While solving online code exercises, I came across this one:
Given a 1-dimensional array of numbers and the number of queries, where each query has start index a, end index b and a number c, find the sum of numbers between indexes a and b (inclusive). For each occurrence of zero within the range [a,b], add the value of c to the sum. For example, numbers = [4,6,0,10], queries = [1,3,20] => for this example we need to get the sum of [4,6,0] (indexes 1-3), and because [4,6,0] has 0, we also need to add 20.
This is my code so far:
function findSum(numbers, queries) {
//declare empty array that will store the numbers
let arr = []
// declare initial sum
let sum = 0;
// get the last element of queries (c)
let lastElement = queries[0].pop()
// loop through queries and push numbers to arr, to sum them in the end
queries[0].slice(0, 2).forEach(x => {
arr.push(numbers[x - 1])
})
// check if arr has 0
let zero = arr.filter(el => el === 0)
// if arr has 0, according to the instructions we need to add the c of the q
if (zero.length != 0) {
sum = arr.reduce((a, b) => a + b, 0) + lastElement
}
else {
sum = arr.reduce((a, b) => a + b, 0)
}
return sum
}
My code works if queries is an array, but in some test cases queries may be array of arrays like [ [ 2, 2, 20 ], [ 1, 2, 10 ] ]. I don't know know how to check the numbers in case if queries is array of arrays. Any suggestions are greatly appreciated.
in some test cases queries may be array of arrays
I would expect that this would always be the case, not just in some cases. This is also clear from your code:
queries[0].pop()
This assumes a 2-dimensional array! The problem is not that you sometimes get a 1-dimensional array and other times a 2-dimensional array. The problem is that although you always get a 2-dimensional array, your code is only looking at the first query -- the one that sits at queries[0].
Instead, you should loop over all queries.
I also assume that the return value of your function must be an array, having an answer for each of the queries. This means that you probably want to have code like this:
function findSum(numbers, queries) {
return queries.map(query => {
// solve the single query
return sum;
});
}
Note that your code is not making the sum correctly, as your arr will have a length of 2 (arr.push(numbers[x - 1]) is executed exactly twice), yet the query could indicate a range with 100 values and you should derive the sum of those 100 values, not just of two.
But even if you fix all that, you'll end up with an inefficient solution that will have to iterate over many values in the input array multiple times. This needs a smarter approach.
Try to think of a way to analyse the input before processing any queries yet. Would there be something useful you could build that would help to quickly get a sum of a subarray without having to iterate that subsection again?
Here are some hints:
Hint #1
Use the following truth:
sum(numbers.slice(start, end)) == sum(numbers.slice(0, end)) - sum(numbers.slice(0, start - 1))
Hint #2
What if you would know the sum from the start of the array to any given index? Like a running sum... So for numbers=[4, 8, 0, 3] you would know [4, 12, 12, 15]. Would that help in calculating a sum for a certain range of [start, end]?
Hint #3
How could you apply the same principle for the special treatment of zeroes?
I used both methods but I am quite confused regarding the usage of both methods.
Is anything that map can do but reduce can not and vice versa?
Note: I know how to use both methods I am questioning for main difference between these method and when we need to used.
Source
Both map and reduce have as input the array and a function you define. They are in some way complementary: map cannot return one single element for an array of multiple elements, while reduce will always return the accumulator you eventually changed.
map
Using map you iterate the elements, and for each element you return an element you want.
