Videos
function countTheObjects(arr) {
let objects = [];
for (let i = 0; i < arr.length; i++) {
if (!Array.isArray(arr[i]) && typeof arr[i] === 'object' && arr[i] !== null )
objects.push(arr[i]);
}
console.log(objects)
return objects.length;
}
console.log((countTheObjects([1, [], 3, 4, 5, {}, {}, {}, 'foo'])))
You can use Array.filter() and Array.isArray() to do this.
Array.filter() will look at each item in the array and keep if it the return value of a callback function is true.
function countTheObjects(arr) {
return arr.filter((item) => item && typeof item === "object" && !Array.isArray(item)).length;
}
console.log(countTheObjects([1, [], null, '', undefined, 3, 4, 5, {}, {}, {}, 'foo']));That's not an array, is an object literal, you should iterate over the own properties of the object and count them, e.g.:
function objectLength(obj) {
var result = 0;
for(var prop in obj) {
if (obj.hasOwnProperty(prop)) {
// or Object.prototype.hasOwnProperty.call(obj, prop)
result++;
}
}
return result;
}
objectLength(member); // for your example, 3
The hasOwnProperty method should be used to avoid iterating over inherited properties, e.g.
var obj = {};
typeof obj.toString; // "function"
obj.hasOwnProperty('toString'); // false, since it's inherited
You can try this code, it works perfectly in a browser:
Object.keys(member).length;
Learning JS and I feel pretty stumped about this reduce method exercise. The video I saw counts the occurrence of every element in an array.
What does total[item] = (total[item] || 0) +1
From my understanding, the first argument in a reduce method is the previous value and the second argument is the current value. How can we have previousValue[currentValue]?
what does this mean: previousValue[currentValue] = (previousValue[currentValue] || 0) +1
Thank you!
--------------- Exercise -------------
const toys = ['car', 'dog', 'car', 'truck', 'boat']
let countToys = toys.reduce((total, item) => {total[item] = (total[item] || 0) +1
return total}, {})
console.log(countToys)
[this answer is a bit dated: read the edits, in the notion of 'equal' in javascript is ambiguous]
Say hello to your friends: map and filter and reduce and forEach and every etc.
(I only occasionally write for-loops in javascript, because of block-level scoping is missing, so you have to use a function as the body of the loop anyway if you need to capture or clone your iteration index or value. For-loops are more efficient generally, but sometimes you need a closure.)
The most readable way:
[....].filter(x => x==2).length
(We could have written .filter(function(x){return x==2}).length instead)
The following is more space-efficient (O(1) rather than O(N)), but I'm not sure how much of a benefit/penalty you might pay in terms of time (not more than a constant factor since you visit each element exactly once):
[....].reduce((total,x) => (x==2 ? total+1 : total), 0)
or as a commenter kindly pointed out:
[....].reduce((total,x) => total+(x==2), 0)
(If you need to optimize this particular piece of code, a for loop might be faster on some browsers... you can test things on jsperf.com.)
You can then be elegant and turn it into a prototype function:
[1, 2, 3, 5, 2, 8, 9, 2].count(2)
Like this:
Object.defineProperties(Array.prototype, {
count: {
value: function(value) {
return this.filter(x => x==value).length;
}
}
});
You can also stick the regular old for-loop technique (see other answers) inside the above property definition (again, that would likely be much faster).
2017 edit:
Whoops, this answer has gotten more popular than the correct answer. Actually, just use the accepted answer. While this answer may be cute, the js compilers probably don't (or can't due to spec) optimize such cases. So you should really write a simple for loop:
Object.defineProperties(Array.prototype, {
count: {
value: function(query) {
/*
Counts number of occurrences of query in array, an integer >= 0
Uses the javascript == notion of equality.
*/
var count = 0;
for(let i=0; i<this.length; i++)
if (this[i]==query)
count++;
return count;
}
}
});
You could define a version .countStrictEq(...) which used the === notion of equality. The notion of equality may be important to what you're doing! (for example [1,10,3,'10'].count(10)==2, because numbers like '4'==4 in javascript... hence calling it .countEq or .countNonstrict stresses it uses the == operator.)
Caveat: Defining a common name on the prototype should be done with care. It is fine if you control your code, but bad if everyone wants to declare their own
[].countfunction, especially if they behave differently. You may ask yourself "but.count(query)surely sounds quite perfect and canonical"... but consider perhaps you could do something like[].count(x=> someExpr of x). In that case you define functions likecountIn(query, container)(undermyModuleName.countIn), or something, or[].myModuleName_count().
Also consider using your own multiset data structure (e.g. like python's 'collections.Counter') to avoid having to do the counting in the first place. This works for exact matches of the form [].filter(x=> x==???).length (worst case O(N) down to O(1)), and modified will speed up queries of the form [].filter(filterFunction).length (roughly by a factor of #total/#duplicates).
class Multiset extends Map {
constructor(...args) {
super(...args);
}
add(elem) {
if (!this.has(elem))
this.set(elem, 1);
else
this.set(elem, this.get(elem)+1);
}
remove(elem) {
var count = this.has(elem) ? this.get(elem) : 0;
if (count>1) {
this.set(elem, count-1);
} else if (count==1) {
this.delete(elem);
} else if (count==0)
throw `tried to remove element ${elem} of type ${typeof elem} from Multiset, but does not exist in Multiset (count is 0 and cannot go negative)`;
// alternatively do nothing {}
}
}
Demo:
> counts = new Multiset([['a',1],['b',3]])
Map(2) {"a" => 1, "b" => 3}
> counts.add('c')
> counts
Map(3) {"a" => 1, "b" => 3, "c" => 1}
> counts.remove('a')
> counts
Map(2) {"b" => 3, "c" => 1}
> counts.remove('a')
Uncaught tried to remove element a of type string from Multiset, but does not exist in Multiset (count is 0 and cannot go negative)
sidenote: Though, if you still wanted the functional-programming way (or a throwaway one-liner without overriding Array.prototype), you could write it more tersely nowadays as [...].filter(x => x==2).length. If you care about performance, note that while this is asymptotically the same performance as the for-loop (O(N) time), it may require O(N) extra memory (instead of O(1) memory) because it will almost certainly generate an intermediate array and then count the elements of that intermediate array.
Modern JavaScript:
Note that you should always use triple equals === when doing comparison in JavaScript (JS). The triple equals make sure, that JS comparison behaves like double equals == in other languages (there is one exception, see below). The following solution shows how to solve this the functional way, which will ensure that you will never have out of bounds error:
// Let has local scope
let array = [1, 2, 3, 5, 2, 8, 9, 2]
// Functional filter with an Arrow function
// Filter all elements equal to 2 and return the length (count)
array.filter(x => x === 2).length // -> 3
The following anonymous Arrow function (lambda function) in JavaScript:
(x) => {
const k = 2
return k * x
}
may be simplified to this concise form for a single input:
x => 2 * x
where the return is implied.
Always use triple equals: === for comparison in JS, with the exception of when checking for nullability: if (something == null) {} as it includes a check for undefined, if you only use double equals as in this case.