I found this super helpful gist.
In reduce function implementation, a long list of arguments has been passed: undefined, accumulator, this[i], this, i. I do not understand why. In my understanding that line should be, accumulator = callback.call(accumulator, this[i]). But if I remove any of these arguments, the reduce function works incorrectly. Please help me fill gaps in my understanding!
Array.prototype.myReduce = function(callback, initialVal) {
var accumulator = (initialVal === undefined) ? undefined : initialVal;
for (var i = 0; i < this.length; i++) {
if (accumulator !== undefined)
accumulator = callback.call(undefined, accumulator, this[i], i, this);
else
accumulator = this[i];
}
return accumulator;
};
//tests
var numbers3 = [20, 20, 2, 3];
var total = numbers3.myReduce(function(a, b) {
return a + b;
}, 10);
console.log(total); // 55
var flattened = [
[0, 1],
[2, 3],
[4, 5]
].reduce(function(a, b) {
return a.concat(b);
});
console.log(flattened); //[ 0, 1, 2, 3, 4, 5 ]
The first argument to call is the this value to use inside the callback. But this implementation is not using any this inside the callback, so the first argument must be undefined. If you do
accumulator = callback.call(accumulator, this[i]).
then you're calling the callback with a single argument, the this[i] (where this inside the callback will be accumulator, which definitely isn't what you want).
That said, the implementation in the gist is incorrect - it does not call the callback when the last callback returned undefined. It also does not properly check the initial value (which may have been passed as an argument, but is undefined):
// Unaltered function below:
Array.prototype.myReduce = function(callback, initialVal) {
var accumulator = (initialVal === undefined) ? undefined : initialVal;
for (var i = 0; i < this.length; i++) {
if (accumulator !== undefined)
accumulator = callback.call(undefined, accumulator, this[i], i, this);
else
accumulator = this[i];
}
return accumulator;
};
// My test:
[1, 2, 3].reduce((a, b, i) => {
console.log('native reduce iteration!');
return i === 0 ? b : undefined;
}, 0);
[1, 2, 3].myReduce((a, b, i) => {
console.log('CUSTOM reduce iteration!');
return i === 0 ? b : undefined;
}, 0);
To fix it, call the callback unconditionally (and remember to throw an error if called on an empty array with no initial value):
// Unaltered function below:
Array.prototype.myReduce = function(...args) {
const [callback, initialVal] = args;
let i;
let accumulator;
if (args.length >= 2) {
accumulator = initialVal;
i = 0;
} else if (this.length === 0) {
throw new TypeError('Reduce called on an empty array with no initial value');
} else {
accumulator = this[0];
i = 1;
}
for (; i < this.length; i++) {
accumulator = callback.call(undefined, accumulator, this[i], i, this);
}
return accumulator;
};
// My test:
[1, 2, 3].reduce((a, b, i) => {
console.log('native reduce iteration!');
return i === 0 ? b : undefined;
}, 0);
[1, 2, 3].myReduce((a, b, i) => {
console.log('CUSTOM reduce iteration!');
return i === 0 ? b : undefined;
}, 0);
console.log(
[2, 3, 4].myReduce((a, b) => a + b)
);
Working on a function that verifies if each number in an array is true or false and returns the first true number.
Have a working solution with a for loop as follows:
function findElement(arr, func) {
var num;
for (var a = 0; a < arr.length; a++) {
if (func(arr[a])) {
num = arr[a];
return num;
}
}
return num;
}
findElement([1, 3, 5, 8, 9, 10], function(num) {
return num % 2 === 0;
})
// Should return 8
But I'm trying (in order to get my head around forEach better) to convert it into a forEach loop.
This is where I am so far, but I don't see how to actually return the num out of the loop after it's been established that the function result is true:
function findElement(arr, func) {
if (arr.forEach(func) === true) {
return num;
}
}
findElement([1, 2, 3, 4], num => num % 2 === 0);
Not sure how to use forEach but you could use array.prototype.filter:
function findElement(arr, func) {
return arr.filter(func)[0];
}
#cdhowie suggested the use array.prototype.find in order to need the usage of [0].
function findElement(arr, func) {
return arr.find(func);
}
Obviously, that is just a guess because it may raise an error if no item in the array meets the requirements of func.
