Creates a function that invokes iteratees with the arguments it receives and returns their results.

  It is so simple: _.over(['a', 'b', 'd'])({a: 1, b: 2, c: 3}) == [1, 2, undefined]