My goal is to count iterations of 1 in an array.
I have this code:
var array = [10, 3, 22, 40, 1, 40, 1, 22, 1, 0, 3, 53, 1, 1];
var count = 0;
for(var i = 0; i < array.length; ++i){
if(array[i] === 1)
count++;
}
console.log(count);
Right now, it logs 5, but including "10" it should log 6, since "10" also contains a 1. What code can I use for this?
What about:
var count = array.toString().split('1').length - 1;
You can use filter method in combination with length property in order to write a more cleaner solution.
Also, use join and split methods in order to achieve your requirement.
var array = [10, 3, 22, 40, 1, 40, 1, 22, 1, 0, 3, 53, 1, 1];
console.log(array.join('').split('').filter(a=>a==1).length);
or using reduce method.
var array = [10, 3, 22, 40, 1, 40, 1, 22, 1, 0, 3, 53, 1, 1];
console.log(array.join('').split('').reduce((c,i)=> i==1 ? c+1 :c).length);
Just convert every item to string, and then check if 1 belongs to that number or not.
So for number 10, the string is '10' and then '10'.indexOf('1') equals to 0. So every time 1 is found within the string, you increment the counter.
var array = [10, 3, 22, 40, 1, 40, 1, 22, 1, 0, 3, 53, 1, 1];
var count = 0;
for (var i = 0; i < array.length; ++i) {
if (array[i].toString().indexOf('1') >= 0)
count++;
}
console.log(count);
You need to split your numbers into digits and iterate over each digit (which is a character) and compare with it. Use Object#toString to parse the number into the string, split into characters by String#split and using Array#forEach iterate over characters and do the same comparison there.
var array = [10, 3, 22, 40, 1, 40, 1, 22, 1, 0, 3, 53, 1, 1];
var count = 0;
for(var i = 0; i < array.length; ++i) {
var numberAsString = array[i].toString();
numberAsString.split('').forEach(char => {
if(char === '1') {
count++
}
});
}
console.log(count);
You'll need to convert each number to a string and use indexOf
var array = [10, 3, 22, 40, 1, 40, 1, 22, 1, 0, 3, 53, 1, 1];
var count = array.reduce(function(p,c){
if(c.toString().indexOf("1")>-1)
p++;
return p;
},0);
console.log(count);
You can Array#join the array items to a string and count using String#match with a regular expression:
var array = [10, 3, 22, 40, 1, 40, 1, 22, 1, 0, 3, 53, 1, 1];
// the || [] is a fallback in case there is no 1, and match returns null
var result = (array.join('').match(/1/g) || []).length;
console.log(result);
You could check every character of the joined string.
var array = [10, 3, 22, 40, 1, 40, 1, 22, 1, 0, 3, 53, 1, 1],
count = 0,
i,
temp = array.join('');
for (i = 0; i < temp.length; ++i) {
if (temp[i] === '1') {
count++;
}
}
console.log(count);
var array = [10, 3, 22, 40, 1, 40, 1, 22, 1, 0, 3, 53, 1, 1];
var count = 0;
for(var i = 0; i < array.length; ++i){
if( array[i] === 1 || array[i].toString().includes("1") )
count++;
}
console.log( count );
Related
It's a bit of a tricky situation I'm in, but I have an array like this:
const nums = [32, -3, 62, 8, 121, -231, 62, 13];
and need to replace them by their corresponding ascending index. The above example should yield:
[4, 1, 5, 2, 7, 0, 6, 3]
The solution I've come up with is this: TS Playground
const nums = [32, -3, 62, 8, 121, -231, 62, 13];
const numsCopy = nums.map(e => e);
// Basic sorting
for (let i = 0; i < numsCopy.length; i++) {
for (let j = 0; j < numsCopy.length; j++) {
if (numsCopy[i] < numsCopy[j]) {
let t = numsCopy[j];
numsCopy[j] = numsCopy[i];
numsCopy[i] = t;
}
}
}
for (let i = 0; i < numsCopy.length; i++) {
let sortedValue = numsCopy[i];
nums[nums.indexOf(sortedValue)] = i;
}
Problems arise however when I change nums to include a value nums.length > n >= 0. The call nums.indexOf(...) may return a faulty result, as it may have already sorted an index, even though it exists somewhere in the array.
If you replace nums with these values, -231 will have an index of 2 for some reason...
const nums = [32, -3, 62, 7, 121, -231, 62, 13, 0];
> [5, 1, 6, 3, 8, 2, 7, 4, 0]
Is there a better approach to this problem, or a fix to my solution?
