This question already has answers here:
Why does javascript turn array indexes into strings when iterating?
(6 answers)
Is a JavaScript array index a string or an integer?
(5 answers)
Why is key a string in for ... in
(3 answers)
When iterating over values, why does typeof(value) return "string" when value is a number? JavaScript
(1 answer)
Closed 1 year ago.
I've simplified my program down to this, and it's still misbehaving:
var grid = [0, 1, 2, 3];
function moveUp(moveDir) {
for (var row in grid) {
console.log('row:');
console.log(row + 5);
}
}
It seems that row is a string instead of an integer, for example the output is
row:
05
row:
15
row:
25
row:
35
rather than 5, 6, 7, 8, which is what I want. Shouldn't the counter in the for loop be a string?
Quoting from MDN Docs of for..in,
for..in should not be used to iterate over an Array where index order
is important. Array indexes are just enumerable properties with
integer names and are otherwise identical to general Object
properties. There is no guarantee that for...in will return the
indexes in any particular order and it will return all enumerable
properties, including those with non–integer names and those that are
inherited.
Because the order of iteration is implementation dependent, iterating
over an array may not visit elements in a consistent order. Therefore
it is better to use a for loop with a numeric index (or Array.forEach
or the non-standard for...of loop) when iterating over arrays where
the order of access is important.
You are iterating an array with for..in. That is bad. When you iterate with for..in, what you get is the array indices in string format.
So on every iteration, '0' + 5 == '05', '1' + 5 == '15'... is getting printed
What you should be doing is,
for (var len = grid.length, i = 0; i < len; i += 1) {
console.log('row:');
console.log(grid[i] + 5);
}
For more information about why exactly array indices are returned in the iteration and other interesting stuff, please check this answer of mine
You should use a normal for loop rather than a for...in loop for arrays.
for (var row = 0, l = grid.length; row < l; row++) {
console.log('row:');
console.log(5 + row);
}
I think this is what your expected output should be.
Fiddle
Try with parseInt(..) method to force int value
console.log(parseInt(row,10) + 5);
second param 10 is to be parsed as decimal value.
See the answer here How do I add an integer value with javascript (jquery) to a value that's returning a string?
Related
What is the difference between the two?
So I know that array.size() is a function while array.length is a property. Is there a usecase for using one over the other? Is one more efficient? (I would imagine .length to be significantly faster as it is a property rather then a method call?) Why would one ever use the slower option? Are there some browsers that are incompatible with one or the other?
var x = [];
console.log(x.size());
console.log(x.length);
console.log(x.size()==x.length);
x =[1,2,3];
console.log(x.size());
console.log(x.length);
console.log(x.size()==x.length);
Will print:
0, 0, true
3, 3, true
Array.size() is not a valid method
Always use the length property
There is a library or script adding the size method to the array prototype since this is not a native array method. This is commonly done to add support for a custom getter. An example of using this would be when you want to get the size in memory of an array (which is the only thing I can think of that would be useful for this name).
Underscore.js unfortunately defines a size method which actually returns the length of an object or array. Since unfortunately the length property of a function is defined as the number of named arguments the function declares they had to use an alternative and size was chosen (count would have been a better choice).
.size() is not a native JS function of Array (at least not in any browser that I know of).
.length should be used.
If
.size() does work on your page, make sure you do not have any extra libraries included like prototype that is mucking with the Array prototype.
or
There might be some plugin on your browser that is mucking with the Array prototype.
The .size() function is available in Jquery and many other libraries.
The .length property works only when the index is an integer.
The length property will work with this type of array:
var nums = new Array();
nums[0] = 1;
nums[1] = 2;
print(nums.length); // displays 2
The length property won't work with this type of array:
var pbook = new Array();
pbook["David"] = 1;
pbook["Jennifer"] = 2;
print(pbook.length); // displays 0
So in your case you should be using the .length property.
.size() is jQuery's, much probably you're either confusing with or took from someone else who had imported the jQuery library to his project.
If you'd have jQuery imported and you'd write like $(array).size(), it would return the array length.
array.length isn't necessarily the number of items in the array:
var a = ['car1', 'car2', 'car3'];
a[100] = 'car100';
a.length; // 101
The length of the array is one more than the highest index.
As stated before Array.size() is not a valid method.
More information
The property 'length' returns the (last_key + 1) for arrays with numeric keys:
var nums = new Array();
nums [ 10 ] = 10 ;
nums [ 11 ] = 11 ;
log.info( nums.length );
will print 12!
