I'm trying to do a check that the first array contains the same values of the second array.
However I'm confused about my code.
First question is: why is my code running my else statement if all letters in the first array are contained in the second? it will run 2 lines of "this is not valid"
Second question is: if my first array contains a duplicate letter it will still pass the check e.g
["a", "b" , "a", "d", "e", "f"]; even though there is two a's in the first it will see the same "a" again. Anyone know a way around this.
Sorry for my long winded questions but I hope it makes sense. Thanks :)
var letters = ["a", "b" , "c", "d", "e", "f"];
var otherLetters = ["a","b", "c" , "d", "e", "f"];
var i = -1;
while(i<=letters.length){
i++;
if(otherLetters.includes(letters[i])){
console.log("This is valid");
}
else
console.log("This is not valid");
}
You didn't close the brackets. And your loop is very confusing, please use foreach. Here is a working example:
const letters = ["a", "b" , "c", "d", "e", "f"];
const otherLetters = ["a","b", "c" , "d", "e", "f"];
letters.forEach(el => {
if (otherLetters.includes(el)) {
console.log(el + 'is valid');
} else {
console.log(el + 'is not valid');
}
});
You are trying to access array elements which are out of bounds. The script runs 8 iterations over an array with 6 elements.
Nothing to worry, cpog90.
Try this solution.
var letters = ["a", "b" , "c", "d", "e", "f"];
var otherLetters = ["a","b", "c" , "d", "e", "f"];
var i = 0;
while(i<letters.length){
if(otherLetters.includes(letters[i])){
console.log("This is valid");
}
else {
console.log("This is not valid "+i);
}
i++;
}
What went wrong in your logic?
If you declare i = -1 and while(i<=letters.length), As 6 is length of letters, 8 iterations will be done as follows.
For first iteration (i = -1), 'while' condition returns true and checks for 'a'
output: This is valid
For second iteration (i = 0), 'while' condition returns true and checks for 'b'
output: This is valid
For third iteration (i = 1), 'while' condition returns true and checks for 'c'
output: This is valid
For fourth iteration (i = 2), 'while' condition returns true and checks for 'd'
output: This is valid
For fifth iteration (i = 3), 'while' condition returns true and checks for 'e'
output: This is valid
For sixth iteration (i = 4), 'while' condition returns true and checks for 'f'
output: This is valid
For seventh iteration (i = 5), 'while' condition returns true and checks for undefined value.
output: This is not valid
For eighth iteration (i = 6), 'while' condition returns true and checks for undefined value.
output: This is not valid
First of all you have set i = -1 which is confusing since array start position is 0.
The reason your loop is running two extra times is because loop started at -1 instead of 0 and next the condition i <= length.
Since [array length = last index + 1] your loop runs extra two times.
Just to make your code work assign var i = 0 and while condition i < letters.length
Simplest solution is using lodash. It has all optimizations out-of-the-box:
var letters = ["a", "b" , "c", "d", "e", "f"];
var otherLetters = ["f", "a","b", "c" , "d", "e"];
const finalLetters = _.sortBy(letters);
const finalOtherLetters = _.sortBy(otherLetters);
if (_.isEqual(finalLetters, finalOtherLetters)) {
console.log('Two arrays are equal.');
} else {
console.log('Two arrays are not equal.');
}
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.15/lodash.min.js"></script>
Arrays are index based and starts from 0. So, the -1 and the less than letters.length check puts the code into out of bounds.
var letters = ["a", "b" , "c", "d", "e", "f"];
var otherLetters = ["a","b", "c" , "d", "e", "f"];
var i = 0;
while(i<letters.length)
{
if(otherLetters.includes(letters[i]))
{
console.log("This is valid");
}
else
{
console.log("This is not valid");
}
i++;
}
You can use a combination of Array.prototype.every with Array.prototype.includes, along with some extra guard clauses.
const areSequenceEqual = (arr1, arr2) => {
if (!arr1 || !arr2) {
return false;
}
if (arr1.length !== arr2.length) {
return false;
}
return arr1.every(x => arr2.includes(x));
};
const letters = ["a", "b", "c", "d", "e", "f"];
const otherLetters = ["a", "b", "c", "d", "e", "f"];
const someOtherLetters = ["a", "b", "c", "d", "e", "f", "g"];
console.log(areSequenceEqual(letters, otherLetters));
console.log(areSequenceEqual(letters, undefined));
console.log(areSequenceEqual(letters, someOtherLetters));
Related
Let's say I have an array
var array = ["A", "A", "A", "B", "B", "A", "C", "B", "C"];
And I want to do a check, and get true if there are 4 identical elements "A"
If I use array.includes("A"), then it looks for only one such among all and in any case returns true, but I need when there are exactly 4 such elements
Or let's say in the same array I want to find 3 elements "B" and 2 elements "C", and return true only if there are as many of them as I'm looking for, if less, then return false
How can I do this?