For example, if you have an array of numbers and want to get their squares, you can do this:
// A function which calculates the square
const square = x => x * x
// Use `map` to get the square of each number
console.log([1, 2, 3, 4, 5].map(square))
reduce
Using an array as an input, you can get one single element (let's say an Object, or a Number, or another Array) based on the callback function (the first argument) which gets the accumulator and current_element parameters:
const numbers = [1, 2, 3, 4, 5]
// Calculate the sum
console.log(numbers.reduce(function (acc, current) {
return acc + current
}, 0)) // < Start with 0
// Calculate the product
console.log(numbers.reduce(function (acc, current) {
return acc * current
}, 1)) // < Start with 1
Which one should you choose when you can do the same thing with both? Try to imagine how the code looks. For the example provided, you can compute the squares array like you mentioned, using reduce:
// Using reduce
[1, 2, 3, 4, 5].reduce(function (acc, current) {
acc.push(current*current);
return acc;
}, [])
// Using map
[1, 2, 3, 4, 5].map(x => x * x)
Now, looking at these, obviously the second implementation looks better and it's shorter. Usually you'd choose the cleaner solution, which in this case is map. Of course, you can do it with reduce, but in a nutshell, think which would be shorter and eventually that would be better.
I think this picture will answer you about the difference between those Higher Order Functions
Generally "map" means converting a series of inputs to an equal length series of outputs while "reduce" means converting a series of inputs into a smaller number of outputs.
What people mean by "map-reduce" is usually construed to mean "transform, possibly in parallel, combine serially".
When you "map", you're writing a function that transforms x with f(x) into some new value x1. When you "reduce" you're writing some function g(y) that takes array y and emits array y1.
They produce different results in terms of data structure.
The map() function returns a new array through passing a function over each element in the input array.
This is different to reduce() which takes an array and a function in the same way, but the function takes 2 inputs - an accumulator and a current value.
So reduce() could be used like map() if you always .concat onto the accumulator the next output from a function. However it is more commonly used to reduce the dimensions of an array so either taking a one dimensional and returning a single value or flattening a two dimensional array etc.
Let's take a look of these two one by one.
Map
Map takes a callback and run it against every element on the array but what's
makes it unique is it generate a new array based on your existing array.
var arr = [1, 2, 3];
var mapped = arr.map(function(elem) {
return elem * 10;
})
console.log(mapped); // it genrate new array
Reduce
Reduce method of the array object is used to reduce the array to one single value.
var arr = [1, 2, 3];
var sum = arr.reduce(function(sum, elem){
return sum + elem;
})
console.log(sum) // reduce the array to one single value
I think this question is a very good question and I can't disagree with the answers but I have the feeling we are missing the point entirely.
Thinking of map and reduce more abstractly can provide us with a LOT of very good insights.
This answer is divided in 3 parts:
Defining and deciding between map and reduce (7 minutes)
Using reduce intentionally (8 minutes)
Bridging map and reduce with transducers (5 minutes)
map or reduce
Common traits
map and reduce are implemented in a meaningful and consistent way on a wide range of objects which are not necessarily collections.
They return a value useful to the surrounding algorithm, and they only care about this value.
Their major role is conveying intent regarding transformation or preservation of structure.
Structure
By "structure" I mean a set of conceptual properties which characterise abstract objects, such as an unordered list or a 2D matrix, and their concretion in data structures.
Note that there can be a disconnect between the two:
an unordered list can be stored as an array, which has the concept of ordering carried by indexed keys;
a 2D matrix can be stored as a TypedArray, which lacks the concept of dimension (or nesting).
map
map is a strict structure-preserving transformation.
It is useful to implement it on other kinds of objects to grasp its semantic value:
class A {
constructor (value) {
this.value = value
}
map (f) {
return new A(f(this.value))
}
}
new A(5).map(x => x * 2); // A { value: 10 }
Objects implementing map can have all kinds of behaviours, but they always return the same kind of object you started with while transforming the values with the supplied callback.
Array.map returns an array of the same length and the same ordering as the original.
On the callback arity
Because it preserves structure, map is viewed as a safe operation, but not every callback is equal.
With a unary callback: map(x => f(x)), each value of the array is totally indifferent to the presence of other values.
Using the other two parameters on the other hand introduces coupling, which may not be true to the original structure.
Imagine removing or reordering the second item in the arrays bellow: doing it before or after the map would not yield the same result.