This will work:
var nums = new Array();
nums [ 10 ] = 10 ;
nums [ 11 ] = 11 ;
nums [ 12 ] = 12 ;
log.info( nums.length + ' / '+ Object.keys(nums).length );
The .size() method is deprecated as of jQuery 1.8. Use the .length property instead
See: https://api.jquery.com/size/
Size detects duplicates, it will return the number of unique values
const set1 = new Set([1, 2, 3, 4, 5, 5, 5, 6]);
console.log(set1.size);
// expected output: 6
Actually, .size() is not pure JavaScript method, there is a accessor property .size of Set object that is a little looks like .size() but it is not a function method, just as I said, it is an accessor property of a Set object to show the unique number of (unique) elements in a Set object.
The size accessor property returns the number of (unique) elements in a Set object.
const set1 = new Set();
const object1 = new Object();
set1.add(42);
set1.add('forty two');
set1.add('forty two');
set1.add(object1);
console.log(set1.size);
// expected output: 3
And length is a property of an iterable object(array) that returns the number of elements in that array. The value is an unsigned, 32-bit integer that is always numerically greater than the highest index in the array.
const clothing = ['shoes', 'shirts', 'socks', 'sweaters'];
console.log(clothing.length);
// expected output: 4
we can you use .length property to set or returns number of elements in an array. return value is a number
> set the length: let count = myArray.length;
> return lengthof an array : myArray.length
we can you .size in case we need to filter duplicate values and get the count of elements in a set.
const set = new set([1,1,2,1]);
console.log(set.size) ;`
Im just wondering who can explain the algorithm of this solution step by step. I dont know how hashmap works. Can you also give a basic examples using a hashmap for me to understand this algorithm. Thank you!
var twoSum = function(nums, target) {
let hash = {};
for(let i = 0; i < nums.length; i++) {
const n = nums[i];
if(hash[target - n] !== undefined) {
return [hash[target - n], i];
}
hash[n] = i;
}
return [];
}
Your code takes an array of numbers and a target number/sum. It then returns the indexes in the array for two numbers which add up to the target number/sum.
Consider an array of numbers such as [1, 2, 3] and a target of 5. Your task is to find the two numbers in this array which add to 5. One way you can approach this problem is by looping over each number in your array and asking yourself "Is there a number (which I have already seen in my array) which I can add to the current number to get my target sum?".
Well, if we loop over the example array of [1, 2, 3] we first start at index 0 with the number 1. Currently, there are no numbers which we have already seen that we can add with 1 to get our target of 5 as we haven't looped over any numbers yet.
So, so far, we have met the number 1, which was at index 0. This is stored in the hashmap (ie object) as {'1': 0}. Where the key is the number and the value (0) is the index it was seen at. The purpose of the object is to store the numbers we have seen and the indexes they appear at.
Next, the loop continues to index 1, with the current number being 2. We can now ask ourselves the question: Is there a number which I have already seen in my array that I can add to my current number of 2 to get the target sum of 5. The amount needed to add to the current number to get to the target can be obtained by doing target-currentNumber. In this case, we are currently on 2, so we need to add 3 to get to our target sum of 5. Using the hashmap/object, we can check if we have already seen the number 3. To do this, we can try and access the object 3 key by doing obj[target-currentNumber]. Currently, our object only has the key of '1', so when we try and access the 3 key you'll get undefined. This means we haven't seen the number 3 yet, so, as of now, there isn't anything we can add to 2 to get our target sum.
So now our object/hashmap looks like {'1': 0, '2': 1}, as we have seen the number 1 which was at index 0, and we have seen the number 2 which was at index 1.
Finally, we reach the last number in your array which is at index 2. Index 2 of the array holds the number 3. Now again, we ask ourselves the question: Is there a number we have already seen which we can add to 3 (our current number) to get the target sum?. The number we need to add to 3 to get our target number of 5 is 2 (obtained by doing target-currentNumber). We can now check our object to see if we have already seen a number 2 in the array. To do so we can use obj[target-currentNumber] to get the value stored at the key 2, which stores the index of 1. This means that the number 2 does exist in the array, and so we can add it to 3 to reach our target. Since the value was in the object, we can now return our findings. That being the index of where the seen number occurred, and the index of the current number.
In general, the object is used to keep track of all the previously seen numbers in your array and keep a value of the index at which the number was seen at.
Here is an example of running your code. It returns [1, 2], as the numbers at indexes 1 and 2 can be added together to give the target sum of 5:
const twoSum = function(nums, target) {
const hash = {}; // Stores seen numbers: {seenNumber: indexItOccurred}
for (let i = 0; i < nums.length; i++) { // loop through all numbers
const n = nums[i]; // grab the current number `n`.
if (hash[target - n] !== undefined) { // check if the number we need to add to `n` to reach our target has been seen:
return [hash[target - n], i]; // grab the index of the seen number, and the index of the current number
}
hash[n] = i; // update our hash to include the. number we just saw along with its index.