Take a shot like this.
const myArr = ["A", "A", "A", "B", "B", "A", "C", "B", "C"];
const findExactly = (arr, val, q) => arr.filter(x => x == val).length == q;
// logs "true" as expected :)
console.log(findExactly(myArr, 'A', 4));
So the function findExactly receives an array, a value and a number X as the quantity. Returns a boolean if the array contains the value X times. So the example above works for the example you gave on the question "And I want to do a check, and get true if there are 4 identical elements "A"".
Pulling my answer from this on stackflow.
var array = ["A", "A", "A", "B", "B", "A", "C", "B", "C"];
const counts = {};
for (const el of arr) {
counts[el] = counts[el] ? counts[el] + 1 : 1;
}
console.log(counts["A"] > 4) // 4 or more "A"s have been found
console.log(counts["B"] === 3 && counts["C"] === 3) // exactly 3 "B"s and 2 "C"s have been found
I am creating a roulette game that displays random items from different arrays when the wheel lands on a specific category. So far everything works except when the wheel lands on a category, it selects the same random item from the correct array over and over again. I am trying to use math.random and the splice method to randomly select an item from an array, and remove that item so only new, random items from the array can be displayed after, but it hasn't worked.
I rearranged the input arrays into an array of arrays (5 arrays of 9 elements = 45). I'm guessing that you want the whole thing shuffled.
const log = data => console.log(JSON.stringify(data));
let zones = [
["🎁", "🎆", "🎯", "🌈", "🌛", "💩", "💰", "🍒", "🌴"],
["🌞", "🍀", "💀", "💘", "💣", "🎲", "🎰", "🎈", "🗿"],
["🎁", "🎆", "🎯", "🌈", "🌛", "💩", "💰", "🍒", "🌴"],
["🌞", "🍀", "💀", "💘", "💣", "🎲", "🎰", "🎈", "🗿"],
["🎁", "🎆", "🎯", "🌈", "🌛", "💩", "💰", "🍒", "🌴"]
];
let zoneSize = zones.length * zones[0].length;
let symbolZones = [];
for (let i = zoneSize; i > 0; i--) {
let deg = zones.flatMap(z => z.splice(Math.floor(Math.random() * z.length), 1));
if(deg.length > 0) {
symbolZones.push(deg);
}
}
log(`Original Array zones`);
log(zones);
log(`New Array symbolZones`);
log(symbolZones);
.as-console-row-code {
display: block;
width: 100%;
overflow-wrap: anywhere;
}
.as-console-row::after {
content: ''!important
}
I don't know how the rest of the code is, but if you always get the same value over and over again it could be that you're not re-running the code that gets the values.
Try wrapping the logic for retrieving the symbolZones in a function:
function getSymbolZones() {
const symbolZones = [];
symbolZones[1] = a[Math.floor(Math.random()*a.length)];
symbolZones[2] = b[Math.floor(Math.random()*b.length)];
symbolZones[3] = c[Math.floor(Math.random()*c.length)];
symbolZones[4] = d[Math.floor(Math.random()*d.length)];
symbolZones[5] = e[Math.floor(Math.random()*e.length)];
symbolZones[6] = f[Math.floor(Math.random()*f.length)];
symbolZones[7] = g[Math.floor(Math.random()*g.length)];
symbolZones[8] = h[Math.floor(Math.random()*h.length)];
return symbolZones;
}
and then use it in the handleWin function.
const handleWin = (actualDeg) => {
const symbolZones = getSymbolZones();
const winningSymbolNr = Math.ceil(actualDeg / zoneSize);
display.innerHTML = symbolZones[winningSymbolNr];
}
P.S.
I understand that you start the array at index 1 because that's the smallest zone you can get from the operation 45 / 45.
But if I were you I'd start the index from 0 and just subtract 1 when accessing the array: symbolZones[winningSymbolNr - 1].
This way you don't get weird bugs when trying to loop through an array that has an empty first index.
Other than using array, I would suggest you use object and a for loop to make the code easier to execute. This code should work:
let deg = 0;
// The 360 degree wheel has 8 zones, so each zone is 45 degrees
let zoneSize = 45;
let symbol = {
a: ["a", "b", "c", "d", "e"],
b: ["f", "g", "h", "i", "j"],
c: ["a", "b", "c", "d", "e"],
d: ["f", "g", "h", "i", "j"],
e: ["a", "b", "c", "d", "e"],
f: ["f", "g", "h", "i", "j"],
g: ["a", "b", "c", "d", "e"],
h: ["f", "g", "h", "i", "j"],
}
// zones for each 8 categories
let ran = Math.floor(Math.random() * Object.keys(symbol).length)
console.log(Object.values(symbol)[ran])
//Random select a zone from above 8
let selectedzone = Object.values(symbol)[ran]
//Random select an item from selected zone
let index = Math.floor(Math.random() * selectedzone.length)
let symbolZones = selectedzone[index]
console.log(symbolZones)
const handleWin = (actualDeg) => {
const winningSymbolNr = Math.ceil(actualDeg / zoneSize);
display.innerHTML = symbolZones[winningSymbolNr];
}
i have array javascriptarray1 = ["a", "b", "c", "d"]; and array2 = ["a", "b", "c", "d", "e", "f"];
and how to convert array1 to [{"a", "c"}, {"b,"d"}]; and how to convert array2 to [{"a", "c", "e"}, {"b", "d", "f"}];
For loop which iterates index + 2 and appends value to x array
For loop which iterates index + 1 and appends value to y array
Then do whatever you need with those arrays :)
Hope that helps
a,c are the odd indices and b,d are the even indices. So,
a and c can be separated with the increment of indices by 2.