Coupling with array size:
[6, 3, 12].map((x, _, a) => x/a.length);
// [2, 1, 4]
Coupling with ordering:
['foo', 'bar', 'baz'].map((x, i) => [i, x]);
// [[0, 'foo'], [1, 'bar'], [2, 'baz']]
Coupling with one specific value:
[1, 5, 3].map((x, _, a) => x/Math.max(...a));
//[ 0.2, 1, 0.6]
Coupling with neighbours:
const smooth = (x, i, a) => {
const prev = a[i - 1] ?? x;
const next = a[i + 1] ?? x;
const average = (prev + x + next) / 3;
return Math.round((x + average) / 2);
};
[1, 10, 50, 35, 40, 1].map(smoothh);
// [ 3, 15, 41, 38, 33, 8 ]
I recommend making it explicit on the call site whether or not these parameters are used.
const transfrom = (x, i) => x * i;
❌ array.map(transfrom);
⭕ array.map((x, i) => transfrom(x, i));
This has other benefits when you use variadic functions with map.
❌ ["1", "2", "3"].map(parseInt);
// [1, NaN, NaN]
⭕ ["1", "2", "3"].map(x => parseInt(x));
// [1, 2, 3]
reduce
reduce sets a value free from its surrounding structure.
Again, let's implement it on a simpler object:
class A {
constructor (value) {
this.value = value
}
reduce (f, init) {
return init !== undefined
? f(init, this.value)
: this.value
}
}
new A(5).reduce(); // 5
const concat = (a, b) => a.concat(b);
new A(5).reduce(concat, []); // [ 5 ]
Whether you leave the value alone or you put it back into something else, the output of reduce can be of any shape. It is literally the opposite of map.
Implications for arrays
Arrays can contain multiple or zero values, which gives rise to two, sometimes conflicting, requirements.
The need to combine
How can we return multiple values with no structure around them?
It is impossible. In order to return only one value, we have two options:
summarising the values into one value;
moving the values into a different structure.
Doesn't it make more sense now?
The need to initialise
What if there is no value to return?
If reduce returned a falsy value, there would be no way to know if the source array was empty or if it contained that falsy value, so unless we provide an initial value, reduce has to throw.
The true purpose of the reducer
You should be able to guess what the reducer f does in the following snippet:
[a].reduce(f);
[].reduce(f, a);
Nothing. It is not called.
It is the trivial case: a is the single value we want to return, so f is not needed.
This is by the way the reason why we didn't make the reducer mandatory in our class A earlier: because it contained only one value. It is mandatory on arrays because arrays can contain multiple values.
Since the reducer is only called when you have 2 or more values, saying its sole purpose is to combine them is only a stone throw away.
On transforming values
On arrays of variable lengths, expecting the reducer to transform the values is dangerous because, as we discovered, it may not be called.
I encourage you to map before you reduce when you need to both transform values and change shape.
It is a good idea to keep these two concerns separate for readability anyway.
When not to use reduce
Because reduce is this general-purpose tools for achieving structure transformation, I advise you to avoid it when you want an array back if there exists another more focussed method which does what you want.
Specifically, if you struggle with nested arrays in a map, think of flatMap or flat before reaching for reduce.
At the heart of reduce
a recursive binary operation
Implementing reduce on arrays introduces this feedback loop where the reducer's first argument is the return value of the previous iteration.
Needless to say it looks nothing like map's callback.
We could implement Array.reduce recursively like so:
const reduce = (f, acc, [current, ...rest]) =>
rest.length == 0
? f(acc, current)
: reduce(f, f(acc, current), rest)
This highlights the binary nature of the reducer f and how its return value becomes the new acc in the next iteration.
