Snippet

partition()

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The other day I wanted to filter an array into two separate arrays. I’ll make a trivial example:

const nums = [1, 2, 3, 4, 5, 6]

// The 1s and 0s are coerced into trues and falses respectively
const odds = nums.filter(n => n % 2)
const evens = nums.filter(n => !(n % 2))

From a big-O notation perspective, it’s more than likely fine to do this loop twice. It would take a rather large array to make this very problematic. That said, it’s undeniably frustrating that we can’t already get the other items from .filter(). Hence, partition.

I didn’t know this was in lodash before I set out to write the function (but that’s where I got the name), so feel free to use that version instead. My version is curried for funsies. Slap the arguments together if you prefer.

const partition = predicate => array =>
  array.reduce(
    (acc, cur) => {
      predicate(cur) ? acc[0].push(cur) : acc[1].push(cur)
      return acc
    },
    [[], []]
  )

Using reduce, we can do a single pass through the array, calling the predicate function for each item . As we loop through our items, we build up a tuple. The first array in the tuple are the trues, the second, the falses. Using our example from before:

const nums = [1, 2, 3, 4, 5, 6]

const [odds, evens] = partition(n => n % 2)(nums)

And there you have it.


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