If you still are looking about use forEach maybe you could do something like this:
function findElement(arr, func) {
matches = []
arr.forEach((item) => {
if (func(item)) {
matches.push(item)
}
});
return matches[0];
}
Or:
function findElement(arr, func) {
match = null
arr.forEach((item) => {
match = (match == null && func(item)) ? item : match
});
return match;
}
Again, you will have to check how to handle the error if no item in the array meets the requirements f your func.
Any of both codes produce
console.log(findElement([1, 3, 5, 8, 9, 10], function(num) { return num % 2 === 0; })))
8
function findElement(arr, func) {
var num;
arr.forEach(function(element) {
if (!num && func(element)) {
num = element;
}
});
return num;
}
For more options you can check this question: how to stop Javascript forEach?
As has already been mentioned, this is not how forEach should be used.
However, you can still get the behavior you want:
function findElement(arr, func) {
var num;
arr.forEach(item =>{
if (func(item)) {
num = item;
// .forEach does not have an early return
// but you can force it to skip elements by removing them
while (true) {
// Remove all elements
var removedItem = arr.shift();
if (removedItem === undefined) {
// All elements removed
break;
}
}
}
return num;
}
This is even mentioned in the documentation
maybe like this
function findElement(arr, func) {
var num = null;
for (var a = 0; a < arr.length && num === null; a++) {
var val = arr[a];
num = func(val) ? val : null;
}
return num;
}
console.log(findElement([1, 3, 5, 8, 9, 10], function(num) {
return num % 2 === 0;
}));
here is your findElement method with foreach
function findElement(arr, func) {
var num = 0;
arr.forEach(function(item){
if (func(item))
return item;
})
return num;
}
The problem is to try and remove nested arrays:
steamrollArray([1, [2], [3, [[4]]]]); // should return [1, 2, 3, 4]
I have tried this but the recursion is failing when a nested array appears.
function checkElement(el) {
if (Array.isArray(el)) {
if (el.length === 1) {
return checkElement(el[0]);
} else {
for (var i=0; i < el.length; i++){
checkElement(el[i]);
}
}
} else {
return (el);
}
}
function steamrollArray(arr) {
var finalArr = [];
for (var i=0; i < arr.length; i++){
finalArr.push(checkElement(arr[i]));
}
return (finalArr);
}
A proposal for the first part:
You could change the return value to array and use concat instead of push.
function checkElement(el) {
// collect the values of the checked array
var temp = [];
if (Array.isArray(el)) {
if (el.length === 1) {
return checkElement(el[0]);
} else {
for (var i = 0; i < el.length; i++) {
// collect the values
temp = temp.concat(checkElement(el[i]));
}
// return flat values
return temp;
}
} else {
return el;
}
}
// this can be shorten to
function steamrollArray(arr) {
return checkElement(arr);
}
console.log(steamrollArray([1, [2], [3, [[4]]]]));
Part two, a bit shorter:
function steamrollArray(arr) {
return arr.reduce(function flat(r, a) {
return Array.isArray(a) && a.reduce(flat, r) || r.concat(a);
}, []);
}
console.log(steamrollArray([1, [2], [3, [[4]]]]));
You could use reduce:
function flatten( array ){
return array.reduce( function (a, b){
return a.concat( Array.isArray(b) ? flatten(b) : b );
}, [] );
}
I think this would be the funniest way to do this and also it's one line no more. Also it leaves extraction to native code which is much faster than Scripting.
var nestedArray = [1, [2], [3, [[4]]]];
var flatten = nestedArray.toString().split(',').map(Number);
console.log(flatten);
You can use recursion like this:
function flatten(array) {
var flat = []; //The result array
//An IIFE that will perform the recursion,
//is equivalent to: function rec(param) {.....}; rec(param);
(function rec(a) {
//For each element in the array:
//If the element is an array then call the 'rec' function.
//Else, push it to the result array.