I let you convince yourself that the following is true:
reduce(f, a, [b, c, d])
// is equivalent to
f(f(f(a, b), c), d)
// or if you squint a little
((a ❋ b) ❋ c) ❋ d
This should seem familiar: you know arithmetic operations obey rules such as "associativity" or "commutativity". What I want to convey here is that the same kind of rules apply.
reduce may strip out the surrounding structure, values are still bound together in an algebraic structure for the time of the transformation.
the algebra of reducers
Algebraic structures are way out of the scope of this answer, so I will only touch on how they are relevant.
((a ❋ b) ❋ c) ❋ d
Looking at the expression above, it is self-evident that there is a constraint that ties all the values together : ❋ must know how to combine them the same way + must know how to combine 1 + 2 and just as importantly (1 + 2) + 3.
Weakest safe structure
One way to ensure this is to enforce that these values belong to a same set on which the reducer is an "internal" or "closed" binary operation, that is to say: combining any two values from this set with the reducer produces a value which belongs to the same set.
In abstract algebra this is called a magma. You can also look up semi-groups which are more talked about and are the same thing with associativity (no braces required), although reduce doesn't care.
Less safe
Living in a magma is not absolutely necessary : we can imagine a situation where ❋ can combine a and b but not c and b.
An example of this is function composition. One of the following functions returns a string, which constrains the order in which you can combine them:
const a = x => x * 2;
const b = x => x ** 2;
const c = x => x + ' !';
// (a ∘ b) ∘ c
const abc = x => c(b(a(x)));
abc(5); // "100 !"
// (a ∘ c) ∘ b
const acb = x => b(c(a(x)));
acb(5); // NaN
Like many binary operations, function composition can be used as a reducer.
Knowing if we are in a situation where reordering or removing elements from an array could make reduce break is kind of valuable.
So, magmas: not absolutely necessary, but very important.
what about the initial value
Say we want to prevent an exception from being thrown when the array is empty, by introducing an initial value:
array.reduce(f, init)
// which is really the same as doing
[init, ...array].reduce(f)
// or
((init ❋ a) ❋ b) ❋ c...
We now have an additional value. No problem.
"No problem"!? We said the purpose of the reducer was to combine the array values, but init is not a true value: it was forcefully introduced by ourselves, it should not affect the result of reduce.
The question is:
What init should we pick so that f(init, a) or init ❋ a returns a?
We want an initial value which acts as though it was not there. We want a neutral element (or "identity").
You can look up unital magmas or monoids (the same with associativity) which are swear words for magmas equipped with a neutral element.
Some neutral elements
You already know a bunch of neutral elements
numbers.reduce((a, b) => a + b, 0)
numbers.reduce((a, b) => a * b, 1)
booleans.reduce((a, b) => a && b, true)
strings.reduce((a, b) => a.concat(b), "")
arrays.reduce((a, b) => a.concat(b), [])
vec2s.reduce(([u,v], [x,y]) => [u+x,v+y], [0,0])
mat2s.reduce(dot, [[1,0],[0,1]])
You can repeat this pattern for many kinds of abstractions. Note that the neutral element and the computation don't need to be this trivial (extreme example).
Neutral element hardships
We have to accept the fact that some reductions are only possible for non-empty arrays and that adding poor initialisers don't fix the problem.
Some examples of reductions gone wrong:
Only partially neutral
numbers.reduce((a, b) => b - a, 0)
// does not work
numbers.reduce((a, b) => a - b, 0)
Subtracting 0 form b returns b, but subtracting b from 0 returns -b.
We say that only "right-identity" is true.
Not every non-commutative operation lack a symmetrical neutral element but it's a good sign.
Out of range
const min = (a, b) => a < b ? a : b;
// Do you really want to return Infinity?
numbers.reduce(min, Infinity)
Infinity is the only initial value which does not change the output of reduce for non-empty arrays, but it is unlikely that we would want it to actually appear in our program.
The neutral element is not some Joker value we add as a convenience. It has to be an allowed value, otherwise it doesn't accomplish anything.
