This question already has answers here:
Object returning NaN when sum values
(3 answers)
Closed 8 years ago.
I have the following code
var Arr = [-1,3,-4,5,1,-6,2,1];
function solution ( A ) {
var sum;
var len = A.length;
for ( var key in A ) {
sum += +(parseInt(A[key]));
}
return sum;
}
solution( Arr );
and it returns NaN. Can someone help me?
Thanks!
You never initialize sum, so it starts undefined.
undefined + number = NaN
undefined + "any number" is always NaN.
Declare sum with a starting value of 0. (aka: initialize it)
var sum = 0;
Also, it's a good idea to use a radix in parseInt:
parseInt(A[key], 10)
This makes sure parseInt always tries to interpret A[key] as a decimal number.
Here is the working version of your code
function solution ( A ) {
var sum = 0;
var len = A.length;
for ( key in A ) {
sum += parseInt(A[key], 10);
}
return sum;
}
You should initialize sum also your +(parseInt(A[key])) has the same effect as parseInt(A[key])
Related
This question already has answers here:
How do I extract even elements of an Array?
(8 answers)
How to do a script for odd and even numbers from 1 to 1000 in Javascript?
(8 answers)
Closed 2 years ago.
I've spent an embarrassing amount of time on this question only to realize my function is only right 50% of the time. So the goal here is to return only the odd numbers of all the numbers in between the two arguments. (for instance if the arguments are 1 and 5 i'd need to return 2 & 3) the function I wrote is completely dependent on the first argument. if it's even my function will return odds, but if the first number is odd it'll return evens. does anyone know how i can fix this?
function oddNumbers(l, r) {
const arr = [];
const theEvens = [];
for (let i= l; i<r; i++) {
arr.push(i)
}
console.log(arr)
for (let i= 0; i < arr.length; i+= 2 ) {
const evens = arr[0] + i;
theEvens.push(evens);
}
theEvens.forEach(item => arr.splice(arr.indexOf(item), 1));
console.log(arr)
}
oddNumbers(2, 20);
I modified the code a bit to return only odd numbers
We use the % operator that behaves like the remainder operator in math:
so when we say i % 2 if the number is even the result of the operation will be 0
but when the "i" is an odd number the result will be 1
so now we can filter the even from the odd numbers using this operation
function oddNumbers(l, r) {
const arr = [];
for (let i= l; i<r; i++) {
if(i % 2 !== 0) arr.push(i);
}
console.log(arr);
}
oddNumbers(2, 20);
You can loop from initial to end parameters and get odd numbers using modulo, try this:
let result = [];
let returnOdd = (n1, n2) => {
for(i = n1; i < n2; i++){
if(i % 2 != 0){
result.push(i)
}
}
return result;
}
console.log(returnOdd(2, 20));
You could use the filter method.
This method creates a new array based on the condition it has. In this case it will to go through all the numbers in the array, and check if the number is odd (though the remainder operator).
For example:
1 % 2 = 1 ( true, keep in the new array )
2 % 2 = 0 ( false ignore in the new array )
function OddNumbers(start, end) {
// Create an array from the given range
const nums = Array(end - start + 1).fill().map((_, idx) => start + idx);
// Use filter to return the odd numbers via the % operator
return nums.filter(num => num % 2);
}
console.log(OddNumbers(2,20))
This question already has answers here:
How does the Math.max.apply() work?
(5 answers)
Closed 5 years ago.
How does
Math.max.apply(null, arr);
exactly work?
Suppose
var arr = [45,25,4,65] ;
will it compare 'null' and '45' and return maximum number between two adjacent e.g. 45?
After comparing will it again compare returned number and next array number e.g. 25 and return the max number?
And how is the calculation done internally? It is the same way, I think.
The first argument of apply is what will be this for the function.
Here it does not matter as this function does not necessit a specific this value. It could matter in some other cases for instance:
var foo = {
b: true,
func: function() {
return this.b;
}
};
var bar = { b : false };
foo.func(); // true
foo.func.apply(bar); // false
It's equal to
Math.max(...arr);
and that's equal to:
Math.max(45, 25, 4, 65);
How it works internally is then up to the browsers / parsers native implementation. In js it might be:
Math.max = (...args) => {
let max = - Infinity;
for(const number of args)
if(number > max) max = number;
return max;
};
I'm trying to write a function which outputs the correct result when multiplying a number by a negative power of ten using arrays and split() method. For example the following expressions get the right result: 1x10^-2 = 0.01 1x10^-4 = 0.0001.
Problem comes when the number's length is superior to the exponent value (note that my code treats num as a string to split it in an array as shown in code bellow :
//var num is treated as a string to be splited inside get_results() function
//exponent is a number
//Try different values for exponent and different lengths for num to reproduce the problem
//for example var num = 1234 and var exponent = 2 will output 1.234 instead of 12.34
var num = '1';
var sign = '-';
var exponent = 2;
var op = 'x10^'+sign+exponent;
var re = get_result(num);
console.log(num+op +' = '+ re);
function get_result(thisNum) {
if (sign == '-') {
var arr = [];
var splitNum = thisNum.split('');
for (var i = 0; i <= exponent-splitNum.length; i++) {
arr.push('0');
}
for (var j = 0; j < splitNum.length; j++) {
arr.push(splitNum[j]);
}
if (exponent > 0) {
arr.splice(1, 0, '.');
}
arr.join('');
}
return arr.join('');
}
Demo here : https://jsfiddle.net/Hal_9100/c7nobmnj/
I tried different approaches to get the right results with different num lengths and exponent values, but nothing I came with worked and I came to the point where I can't think of anything else.