//I used the conditional (ternary) operator (condition ? expr1 : expr2 )
for(var i in a) Array.isArray(a[i]) ? rec(a[i]) : flat.push(a[i]);
})(array);//Initiate the recursion with the main array
return flat;//Return the final result
};
var a = [1, [2], [3, [[4]]]];
function flatten(array) {
var flat = [];
(function rec(a) {
for(var i in a) Array.isArray(a[i]) ? rec(a[i]) : flat.push(a[i]);
})(array);
return flat;
};
console.log(flatten(a));
Using a generator function allows you to efficiently iterate through nested array elements without allocating unnecessary memory. If you really need the flattened array itself, use [...iterable] or Array.from(iterable):
function* deepIterate(array) {
for (a of array) Array.isArray(a) ? yield* deepIterate(a) : yield a;
}
// Iterate through flattened array:
for (a of deepIterate([1,[2,[3]]])) console.log(a);
// Flatten array:
var flat = Array.from(deepIterate([1,[2,[3]]]));
console.log(flat);
You can't just return the values, or it wouldn't work when you have arrays of length > 1.
Here's a solution:
function steamrollArray(arr, flattened) {
if (!flattened) flattened = [];
for (var i=0; i < arr.length; i++){
if (Array.isArray(arr[i])) {
steamrollArray(arr[i], flattened);
} else {
flattened.push(arr[i]);
}
}
return flattened;
}
console.log(steamrollArray([1, [2], [3, [[4]]]])); // should return [1, 2, 3, 4]
Try This if it work for you
function steamrollArray(unflatenArr) {
var flatenArr = [];
if (Array.isArray(unflatenArr)) {
for (var i = 0; i < unflatenArr.length; i++)
arrFlat(unflatenArr[i], flatenArr);
}
return flatenArr;
}
function arrFlat(arr, refArr) {
if (Array.isArray(arr)) {
for (var i = 0; i < arr.length; i++) {
arrFlat(arr[i], refArr);
}
}
else {
refArr.push(arr);
}
}
A simpler solution without using any recursion is by using splice method of Arrays. It works for any level of nesting.
function flattenArray(arr){
for(var i=0;i<arr.length;i++){
if(arr[i] instanceof Array){
Array.prototype.splice.apply(arr,[i,1].concat(arr[i]))
i--;
}
}
return arr;
}
Try this:
function steamrollArray(unflatenArr){
return eval("["+(JSON.stringify(unflatenArr).replace(/\[/g,'').replace(/\]/g,''))+"]")
}
steamrollArray([1, [2], [3, [[4]]]]);
How can I break the iteration of reduce() method?
for:
for (var i = Things.length - 1; i >= 0; i--) {
if(Things[i] <= 0){
break;
}
};
reduce()
Things.reduce(function(memo, current){
if(current <= 0){
//break ???
//return; <-- this will return undefined to memo, which is not what I want
}
}, 0)
You CAN break on any iteration of a .reduce() invocation by mutating the 4th argument of the reduce function: "array". No need for a custom reduce function. See Docs for full list of .reduce() parameters.
Array.prototype.reduce((acc, curr, i, array))
The 4th argument is the array being iterated over.
const array = ['apple', '-pen', '-pineapple', '-pen'];
const x = array
.reduce((acc, curr, i, arr) => {
if(i === 2) arr.splice(1); // eject early
return acc += curr;
}, '');
console.log('x: ', x); // x: apple-pen-pineapple
WHY?:
The one and only reason I can think of to use this instead of the many other solutions presented is if you want to maintain a functional programming methodology to your algorithm, and you want the most declarative approach possible to accomplish that. If your entire goal is to literally REDUCE an array to an alternate non-falsey primitive (string, number, boolean, Symbol) then I would argue this IS in fact, the best approach.
WHY NOT?
There's a whole list of arguments to make for NOT mutating function parameters as it's a bad practice.
UPDATE
Some of the commentators make a good point that the original array is being mutated in order to break early inside the .reduce() logic.
Therefore, I've modified the answer slightly by adding a .slice(0) before calling a follow-on .reduce() step, yielding a copy of the original array.
NOTE: Similar ops that accomplish the same task are slice() (less explicit), and spread operator [...array] (slightly less performant). Bear in mind, all of these add an additional constant factor of linear time to the overall runtime ... + O(n).
The copy, serves to preserve the original array from the eventual mutation that causes ejection from iteration.
const array = ['apple', '-pen', '-pineapple', '-pen'];
const x = array
.slice(0) // create copy of "array" for iterating
.reduce((acc, curr, i, arr) => {
if (i === 2) arr.splice(1); // eject early by mutating iterated copy
return (acc += curr);
}, '');
console.log("x: ", x, "\noriginal Arr: ", array);
// x: apple-pen-pineapple
// original Arr: ['apple', '-pen', '-pineapple', '-pen']
Don't use reduce. Just iterate on the array with normal iterators (for, etc) and break out when your condition is met.