Nonsensical
The reductions bellow rely on position, but adding an initialiser naturally shifts the first element to the second place, which requires messing with the index in the reducer to maintain the behaviour.
const first = (a, b, i) => !i ? b : a;
things.reduce(first, null);
const camelCase = (a, b, i) => a + (
!i ? b : b[0].toUpperCase() + b.slice(1)
);
words.reduce(camelCase, '');
It would have been a lot cleaner to embrace the fact the array can't be empty and simplify the definition of the reducers.
Moreover, the initials values are degenerate:
null is not the first element of an empty array.
an empty string is by no means a valid identifier.
There is no way to preserve the notion of "firstness" with an initial value.
conclusion
Algebraic structures can help us think of our programs in a more systematic way. Knowing which one we are dealing with can predict exactly what we can expect from reduce, so I can only advise you to look them up.
One step further
We have seen how map and reduce were so different structure-wise, but it is not as though they were two isolated things.
We can express map in terms of reduce, because it is always possible to rebuild the same structure we started with.
const map = f => (acc, x) =>
acc.concat(f(x))
;
const double = x => x * 2;
[1, 2, 3].reduce(map(double), []) // [2, 4, 6]
Pushing it a little further has led to neat tricks such as transducers.
I will not go into much detail about them, but I want you to notice a couple of things which will echo what we have said before.
Transducers
First let's see what problem we are trying to solve
[1, 2, 3, 4].filter(x => x % 2 == 0)
.map(x => x ** 2)
.reduce((a, b) => a + b)
// 20
We are iterating 3 times and creating 2 intermediary data structures. This code is declarative, but not efficient. Transducers attempt to reconcile the two.
First a little util for composing functions using reduce, because we are not going to use method chaining:
const composition = (f, g) => x => f(g(x));
const identity = x => x;
const compose = (...functions) =>
functions.reduce(composition, identity)
;
// compose(a, b, c) is the same as x => a(b(c(x)))
Now pay attention to the implementation of map and filter bellow. We are passing in this reducer function instead of concatenating directly.
const map = f => reducer => (acc, x) =>
reducer(acc, f(x))
;
const filter = f => reducer => (acc, x) =>
f(x) ? reducer(acc, x) : acc
;
look at this more specifically:
reducer => (acc, x) => [...]
after the callback function f is applied, we are left with a function which takes a reducer as input and returns a reducer.
These symmetrical functions is what we pass to compose:
const pipeline = compose(
filter(x => x % 2 == 0),
map(x => x ** 2)
);
Remember compose is implemented with reduce: our composition function defined earlier combines our symmetrical functions.
The output of this operation is a function of the same shape: something which expects a reducer and returns a reducer, which means
we have a magma. We can keep composing transformations as long as they have this shape.
we can consume this chain by applying the resulting function with a reducer, which will return a reducer that we can use with reduce
I let you expand the whole thing if you need convincing. If you do so you will notice that transformations will conveniently be applied left to right, which is the opposite direction of compose.
Alright, lets use this weirdo:
const add = (a, b) => a + b;
const reducer = pipeline(add);
const identity = 0;
[1, 2, 3, 4].reduce(reducer, identity); // 20
We have composed operations as diverse as map, filter and reduce into a single reduce, iterating only once with no intermediary data-structure.
This is no small achievement! And it is not a scheme you can come up with by deciding between map and reduce merely on the basis of the conciseness of the syntax.
Also notice that we have full control over the initial value and the final reducer. We used 0 and add, but we could have used [] and concat (more realistically push performance-wise) or any other data-structure for which we can implement a concat-like operation.
To understand the difference between map, filter and reduce, remember this:
All three methods are applied on array so anytime you want to make any operation on an array, you will be using these methods.
All three follow functional approaches and therefore the original array remains the same. Original array doesn't change instead a new array/value is returned.