You can see my latest try here : https://jsfiddle.net/Hal_9100/vq1hrru5/
Any idea how I could solve this problem ?
PS: I know most of the rounding errors due to javascript floating point conversion are pretty harmless and can be fixed using toFixed(n) or by using specialized third-party librairies, but my only goal here is to get better at writing pure javascript functions.
I am not sure if you want to keep going with the array approach to a solution, but it seems like this could be solved with using the Math.pow() method that already exists.
function computeExponentExpression ( test ) {
var base;
var multiplier;
var exponent;
test.replace(/^(\d+)(x)(\d+)([^])([-]?\d+)$/, function() {
base = parseInt(arguments[1], 10);
multiplier = parseInt(arguments[3], 10);
exponent = parseInt(arguments[5], 10);
return '';
} );
console.log( base * Math.pow(multiplier, exponent));
}
computeExponentExpression('1x10^-4');
computeExponentExpression('1x10^2');
computeExponentExpression('4x5^3');
The problem is where you push the decimal point .
instead of
arr.splice(1, 0, '.');
try this:
arr.splice(-exponent, 0, '.');
See fiddle: https://jsfiddle.net/free_soul/c7nobmnj/1/
I'm trying to solve a Coderbyte challenge, and I'm still trying to fully understand recursion.
Here's the problem: Using the JavaScript language, have the function AdditivePersistence(num) take the num parameter being passed which will always be a positive integer and return its additive persistence which is the number of times you must add the digits in num until you reach a single digit. For example: if num is 2718 then your program should return 2 because 2 + 7 + 1 + 8 = 18 and 1 + 8 = 9 and you stop at 9.
Here's the solution I put into jsfiddle.net to try out:
function AdditivePersistence(num) {
var count=0;
var sum=0;
var x = num.toString().split('');
for(var i=0; i<x.length; i++) {
sum += parseInt(x[i]);
}
if(sum.length == 1) {
return sum;
}
else {
return AdditivePersistence(sum);
}
}
alert(AdditivePersistence(19));
It tells me that there's too much recursion. Is there another "else" I could put that would basically just re-run the function until the sum was one digit?
One of the problems is that your if statement will never evaluate as 'true'. The reason being is that the sum variable is holding a number, and numbers don't have a length function. Also, as 'Barmar' pointed out, you haven't incremented the count variable, and neither are you returning the count variable.
Here's a solution that works using recursion.
function AdditivePersistence(num) {
var result = recursive(String(num).split('').reduce(function(x,y){return parseInt(x) + parseInt(y)}), 1);
function recursive(n, count){
c = count;
if(n < 10)return c;
else{
count += 1
return recursive(String(n).split('').reduce(function(x,y){return parseInt(x) + parseInt(y)}), count)
}
}
return num < 10 ? 0 : result
}
To fix the 'too much recursion problem',
if(sum.toString().length == 1)
However, as the others have said, your implementation does not return the Additive Persistence. Use James Farrell's answer to solve the Coderbyte challenge.
I have a function that I'm using to remove unwanted characters (defined as currency symbols) from strings then return the value as a number. When returning the value, I am making the following call:
return parseFloat(x);
The problem I have is that when x == "0.00" I expect to get 0.00 (a float with two decimals) back. What I get instead is simply 0.
I've also tried the following:
return parseFloat(x).toFixed(2);
and still get simply 0 back. Am I missing something? Any help would be greatly appreciated.
Thank you!!
parseFloat() turns a string into a floating point number. This is a binary value, not a decimal representation, so the concept of the number of zeros to the right of the decimal point doesn't even apply; it all depends on how it is formatted back into a string. Regarding toFixed, I'd suggest converting the floating point number to a Number:
new Number(parseFloat(x)).toFixed(2);
this should work:
return parseFloat(x).toFixed(2);
you can test it by running this in firebug:
var x = '0.00';
alert(parseFloat(x).toFixed(2));
simple:
function decimalPlaces(float, length) {
ret = "";
str = float.toString();
array = str.split(".");
if (array.length == 2) {
ret += array[0] + ".";
for (i = 0; i < length; i++) {
if (i >= array[1].length) ret += '0';
else ret += array[1][i];
}
} else if (array.length == 1) {
ret += array[0] + ".";
for (i = 0; i < length; i++) {
ret += '0'
}
}
return ret;
}
console.log(decimalPlaces(3.123, 6));
For future readers, I had this issue as I wanted to parse the onChange value of a textField into a float, so as the user typed I could update my model.
The problem was with the decimal place and values such as 12.120 would be parsed as 12.12 so the user could never enter a value like 12.1201.
The way I solved it was to check to see if the STRING value contained a decimal place and then split the string at that decimal and then count the number of characters after the place and then format the float with that specific number of places.
To illustrate:
const hasDecimal = event.target.value.includes(".");
const decimalValue = (hasDecimal ? event.target.value.split(".") : [event.target.value, ""])[1];
const parsed = parseFloat(event.target.value).toFixed(decimalValue.length);
const value = isNaN(parsed) ? "" : parsed;
onEditValue(value);
Here is dynamic version of floatParser for those who need
function customParseFloat(number){
if(isNaN(parseFloat(number)) === false){
let toFixedLength = 0;
let str = String(number);
// You may add/remove seperator according to your needs
[".", ","].forEach(seperator=>{
let arr = str.split(seperator);
if( arr.length === 2 ){
toFixedLength = arr[1].length;
}
})
return parseFloat(str).toFixed(toFixedLength);
}
return number; // Not a number, so you may throw exception or return number itself
}