}
return []; // If no numbers add up to equal the `target`, we can return an empty array
}
console.log(twoSum([1, 2, 3], 5)); // [1, 2]
A solution like this might seem over-engineered. You might be wondering why you can't just look at one number in the array, and then look at all the other numbers and see if you come across a number that adds up to equal the target. A solution like that would work perfectly fine, however, it's not very efficient. If you had N numbers in your array, in the worst case (where no two numbers add up to equal your target) you would need to loop through all of these N numbers - that means you would do N iterations. However, for each iteration where you look at a singular number, you would then need to look at each other number using a inner loop. This would mean that for each iteration of your outer loop you would do N iterations of your inner loop. This would result in you doing N*N or N2 work (O(N2) work). Unlike this approach, the solution described in the first half of this answer only needs to do N iterations over the entire array. Using the object, we can find whether or not a number is in the object in constant (O(1)) time, which means that the total work for the above algorithm is only O(N).
For further information about how objects work, you can read about bracket notation and other property accessor methods here.
You may want to check out this method, it worked so well for me and I have written a lot of comments on it to help even a beginner understand better.
let nums = [2, 7, 11, 15];
let target = 9;
function twoSums(arr, t){
let num1;
//create the variable for the first number
let num2;
//create the variable for the second number
let index1;
//create the variable for the index of the first number
let index2;
//create the variable for the index of the second number
for(let i = 0; i < arr.length; i++){
//make a for loop to loop through the array elements
num1 = arr[i];
//assign the array iteration, i, value to the num1 variable
//eg: num1 = arr[0] which is 2
num2 = t - num1;
//get the difference between the target and the number in num1.
//eg: t(9) - num1(2) = 7;
if(arr.includes(num2)){
//check to see if the num2 number, 7, is contained in the array;
index1 = arr.indexOf(num2);
//if yes get the index of the num2 value, 7, from the array,
// eg: the index of 7 in the array is 1;
index2 = arr.indexOf(num1)
//get the index of the num1 value, which is 2, theindex of 2 in the array is 0;
}
}
return(`[${index1}, ${index2}]`);
//return the indexes in block parenthesis. You may choose to create an array and push the values into it, but consider space complexities.
}
console.log(twoSums(nums, target));
//call the function. Remeber we already declared the values at the top already.
//In my opinion, this method is best, it considers both time complexity and space complexityat its lowest value.
//Time complexity: 0(n)
function twoSum(numbers, target) {
for (let i = 0; i < numbers.length; i++) {
for (let j = i + 1; j < numbers.length; j++) {
if (numbers[i] + numbers[j] === target) {
return [numbers.indexOf(numbers[i]), numbers.lastIndexOf(numbers[j])];
}
}
}
}
This question already has answers here:
forEach on array of undefined created by Array constructor
(5 answers)
Closed 6 years ago.
Array(3) yields [ , , ], which has a length of 3.
[1, 2, 3].forEach loops 3 times, as expected.
Neither Array(3).forEach nor [ , , ].forEach loops at all, however.
Why is this? I thought I'd discovered a way of doing something n times without using for loops, and am disappointed to find it doesn't work!
forEach() executes the provided callback once for each element present
in the array in ascending order. It is not invoked for index
properties that have been deleted or are uninitialized (i.e. on sparse
arrays)
Example from MDN:
Fiddle
function logArrayElements(element, index, array) {
console.log('a[' + index + '] = ' + element);
}
// Notice that index 2 is skipped since there is no item at
// that position in the array.
[2, 5, , 9].forEach(logArrayElements);
// logs:
// a[0] = 2
// a[1] = 5
// a[3] = 9
Have a look at MDN article.
.
What is the difference between the two?
So I know that array.size() is a function while array.length is a property. Is there a usecase for using one over the other? Is one more efficient? (I would imagine .length to be significantly faster as it is a property rather then a method call?) Why would one ever use the slower option? Are there some browsers that are incompatible with one or the other?
var x = [];
console.log(x.size());
console.log(x.length);
console.log(x.size()==x.length);
x =[1,2,3];
console.log(x.size());
console.log(x.length);
console.log(x.size()==x.length);
Will print:
0, 0, true
3, 3, true
Array.size() is not a valid method
Always use the length property
There is a library or script adding the size method to the array prototype since this is not a native array method. This is commonly done to add support for a custom getter. An example of using this would be when you want to get the size in memory of an array (which is the only thing I can think of that would be useful for this name).