Map returns a new array with the equal no. of elements as there are in the original array. Therefore, if the original array has 5 elements, the returned array will also have 5 elements. This method is used whenever we want to make some change on every individual element of an array. You can remember that every element of ann array is being mapped to some new value in output array, therefore the name map
For eg,
var originalArr = [1,2,3,4]
//[1,2,3,4]
var squaredArr = originalArr.map(function(elem){
return Math.pow(elem,2);
});
//[1,4,9,16]
Filter returns a new array with equal/less number of elements than the original array. It returns those elements in the array which have passed some condition. This method is used when we want to apply a filter on the original array therefore the name filter. For eg,
var originalArr = [1,2,3,4]
//[1,2,3,4]
var evenArr = originalArr.filter(function(elem){
return elem%2==0;
})
//[2,4]
Reduce returns a single value, unlike a map/filter. Therefore, whenever we want to run an operation on all elements of an array but want a single output using all elements, we use reduce. You can remember an array's output is reduced to a single value therefore the name reduce. For eg,
var originalArr = [1,2,3,4]
//[1,2,3,4]
var sum = originalArr.reduce(function(total,elem){
return total+elem;
},0)
//10
The map function executes a given function on each element but reduce executes a function which reduces the array to a single value. I'll give an example of both:
// map function
var arr = [1, 2, 3, 4];
var mappedArr = arr.map((element) => { // [10, 20, 30, 40]
return element * 10;
})
// reduce function
var arr2 = [1, 2, 3, 4]
var sumOfArr2 = arr2.reduce((total, element) => { // 10
return total + element;
})
It is true that reduce reduces an array to a single value, but since we can pass an object as initialValue, we can build upon it and end up with a more complex object than what we started with, such as this example where we group items by some criteria. Therefore the term 'reduce' can be slightly misleading as to the capabilities of reduce and thinking of it as necessarily reducing information can be wrong since it could also add information.
let a = [1, 2, 3, 4, 5, 6, 7, 8, 9]
let b = a.reduce((prev, curr) => {
if (!prev["divisibleBy2"]) {
prev["divisibleBy2"] = []
}
if (curr % 2 === 0) {
prev["divisibleBy2"].push(curr)
}
if (!prev["divisibleBy3"]) {
prev["divisibleBy3"] = []
}
if (curr % 3 === 0) {
prev["divisibleBy3"].push(curr)
}
if (!prev["divisibleBy5"]) {
prev["divisibleBy5"] = []
}
if (curr % 5 === 0) {
prev["divisibleBy5"].push(curr)
}
return prev
}, {})
console.log(b)
After years of writing loops in C++ the tedious way
for(int i=0; i<N; ++i) {
...
}
it becomes quite nice to use iterators
for(it i=v.begin(); i<v.end(); ++i) {
...
}
and ultimately moving to range iterators
for(auto i:v) {
...
}
In JavaScript also the for can be used, in a style nearly identical
(minus the type declaration and the pre/post increment operator) to
the first one above.
Still, in all of these the for is there. The D3.js
library demonstrates an alternative. One can iterate over an array by writing
d3.select("body")
.selectAll("p")
.data([4, 8, 15, 16, 23, 42])
.enter().append("p")
.text(function(d) { return "I’m number " + d + "!"; });
Here the enter mutates to a for loop. The documentation
explains nicely the client-side view of joins. What I am missing is a
standalone example of the (functional programming?) style of
converting a function call to an iteration.
No doubt this is not unique to D3.js. This is just where I encountered the idiom.
Can you suggest a few lines of standalone JavaScript code that
demonstrate iteration through a function call?
There are at least a couple of built-in functions that come to my mind.
map()
This one is very obvious.
[1, 2, 3]
.map(someNumber => someNumber * someNumber)
.map((powered, index) => index + "::" + powered);
// --> [ "1::1", "2::4", "3::9" ]
Chains well, right? Takes some input and produces the result consisting of elements calculated by applying a function element-wise.
Recommendation: try to use with pure functions whenever possible (produce the same results for same inputs, don't mutate the original collection if possible, nor produce any side effects).
forEach()
This function iterates through all elements of an array too, and applies a function, without returning anything back. Therefore, it can only end a chain of calls, but cannot be used for further chaining.