You could sort the indices by the value and create a new array with index values a sorted positions.
to get the wanted result call the sorting function again and you get the indices sorted by the index order.
const
sort = array => [...array.keys()].sort((a, b) => array[a] - array[b]),
fn = array => sort(sort(array));
console.log(...fn([32, -3, 62, 8, 121, -231, 62, 13])); // 4 1 5 2 7 0 6 3
console.log(...fn([-1, 3, 1, 0, 2, 9, -2, 7])); // 1 5 3 2 4 7 0 6
Copy the array, sort its values, get indexOf, and null the value in the sorted copy:
const sortIndicesByValue = array => {
const sorted = [...array].sort((a, b) => a - b);
return array.map(e => {
const i = sorted.indexOf(e);
sorted[i] = null;
return i;
})
}
console.log(...sortIndicesByValue([32, -3, 62, 8, 121, -231, 62, 13]));
console.log(...sortIndicesByValue([-1, 3, 0, 0, 2, 9, -2, 7]));
What is the method to evenly split an array into two. Then, position the second half underneath the first all for the purpose of comparing whether or not a new final array should receive a zero or one in the slots.
var master_array = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38];
var first_half = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19];
var second_half = [20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38];
var final_usable_array_reduced = [];
So, given a form with 38 check boxes, if the 5th, 19th, 37th, and 38th check boxes were checked, the final_usable_array_reduced would be
final_usable_array_reduced = [0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1];
final_usable_array_reduced[4] = **represents the 5th <strong>or</strong> 24th box being checked**
final_usable_array_reduced[18] = **represents the 19th <strong>or</strong> 38th boxes being checked**
I think you can implement this with a simple for loop over half the master_array, oring the values in the bottom half of the array with those in the top half:
var master_array = new Array(38).fill(0);
master_array[4] = 1;
master_array[18] = 1;
master_array[36] = 1;
master_array[37] = 1;
let len = master_array.length / 2;
final_usable_array_reduced = new Array(len);
for (let i = 0; i < len; i++) {
final_usable_array_reduced[i] = master_array[i] | master_array[i + len];
}
console.log(final_usable_array_reduced);
Check this one:
var array = [0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0];
var firstHalf = array.slice(0,array.length/2);
var secondHalf = array.slice(array.length/2);
var result = firstHalf.map(function(e,i){
return e || secondHalf[i];
});
console.log(result);
What I'm doing here is spliting the array into two halfs with slice (considering that the array elements are even, of course), and after that I map the first one returning 1 (or true, is the same) if firstArray[i] or secondArray[1] is 1.
Consider the following.
$(function() {
function addChecks(count, tObj, cont) {
var i;
if (cont == undefined) {
i = 0;
} else {
i = parseInt($("input:last").val()) - 1;
count = i + count;
}
for (i; i <= count; i++) {
$("<input>", {
type: "checkbox",
value: (i + 1),
title: (i + 1)
}).appendTo(tObj);
}
}
function checksToArray() {
var arr = [];
$("input[type='checkbox']").each(function() {
arr.push($(this).prop("checked"));
});
return arr;
}
function compHalfArr(a) {
var arr = [];
var h = Math.floor(a.length / 2);
for (var i = 0; i <= h; i++) {
arr.push(a[i] || a[(i + h)]);
}
return arr;
}
addChecks(14, $(".section-1"));
addChecks(14, $(".section-2"), true);
$("button").click(function() {
var allChecks = checksToArray();
var compChecks = compHalfArr(allChecks);
$(".results").html("<span>" + compChecks.join("</span><span>") + "</span>").find("span:last").remove();
});
});
.checks input {
margin-right: 20px;
}
.results span {
display: inline-block;
font-family: Arial, sans-serif;
width: 40px;
text-align: center;
}
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div class="checks section-1"></div>
<div class="checks section-2"></div>
<button>Check</button>
<div class="results"></div>
You can do a loop using a halfway marker.
I want to compare each element of array real with each element of array number. And if there is any matches push them in array add so I can see them. In this case add must be 2,3,6,10,14 if code is good.
<!DOCTYPE html>
<html>
<head>
</head>
<body>
<script>
var real=[1,2,3,4,5,6,7,8,10,14,16,233,235,245,2,5,7,236,237];
var number=[2,3,6,10,12,13,14,172,122,234];
var add=[];
for (k=0; k<number.length; k++)
{
for (w=0; w<real.length; w++)
{
if (number[k]==real[w]);
{
add.push(number[k],real[w]);
}
};
};
document.write(add+"<br>");
</script>
Here is a short and simple solution using Array.forEach and Array.indexOf functions:
var real = [1,2,3,4,5,6,7,8,10,14,16,233,235,245,2,5,7,236,237],
number = [2,3,6,10,12,13,14,172,122,234],
add = [];
real.forEach(function(v) {
if (number.indexOf(v) !== -1 && this.indexOf(v) === -1) this.push(v);
}, add);
console.log(add); // [2, 3, 6, 10, 14]
A more elegant and readable solution:
var matched = [];
real.forEach(function(realNum) {
number.forEach(function(numberNum) {
if(realNum == numberNum && matched.indexOf(realNum) === -1) {
matched.push(realNum);
}
});
});
Here's one way you can do it using ES6:
var real = [1, 2, 3, 4, 5, 6, 7, 8, 10, 14, 16, 233, 235, 245, 2, 5, 7, 236, 237];
var number = [2, 3, 6, 10, 12, 13, 14, 172, 122, 234];
var filtered = real.filter(x => number.indexOf(x) > -1);
var unique = new Set(filtered);
document.body.innerHTML = [...unique];
try this way...