You can use functions like some and every as long as you don't care about the return value. every breaks when the callback returns false, some when it returns true:
things.every(function(v, i, o) {
// do stuff
if (timeToBreak) {
return false;
} else {
return true;
}
}, thisArg);
Edit
A couple of comments that "this doesn't do what reduce does", which is true, but it can. Here's an example of using every in a similar manner to reduce that returns as soon as the break condition is reached.
// Soruce data
let data = [0,1,2,3,4,5,6,7,8];
// Multiple values up to 5 by 6,
// create a new array and stop processing once
// 5 is reached
let result = [];
data.every(a => a < 5? result.push(a*6) : false);
console.log(result);
This works because the return value from push is the length of the result array after the new element has been pushed, which will always be 1 or greater (hence true), otherwise it returns false and the loop stops.
There is no way, of course, to get the built-in version of reduce to exit prematurely.
But you can write your own version of reduce which uses a special token to identify when the loop should be broken.
var EXIT_REDUCE = {};
function reduce(a, f, result) {
for (let i = 0; i < a.length; i++) {
let val = f(result, a[i], i, a);
if (val === EXIT_REDUCE) break;
result = val;
}
return result;
}
Use it like this, to sum an array but exit when you hit 99:
reduce([1, 2, 99, 3], (a, b) => b === 99 ? EXIT_REDUCE : a + b, 0);
> 3
Array.every can provide a very natural mechanism for breaking out of high order iteration.
const product = function(array) {
let accumulator = 1;
array.every( factor => {
accumulator *= factor;
return !!factor;
});
return accumulator;
}
console.log(product([2,2,2,0,2,2]));
// 0
You can break every code - and thus every build in iterator - by throwing an exception:
function breakReduceException(value) {
this.value = value
}
try {
Things.reduce(function(memo, current) {
...
if (current <= 0) throw new breakReduceException(memo)
...
}, 0)
} catch (e) {
if (e instanceof breakReduceException) var memo = e.value
else throw e
}
You can use try...catch to exit the loop.
try {
Things.reduce(function(memo, current){
if(current <= 0){
throw 'exit loop'
//break ???
//return; <-- this will return undefined to memo, which is not what I want
}
}, 0)
} catch {
// handle logic
}
As the promises have resolve and reject callback arguments, I created the reduce workaround function with the break callback argument. It takes all the same arguments as native reduce method, except the first one is an array to work on (avoid monkey patching). The third [2] initialValue argument is optional. See the snippet below for the function reducer.
var list = ["w","o","r","l","d"," ","p","i","e","r","o","g","i"];
var result = reducer(list,(total,current,index,arr,stop)=>{
if(current === " ") stop(); //when called, the loop breaks
return total + current;
},'hello ');
console.log(result); //hello world
function reducer(arr, callback, initial) {
var hasInitial = arguments.length >= 3;
var total = hasInitial ? initial : arr[0];
var breakNow = false;
for (var i = hasInitial ? 0 : 1; i < arr.length; i++) {
var currentValue = arr[i];
var currentIndex = i;
var newTotal = callback(total, currentValue, currentIndex, arr, () => breakNow = true);
if (breakNow) break;
total = newTotal;
}
return total;
}
And here is the reducer as an Array method modified script:
Array.prototype.reducer = function(callback,initial){
var hasInitial = arguments.length >= 2;
var total = hasInitial ? initial : this[0];
var breakNow = false;
for (var i = hasInitial ? 0 : 1; i < this.length; i++) {
var currentValue = this[i];
var currentIndex = i;
var newTotal = callback(total, currentValue, currentIndex, this, () => breakNow = true);
if (breakNow) break;
total = newTotal;
}
return total;
};
var list = ["w","o","r","l","d"," ","p","i","e","r","o","g","i"];
var result = list.reducer((total,current,index,arr,stop)=>{
if(current === " ") stop(); //when called, the loop breaks
return total + current;
},'hello ');
console.log(result);
Reduce functional version with break can be implemented as 'transform', ex. in underscore.