"initialize the 'i' value with '0' ".
for (i = 0; i < array length; i=i+2) {
// Your code here
}
b and d can be separated with the increment of indices by 2 .
"initialize the 'i' value with '1' ".
for (i = 1; i < array length; i=i+2) {
//Your code here
}
I have a list with some items for example
["a", "b", "c", ..., "x", "y", "z"]
I would like to iterate it but from the end to beggining and push those items into a new variable, and stop when it has length == 3.
For that simple example I would like to have as result within my new var:
["z", "y", "x"]
I'm thinking of .reverse() my array and then iterate it with .each and push my items, but I believe there is a better way to do that with lodash, that I'm not finding.
Maybe I'm not knowing how to search.
Thanks in advance.
You can do it with the function "_.takeRightWhile" from lodash like the code below:
var arr = ["a", "b", "c", "x", "y", "z"];
var reverseArray = [];
_.takeRightWhile(arr, function(item){
reverseArray.push(item)
return reverseArray.length < 3
});
console.log(reverseArray);
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.4/lodash.min.js"></script>
In plain Javascript you could use Array#slice with a negative count for getting a new array from the end and use Array#reverse for a reversed array.
var array = ["a", "b", "c", "x", "y", "z"],
result = array.slice(-3).reverse();
console.log(result);
For processing items, you could use Array#reduceRight.
var array = ["a", "b", "c", "x", "y", "z"],
result = array.slice(-3).reduceRight((r, a) => r.concat(a), []);
console.log(result);
Another solution that iterates the original array:
var arr = ["a", "b", "c", "x", "y", "z"], res=[], count=3;
if (count <= arr.length)
for (var i=0; i<count; i++) res.push(arr[arr.length-1-i]);
console.log(res);
I'm using regex to test certain elements in an array of arrays. If an inner array doesn't follow the desired format, I'd like to remove it from the main/outer array. The regex I'm using is working correctly. I am not sure why it isn't removing - can anyone advise or offer any edits to resolve this problem?
for (var i = arr.length-1; i>0; i--) {
var a = /^\w+$/;
var b = /^\w+$/;
var c = /^\w+$/;
var first = a.test(arr[i][0]);
var second = b.test(arr[i][1]);
var third = c.test(arr[i][2]);
if ((!first) || (!second) || (!third)){
arr.splice(i,1);
}
When you cast splice method on an array, its length is updated immediately. Thus, in future iterations, you will probably jump over some of its members.
For example:
var arr = ['a','b','c','d','e','f','g']
for(var i = 0; i < arr.length; i++) {
console.log(i, arr)
if(i%2 === 0) {
arr.splice(i, 1) // remove elements with even index
}
}
console.log(arr)
It will output:
0 ["a", "b", "c", "d", "e", "f", "g"]
1 ["b", "c", "d", "e", "f", "g"]
2 ["b", "c", "d", "e", "f", "g"]
3 ["b", "c", "e", "f", "g"]
4 ["b", "c", "e", "f", "g"]
["b", "c", "e", "f"]
My suggestion is, do not modify the array itself if you still have to iterate through it. Use another variable to save it.
var arr = ['a','b','c','d','e','f','g']
var another = []
for(var i = 0; i < arr.length; i++) {
if(i%2) {
another.push(arr[i]) // store elements with odd index
}
}
console.log(another) // ["b", "d", "f"]
Or you could go with Array.prototype.filter, which is much simpler:
arr.filter(function(el, i) {
return i%2 // store elements with odd index
})
It also outputs:
["b", "d", "f"]
Your code seems to work to me. The code in your post was missing a } to close the for statement but that should have caused the script to fail to parse and not even run at all.
I do agree with Leo that it would probably be cleaner to rewrite it using Array.prototype.filter though.
The code in your question would look something like this as a filter:
arr = arr.filter(function (row) {
return /^\w+$/.test(row[0]) && /^\w+$/.test(row[1]) && /^\w+$/.test(row[2]);
});
jsFiddle
I'm assuming it is 3 different regular expressions in your actual code, if they are all identical in your code you can save a little overhead by defining the RegExp literal once:
arr = arr.filter(function (row) {
var rxIsWord = /^\w+$/;
return rxIsWord.test(row[0]) && rxIsWord.test(row[1]) && rxIsWord.test(row[2]);
});