Immutable.js - lazy sequence

Question

I recently heard about the Facebook library Immutable.js ( https://github.com/facebook/immutable-js ). I am confused because of their documentation:

var oddSquares = Immutable.Sequence(1,2,3,4,5,6,7,8) .filter(x => x % 2).map(x => x * x); console.log(oddSquares.last()); In this example, no intermediate arrays are ever created, filter is only called twice, and map is only called once 

As the filter is called only twice, name the card?

Answer 1

Due to the logic of the card and the filter regarding lazy sequence estimation

map

last (), called on the displayed sequence, returns the last () of the original sequence processed by the matching function.

So for example:

  var mappedSequence = Immutable.Sequence(1,2,3,4,5,6,7,8).map(x => x * x); console.log(mappedSequence.last()); 

It will output 64 and output the card only once , because the only thing it does is get the last element of the original sequence (8) and map it to x => x * x (resulting in 64)

filter

last (), called by the filtered sequence, will reverse the sequence until it finds a value in sequnce that matches the criteria. So for example

  var mappedSequence = Immutable.Sequence(1,2,3,4,5,6,7,8).filter(x => x % 2); console.log(mappedSequence.last()); 

The output is WIll 7 and will only call the filter twice because it first calls the filter (x => x% 2) for 8, it returns 0, which means false for javascript (so it should be filtered), and then call the filter function again. to get 7% 2 = 1, which is true for javascript, and return this value as the last without the functoin filter function anymore.

As an additional example, to help understand:

  var mappedSequence = Immutable.Sequence(1,2,3,4,6,8).filter(x => x % 2); console.log(mappedSequence.last()); 

Call the filter function four times , one for 8 (total false), one for 6 (false again), one for 4 (false again) and finally for 3 (finally, the result is true)

Combining both parts

Your example:

 var oddSquares = Immutable.Sequence(1,2,3,4,5,6,7,8) .filter(x => x % 2).map(x => x * x); console.log(oddSquares.last()); 

• To get the last () value of the displayed sequence, you first need to get the last () value of the filtered sequence
• To get the last () value of the filtered sequence, first get the last () value of the original sequence (8) and evaluate its filter function again (x => x% 2) , calling it for the first time
• Since 8% 2 = 0, it is incorrect for JS and should be filtered out, so we go to the next value (7) and call the filter function with this value again
• 7% 2 = 1, this is true for JS and should NOT be filtered, so this is the last value of the filtered sequence
• We have the last value (7) needed for the displayed sequence, so we call the display function (x => x * x) only once to get 49, the final result

In the end, we get the filter function two times , and the display function only once