I tried to implement it with a config flag to stop it so that the implementation reduce doesn't have to change the data structure that you are currently using.
const transform = (arr, reduce, init, config = {}) => {
const result = arr.reduce((acc, item, i, arr) => {
if (acc.found) return acc
acc.value = reduce(config, acc.value, item, i, arr)
if (config.stop) {
acc.found = true
}
return acc
}, { value: init, found: false })
return result.value
}
module.exports = transform
Usage1, simple one
const a = [0, 1, 1, 3, 1]
console.log(transform(a, (config, acc, v) => {
if (v === 3) { config.stop = true }
if (v === 1) return ++acc
return acc
}, 0))
Usage2, use config as internal variable
const pixes = Array(size).fill(0)
const pixProcessed = pixes.map((_, pixId) => {
return transform(pics, (config, _, pic) => {
if (pic[pixId] !== '2') config.stop = true
return pic[pixId]
}, '0')
})
Usage3, capture config as external variable
const thrusts2 = permute([9, 8, 7, 6, 5]).map(signals => {
const datas = new Array(5).fill(_data())
const ps = new Array(5).fill(0)
let thrust = 0, config
do {
config = {}
thrust = transform(signals, (_config, acc, signal, i) => {
const res = intcode(
datas[i], signal,
{ once: true, i: ps[i], prev: acc }
)
if (res) {
[ps[i], acc] = res
} else {
_config.stop = true
}
return acc
}, thrust, config)
} while (!config.stop)
return thrust
}, 0)
You cannot break from inside of a reduce method. Depending on what you are trying to accomplish you could alter the final result (which is one reason you may want to do this)
const result = [1, 1, 1].reduce((a, b) => a + b, 0); // returns 3
console.log(result);
const result = [1, 1, 1].reduce((a, b, c, d) => {
if (c === 1 && b < 3) {
return a + b + 1;
}
return a + b;
}, 0); // now returns 4
console.log(result);
Keep in mind: you cannot reassign the array parameter directly
const result = [1, 1, 1].reduce( (a, b, c, d) => {
if (c === 0) {
d = [1, 1, 2];
}
return a + b;
}, 0); // still returns 3
console.log(result);
However (as pointed out below), you CAN affect the outcome by changing the array's contents:
const result = [1, 1, 1].reduce( (a, b, c, d) => {
if (c === 0) {
d[2] = 100;
}
return a + b;
}, 0); // now returns 102
console.log(result);
Providing you do not need to return an array, perhaps you could use some()?
Use some instead which auto-breaks when you want. Send it a this accumulator. Your test and accumulate function cannot be an arrow function as their this is set when the arrow function is created.
const array = ['a', 'b', 'c', 'd', 'e'];
var accum = {accum: ''};
function testerAndAccumulator(curr, i, arr){
this.tot += arr[i];
return curr==='c';
};
accum.tot = "";
array.some(testerAndAccumulator, accum);
var result = accum.tot;
In my opinion this is the better solution to the accepted answer provided you do not need to return an array (eg in a chain of array operators), as you do not alter the original array and you do not need to make a copy of it which could be bad for large arrays.
So, to terminate even earlier the idiom to use would be arr.splice(0).
Which prompts the question, why can't one just use arr = [] in this case?
I tried it and the reduce ignored the assignment, continuing on unchanged.
The reduce idiom appears to respond to forms such as splice but not forms such as the assignment operator??? - completely unintuitive - and has to be rote-learnt as precepts within the functional programming credo ...
const array = ['9', '91', '95', '96', '99'];
const x = array
.reduce((acc, curr, i, arr) => {
if(i === 2) arr.splice(1); // eject early
return acc += curr;
}, '');
console.log('x: ', x); // x: 99195
The problem is, that inside of the accumulator it is not possible to just stop the whole process. So by design something in the outer scope must be manipulated, which always leads to a necessary mutation.
As many others already mentioned throw with try...catch is not really an approach which can be called "solution". It is more a hack with many unwanted side effects.
The only way to do this WITHOUT ANY MUTATIONS is by using a second compare function, which decides whether to continue or stop. To still avoid a for-loop, it has to be solved with a recursion.