Underscore.js unfortunately defines a size method which actually returns the length of an object or array. Since unfortunately the length property of a function is defined as the number of named arguments the function declares they had to use an alternative and size was chosen (count would have been a better choice).
.size() is not a native JS function of Array (at least not in any browser that I know of).
.length should be used.
If
.size() does work on your page, make sure you do not have any extra libraries included like prototype that is mucking with the Array prototype.
or
There might be some plugin on your browser that is mucking with the Array prototype.
The .size() function is available in Jquery and many other libraries.
The .length property works only when the index is an integer.
The length property will work with this type of array:
var nums = new Array();
nums[0] = 1;
nums[1] = 2;
print(nums.length); // displays 2
The length property won't work with this type of array:
var pbook = new Array();
pbook["David"] = 1;
pbook["Jennifer"] = 2;
print(pbook.length); // displays 0
So in your case you should be using the .length property.
.size() is jQuery's, much probably you're either confusing with or took from someone else who had imported the jQuery library to his project.
If you'd have jQuery imported and you'd write like $(array).size(), it would return the array length.
array.length isn't necessarily the number of items in the array:
var a = ['car1', 'car2', 'car3'];
a[100] = 'car100';
a.length; // 101
The length of the array is one more than the highest index.
As stated before Array.size() is not a valid method.
More information
The property 'length' returns the (last_key + 1) for arrays with numeric keys:
var nums = new Array();
nums [ 10 ] = 10 ;
nums [ 11 ] = 11 ;
log.info( nums.length );
will print 12!
This will work:
var nums = new Array();
nums [ 10 ] = 10 ;
nums [ 11 ] = 11 ;
nums [ 12 ] = 12 ;
log.info( nums.length + ' / '+ Object.keys(nums).length );
The .size() method is deprecated as of jQuery 1.8. Use the .length property instead
See: https://api.jquery.com/size/
Size detects duplicates, it will return the number of unique values
const set1 = new Set([1, 2, 3, 4, 5, 5, 5, 6]);
console.log(set1.size);
// expected output: 6
Actually, .size() is not pure JavaScript method, there is a accessor property .size of Set object that is a little looks like .size() but it is not a function method, just as I said, it is an accessor property of a Set object to show the unique number of (unique) elements in a Set object.
The size accessor property returns the number of (unique) elements in a Set object.
const set1 = new Set();
const object1 = new Object();
set1.add(42);
set1.add('forty two');
set1.add('forty two');
set1.add(object1);
console.log(set1.size);
// expected output: 3
And length is a property of an iterable object(array) that returns the number of elements in that array. The value is an unsigned, 32-bit integer that is always numerically greater than the highest index in the array.
const clothing = ['shoes', 'shirts', 'socks', 'sweaters'];
console.log(clothing.length);
// expected output: 4
we can you use .length property to set or returns number of elements in an array. return value is a number
> set the length: let count = myArray.length;
> return lengthof an array : myArray.length
we can you .size in case we need to filter duplicate values and get the count of elements in a set.
const set = new set([1,1,2,1]);
console.log(set.size) ;`
Alright, I'm taking an array, and making another array from it with the only difference being the indexes are displaced by an arbitrary number determined using two reference points (one in each array). Doing this creates negative indexes, which if it didn't stop the script from working, would be useful. Is there any way to have the second array have the negative indexes and work, or am I going to have to use an all-together different method? I rewrote the code to be a simple case.
var firstArray = {
field: [ 1, 2, 3, 4, 5],
referenceIndex : 2
};
var secondArray = {
referenceIndex: 1,
offset: 0,
field : {}
};
// Create secondArray.field by finding the offset.
secondArray.offset = firstArray.referenceIndex - secondArray.referenceIndex;
for (i=0; i < firstArray.field.length; i++){
alert([i - secondArray.offset, firstArray.field[i]].join(" "));
secondArray.field[i - secondArray.offset] = firstArray.field[i]; //creates a negative index.
}
An array can have (in a strict sense) only positive integer as indices. However it is also an object, so it can take any string as a property. So in a sense, it will 'work', but do not trust Array#length to have the right value.
var arr = [1,2,3];
arr.length //3
arr[10] = 10;
arr.length //11
arr["blah"] = 100;
arr.length //still 11
arr[-1] = 200;
arr.length //still 11
I'd also like to point you to this excellent article - http://javascriptweblog.wordpress.com/2010/07/12/understanding-javascript-arrays/
No you can't have negative indices and have it work properly, however, you possibly could save a number and add it to your index to create a positive value. For example you have indeces -2 through 4. In the array this would be 0 - 6 so you would need to add or subtract 2 to get to the value of the index you want.