[1, 2, 3, 4]
.forEach(number => console.info(number));
Recommendation: forEach() is useful when we want to write some code that will result in a side effect per entry in the collection being iterated.
filter()
Filter function uses a predicate is used to sift the wheat from the chaff. The predicate is defining a criteria for the items you want to deal with on the next "stage".
[null, undefined, 0, 1, 2, 3, NaN, "", "You get the idea"]
.filter(Boolean)
.map(filteredElement => filteredElement + "!")
// --> [ "1!", "2!", "3!", "You get the idea!" ]
Recommendation: try to use with pure functions whenever possible. I.e. don't do anything else in filter other than things immediately related to filtration logic itself.
Object.keys() and Object.entries()
These two functions are helpful when we need to iterate over object's keys or key-value pairs, rather than an array's elements.
const targetObject = { a: 1, b: 2, c: 3 };
Object
.keys(targetObject)
.map(key => key + "=" + targetObject[key])
// --> [ "a=1", "b=2", "c=3" ]
same result can be achieved like this
Object
.entries({ a: 1, b: 2, c: 3 })
.map((key, value) => key + "=" + value)
// --> [ "a=1", "b=2", "c=3" ]
Recommendation: you may want to use Object.hasOwnProperty(...) when using working with Object.keys(...). See the documentation for details.
find()
The one is almost trivial. Let's us search for an item that matches a predicate. The search is "left-to-right", and it stops whenever the first "match" is found.
[1, 5, 10, 15]
.find(number >= 7)
// --> 10
findIndex() function can be used when we're looking for a position of an element that matches a predicate.
some() and every()
These functions check whether
a) there is at least one element matching a predicate; or
b) each and every element is matching a predicate.
const arrayOfNumbers = [2, 4, 6, 8, 10];
arrayOfNumbers.every(number => number % 2 === 0); // --> true
arrayOfNumbers.every(number => number % 2 === 1); // --> false
arrayOfNumbers.some(number => number > 1); // --> true
arrayOfNumbers.some(number => number <= 1); // --> false
reduce() and `reduceRight()`
The last one to mention in this quick review is the function that takes a list of things and aggregates it into a single result.
[-1, 0, 1, 2, 3]
.filter(value => value >= 0) // [0, 1, 2, 3]
.map(value => value + 1) // [1, 2, 3, 4]
.reduce((subTotal, currentValue) => subTotal + currentValue, 5);
// --> 15
Recommendation: try to use with pure functions whenever possible.
Universally applicable note on performance. In my benchmarks (don't have them on hand), a hand-written for loop was always faster than forEach, map, and other iterating functions. I do still prefer the functions unless the performance is being severely affected. There two main reasons for that: 1) easier to avoid off-by-one-errors; 2) the code is more readable, since each single function defines an independent step in the data processing flow, thus making code simpler and more maintainable.
I hope, this is an okay overview of some built-in chain-able JavaScript functions. More are described here. Take a look at concat(), sort(), fill(), join(), slice(), reverse() -- I frequently use them too.
If you need something like first() or last(), you will not find them in native functions. Either write your own ones, or use third-party libraries (e.g. lodash, rambda.js).
Here is an example implementation of Array.prototype.forEach:
function foreach(array, cb) {
for (var i = 0; i < array.length; ++i)
cb(array[i], i, array);
}
foreach([2,8,739,9,0], (n, i) =>
console.log("number: %s\nindex: %s\n", n, i));
surely I don't have to spoonfeed you do I?
function array_iterator(array) {
var i = 0;
function next() {
return array[i++];
}
function head() {
return array[i];
}
function tail() {
return array[array.length-1];
}
function more() {
return i < array.length;
}
function done() {
return !more();
}
function reset() {
i = 0;
}
return { next, head, tail, done, more, reset };
}
var nums = [3,34,4];
var iter = array_iterator(nums);
while (iter.more()) {
console.log(iter.next());
}
when using Array.sort()
I would assume it takes the first index but I wanted to verify.