Sorted the main Array, removed duplicates and the find the common elements from both the arrays.
var main = [1,2,3,4,5,6,7,8,10,14,16,233,235,245,2,5,7,236,237];
var compare = [2,3,6,10,12,13,14,172,122,234];
function compareNumbers(a, b) {
return a - b;
}
console.log('Sorted Array :', main.sort(compareNumbers) );
// Sorted Array : [1, 2, 2, 3, 4, 5, 5, 6, 7, 7, 8, 10, 14, 16, 233, 235, 236, 237, 245]
Array.prototype.unique = function() {
var unique = [];
for (var i = 0; i < this.length; i++) {
var current = this[i];
if (unique.indexOf(current) < 0) unique.push(current);
}
return unique;
}
console.log('Unique Array Elements:', main.unique() );
// Unique Array Elements: [1, 2, 3, 4, 5, 6, 7, 8, 10, 14, 16, 233, 235, 236, 237, 245]
function commonElements(arr1, arr2) {
var common = [];
for (var i = 0; i < arr1.length; i++) {
for (var j = 0; j < arr2.length; j++) {
if (arr1[i] == arr2[j] ) {
common.push( arr1[i] );
j == arr2.length; // To break the loop;
}
}
}
return common;
}
console.log('Common Elements from Both Arrays : ', commonElements(main.unique(), compare.unique()) );
//Common Elements from Both Arrays : [2, 3, 6, 10, 14]
What is the best possible way to reduce an array of ranges in javascript.
For example I have
1-3,4-5,10-12,2-4
the result I need for this is
1-5, 10-12
What is the best way to tackle this problem ?
I would first create another array with no duplicates, storing the numbers that are covered by the ranges:
1-3 covers 1, 2, 3 --> [1, 2, 3]
4-5 covers 4, 5 --> [1, 2, 3, 4, 5]
10-12 covers 10, 11, 12 --> [1, 2, 3, 4, 5, 10, 11, 12]
2-4 covers 2, 3, 4 --> [1, 2, 3, 4, 5, 10, 11, 12]
Then, sort the array:
[1, 2, 3, 4, 5, 10, 11, 12] // nothing changed in this example
Finally, rebuild the ranges, depending on the consecutive values:
1-5
10-12
1) parse the input to build a structure like this:
var rlist = [
{min: 1, max: 3},
{min: 4, max: 5},
{min: 10, max: 12}
{min: 2, max: 4}
];
2) marge each interval of that list into a new list:
var olist = [], i, j, r, p, s;
for (i = 0; i < rlist.length; ++i) {
r = rlist[i];
for (j = 0; j < olist.length; ) {
p = olist[j];
if (r.max+1 < p.min) {
// insert here
break;
} else if (p.max+1 >= r.min) {
// intersection
olist.splice(j, 1);
r.min = p.min;
r.max = Math.max(r.max, p.max);
} else {
++j;
}
}
olist.splice(j, 0, r);
}
3) Convert result into a string
s = "";
for (j = 0; j < olist.length; ++j) {
if (j > 0) {
s += ",";
}
s += olist[j].min + "-" + olist[j].max;
}
Fiddle
I have a number I want to conform to the closest value in the following sequence:
2, 5, 9, 12, 15, 19, 22, 25, 29, 32, 35, 39, 42...
If 10 is passed, it would become 12, 13 would become 15, 17 would become 19.
How would I approach this in a function?
If you don't know whether the array is sorted, you could use code like this to find the value in the array that is the closest to the passed in value (higher or lower):
var list = [2, 5, 9, 12, 15, 19, 22, 25, 29, 32, 35, 39, 42];
function findClosestValue(n, list) {
var delta, index, test;
for (var i = 0, len = list.length; i < len; i++) {
test = Math.abs(list[i] - n);
if ((delta === undefined) || (test < delta)) {
delta = test;
index = i;
}
}
return(list[index]);
}
If you want the closest number without going over and the array is sorted, you can use this code:
var list = [2, 5, 9, 12, 15, 19, 22, 25, 29, 32, 35, 39, 42];
function findClosestValue(n, list) {
var delta, index;
for (var i = 0, len = list.length; i < len; i++) {
delta = n - list[i];
if (delta < 0) {
return(index ? list[index] : undefined);
}
index = i;
}
return(list[index]);
}
Here is a jsFiddle with a solution that works for an infinite series of the combination "+3+4+3+3+4+3+3+4+3+3+4....." which seems to be your series. http://jsfiddle.net/9Gu9P/1/ Hope this helps!
UPDATE:
After looking at your answer I noticed you say you want the closes number in the sequence but your examples all go to the next number in the sequence whether it is closest or not compared the the previous number so here is another jsfiddle that works with that in mind so you can choose which one you want :).
http://jsfiddle.net/9Gu9P/2/
function nextInSequence(x){
//sequence=2, 5, 9, 12, 15, 19, 22, 25, 29, 32, 35, 39, 42
var i= x%10;
switch(i){
case 2:case 5:case 9: return x;
case 0:case 1: return x+ 2-i;
case 3:case 4: return x+5-i;
default: return x+9-i;
}
}
alert(nextInSequence(10))