The code:
function reduceCompare(arr, cb, cmp, init) {
return (function _(acc, i) {
return i < arr.length && cmp(acc, arr[i], i, arr) === true ? _(cb(acc, arr[i], i, arr), i + 1) : acc;
})(typeof init !== 'undefined' ? init : arr[0], 0);
}
This can be used like:
var arr = ['a', 'b', 'c', 'd'];
function join(acc, curr) {
return acc + curr;
}
console.log(
reduceCompare(
arr,
join,
function(acc) { return acc.length < 1; },
''
)
); // logs 'a'
console.log(
reduceCompare(
arr,
join,
function(acc, curr) { return curr !== 'c'; },
''
)
); // logs 'ab'
console.log(
reduceCompare(
arr,
join,
function(acc, curr, i) { return i < 3; },
''
)
); // logs 'abc'
I made an npm library out of this, also containing a TypeScript and ES6 version. Feel free to use it:
https://www.npmjs.com/package/array-reduce-compare
or on GitHub:
https://github.com/StefanJelner/array-reduce-compare
You could to write your own reduce method. Invoking it like this, so it follows same logic and you control your own escape / break solution. It retains functional style and allows breaking.
const reduce = (arr, fn, accum) => {
const len = arr.length;
let result = null;
for(let i = 0; i < len; i=i+1) {
result = fn(accum, arr[i], i)
if (accum.break === true) {
break;
}
}
return result
}
const arr = ['a', 'b', 'c', 'shouldnotgethere']
const myResult = reduce(arr, (accum, cur, ind) => {
accum.result = accum.result + cur;
if(ind === 2) {
accum.break = true
}
return accum
}, {result:'', break: false}).result
console.log({myResult})
Or create your own reduce recursion method:
const rcReduce = (arr, accum = '', ind = 0) => {
const cur = arr.shift();
accum += cur;
const isBreak = ind > 1
return arr.length && !isBreak ? rcReduce(arr, accum, ind + 1) : accum
}
const myResult = rcReduce(['a', 'b', 'c', 'shouldngethere'])
console.log({myResult})
Another simple implementation that I came with solving the same issue:
function reduce(array, reducer, first) {
let result = first || array.shift()
while (array.length > 0) {
result = reducer(result, array.shift())
if (result && result.reduced) {
return result.reduced
}
}
return result
}
If you want to chain promises sequentially with reduce using the pattern below:
return [1,2,3,4].reduce(function(promise,n,i,arr){
return promise.then(function(){
// this code is executed when the reduce loop is terminated,
// so truncating arr here or in the call below does not works
return somethingReturningAPromise(n);
});
}, Promise.resolve());
But need to break according to something happening inside or outside a promise
things become a little bit more complicated because the reduce loop is terminated before the first promise is executed, making truncating the array in the promise callbacks useless, I ended up with this implementation:
function reduce(array, promise, fn, i) {
i=i||0;
return promise
.then(function(){
return fn(promise,array[i]);
})
.then(function(result){
if (!promise.break && ++i<array.length) {
return reduce(array,promise,fn,i);
} else {
return result;
}
})
}
Then you can do something like this:
var promise=Promise.resolve();
reduce([1,2,3,4],promise,function(promise,val){
return iter(promise, val);
}).catch(console.error);
function iter(promise, val) {
return new Promise(function(resolve, reject){
setTimeout(function(){
if (promise.break) return reject('break');
console.log(val);
if (val==3) {promise.break=true;}
resolve(val);
}, 4000-1000*val);
});
}
I solved it like follows, for example in the some method where short circuiting can save a lot:
const someShort = (list, fn) => {
let t;
try {
return list.reduce((acc, el) => {
t = fn(el);
console.log('found ?', el, t)
if (t) {
throw ''
}
return t
}, false)
} catch (e) {
return t
}
}
const someEven = someShort([1, 2, 3, 1, 5], el => el % 2 === 0)
console.log(someEven)
UPDATE
Away more generic answer could be something like the following
const escReduce = (arr, fn, init, exitFn) => {
try {
return arr.reduce((...args) => {
if (exitFn && exitFn(...args)) {
throw args[0]
}
return fn(...args)
}, init)
} catch(e){ return e }
}
escReduce(
Array.from({length: 100}, (_, i) => i+1),
(acc, e, i) => acc * e,
1,
acc => acc > 1E9
); // 6227020800
give we pass an optional exitFn which decides to break or not