MDN reference did not say anything:
This SO Post is not relevant.
For this method: ( sorting off the first index will work fine ).
$P.aZindex = function () {
var arr_2d = {},
elements = document.body.getElementsByTagName("*"),
element,
length,
z_index;
// loop through elements and pull information from them
$A.eachIndex(elements, function(val, index){
z_index = win.getComputedStyle(val).getPropertyValue("z-index");
// ignore elements with the auto value
if (z_index !== "auto") {
arr_2d[i] = [val.id, val.tagName, val.className, z_index];
}
});
// sort the array
arr_2d.sort();
return arr_2d;
};
Let's examine the EMCAScript specification for the array sort method. The sort algorithm iteratively compares pairs of elements from the array; those comparisons yield either -1, 1, or 0 (for less than, greater than, or equal to, respectively), and the result of each comparison is used to build a sorted array.
In particular, we are concerned with the "default" comparison case of sort, in which no comparison function is specified. When comparing some pair of x and y:
Let xString be ToString(x).
Let yString be ToString(y).
If xString < yString, return −1.
If xString > yString, return 1.
Return +0.
ECMAScript's ToString, when applied to objects, calls the object's toString method.
To clarify what is meant by x > y and x < y, we can examine ECMAScript's Abstract Relational Comparison Algorithm, which specifies the behavior of the > and < operators. In the event that the operands px and py are both strings:
Let k be the smallest nonnegative integer such that the character at position k within px is different from the character at position k within py...
Let m be the integer that is the code unit value for the character at position k within px.
Let n be the integer that is the code unit value for the character at position k within py.
If m < n, return true. Otherwise, return false.
This is a simple string comparison, based on a comparison of Unicode code unit values of the character at the first differing position within the strings.
As you can see, the sort algorithm stringifies each object element (using the element's toString method) that it contains and then compares those strings to determine ordering. I understand that this seems strange to you: if you have nothing but array elements in your sorting array, why not use the elements of those subarrays to determine ordering? This is simply because the EMCAScript specification chose to keep default element comparison of a potentially heterogeneous array extremely general, since any type of element can be rendered as a string.
However, if array-descent behavior is what you want, it's possible to implement:
var compare_items = function(a,b) {
var result;
if(a instanceof Array && b instanceof Array) {
// iteratively compare items from each array
for(var i=0; i<Math.min(a.length,b.length); ++i) {
result = compare_items(a[i], b[i]);
if(result != 0) { return result; }
}
// if both arrays are equal so far, length is the determining factor
return a.length - b.length;
}
// if the items are both numbers, report their numeric relation
if(typeof a == "number" && typeof b == "number") {
return a - b;
}
// otherwise, fall back to strings
if(a.toString() == b.toString()) { return 0; }
return a.toString() > b.toString() ? 1 : -1;
}
Then, use this comparator function like arr_2d.sort(compare_items);.
This allows you to sort arbitrarily deep N-dimensional arrays. First we compare a[0]...[0][0] against b[0]...[0][0] then a[0]...[0][1] to b[0]...[0][1]; then, if the [0]...[0][*] subarray proves equal, we move up to a[0]...[1][0], etc. Comparing arrays of differing dimension may produce unusual results, because a non-array may be compared to an array, which compares the stringified forms of each operand.
Note that this function has strange results if you have a heterogeneous array: [1, 2, 3, [1,2], [2,3]] sorts to [1, [1,2], 2, [2,3], 3]. The arrays and non-array are relatively sorted correctly, but the arrays are scattered in with the non-arrays in a non-intuitive way.
From the documentation you link to:
the array is sorted lexicographically (in dictionary order) according to the string conversion of each element
This doesn't change just because those elements are also arrays.