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I have an array as follows:
var array = ['Bob','F', 'Nichols'];
I want to detect whether this array contains any values that are a single character long. In other words, I want to know whether there are any initials in this person's name.
var array = ['Bob','F', 'Nichols']; //true
var array = ['B','Freddy', 'Nichols']; //true
var array = ['Bob','Freddy', 'N']; //true
var array = ['B','F', 'N']; //true
var array = ['B','F', 'N']; //true
var array = ['Bob','Freddy', 'Nichols']; //false
if (anyInitials(array)) {
console.log("there are initials");
} else {
console.log("there are no initials");
}
function anyInitials(a) {
var arrayLength = a.length;
var initial = 'no';
for (var i = 0; i < arrayLength; i++) {
if (a[i].length == 1){
initial = 'yes';
}
}
return initial;
}
You can use the function some
let array = ['Bob','F', 'Nichols'];
console.log(array.some(({length}) => length === 1));
let anyInitials = a => a.some(({length}) => length === 1) ? "yes" : "no";
You can use a simple .forEach() loop like below. This loops through the array, and sets isInitial to true if the length is 1.
var array = ['Bob', 'F', 'Nichols'];
function anyInitials(a) {
var isInitial = false;
a.forEach(e => {
if (e.length == 1) {
isInitial = true;
}
});
return isInitial;
}
console.log(anyInitials(array));
You can also use .some() like below. This will return true if any element in the array has a length of 1.
var array = ['Bob', 'F', 'Nichols'];
function anyInitials(a) {
return array.some(e => e.length == 1);
}
console.log(anyInitials(array));
Given an array I want to find the largest sub array by the length i.e
var table = [
["Protein",["Grilled Chicken","Lamb"]],
["Fats",["rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr"]],
["Vegatables",["Carrots","Eggs","Milks","Peppers"]]
];
I want it to return ["Carrots","Eggs","Milks","Peppers"]
Heres my code
function findBiggestSubArray(array){
var biggestArrayIndex = 0;
for(var i=0;i<array.length;i++){
if(i === (array.length-1)){
//We have reached the end of the array then return the array
console.log("Reached the End");
return array[biggestArrayIndex];
} else {
if(!array[biggestArrayIndex][1].length >= array[i][1].length){
biggestArrayIndex = i;
}//End of Inner else block
}//End of Outer else block
}//End of forloop
}
General solution, to find the most largest array in an array-structure:
I would do it with recursion, so the most biggest Array will be found, in any depth..
/**
* array -> The array to check,
* biggestArray -> The most biggestArray found so far
*/
function findBiggestArray(array, biggestArray){
biggestArray = biggestArray || [];
if (array.length > biggestArray.length)
biggestArray = array;
for (var i = 0; i < array.length; i++) {
if (array[i] instanceof Array)
biggestArray = findBiggestArray(array[i],biggestArray)
}
return biggestArray;
}
var multiArray = [
["1", "2", ["234", "334"]],
[1,2,3,4,5, [1,2,3,4,5,6,7,7]]
]
var biggest = findBiggestArray(multiArray)
console.log(biggest)
// This also works!
console.log(findBiggestArray([1, [1,2,3]]))
Oneliner for this special case
// Sort the array by the length of the subarray at position 1, and return the first item
var category = table.sort(function(a, b) { return b[1].length - a[1].length })[0]
// ES6-Syntax
var category = table.sort((a, b) => b[1].length - a[1].length)[0]
category // => ["CategoryName", [ITEMS..]]
I would do this way (see the comments in the code for explanation):
var table = [
["Protein", ["Grilled Chicken", "Lamb"]],
["Fats", ["rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr"]],
["Vegatables", ["Carrots", "Eggs", "Milks", "Peppers"]]
];
function findBiggestSubArray (array) {
// Initialise empty array.
var bigSubArray = ["", []];
// Loop through the given array.
for (var i = 0; i < array.length; i++) {
// Check if the current biggest one is bigger than the saved array.
if (array[i][1].length > bigSubArray[1].length) {
// If bigger, replace it with current array.
bigSubArray = array[i];
}
}
// Return the biggest sub array.
return bigSubArray[1];
}
console.log(findBiggestSubArray(table));
Given an array of words, write a function that returns an array of the words that occur an even number of times.
function even(["hello", "hi", "hello", "elephant", "hi"]);
That output should be:
["hello", "hi"]
This has been a toy problem I have been struggling with recently. I have solved similar problems counting and returning the number of occurrences of elements in an array but am having trouble taking that logic and applying it to this problem.
This is what I have tried so far, but have hit a wall when trying to output just the even occurrences:
function even(collection) {
var results = [];
for(var i = 0; i < collection.length; i++){
var value = collection[i];
if(results[value]){
results[value] = results[value] + 1;
}else{
results[value] = 1;
}
}
return results;
}
You can use reduce to get an actual count of the words, then simply return an array of the ones that have an even count:
function even(wordsArr) {
//Object of words and counts
var wordCounts = wordsArr.reduce(function(counts, word) {
if (!counts.hasOwnProperty(word)) {
counts[word] = 0;
}
counts[word]++;
return counts;
}, {});
//Now filter that out and return
return Object.keys(wordCounts).filter(function(word) {
return wordCounts[word] % 2 === 0
});
}
even(["hello", "hi", "hello", "elephant", "hi"]); //["hello", "hi"]
var arr = ["hello", "hi", "hello", "elephant", "hi"];
function onlyEvens( arr )
{
var countObj = {};
for( var i = 0; i < arr.length; i++ )
{
var item = arr[i];
if( countObj[ item ] !== undefined )
countObj[item]++;
else
countObj[item] = 1;
}//for()
var filteredArray = [];
for(var key in countObj )
{
if( countObj[key] % 2 == 0 )
filteredArray.push( key );
}
return filteredArray;
}//onlyEvens()
console.log( onlyEvens( arr ) );
Issues in your code:
you use collection instead of words
you cannot access array the associative way. You must declare it as object:
results[value]
you return result variable, but it is undeclared.
return result;
results only contains the occurrences of every word. There miss the code that calculates if the occurrences of a word are odd or even.
fixed code:
function even(words) { // <<< in your code was collection
var results = {};
for(var i = 0; i < words.length; i++){
var value = words[i];
if(results[value]){
results[value] = results[value] + 1;
}else{
results[value] = 1;
}
}
var ret = [];
for(var word in results)
if(results[word]%2 !== 0)
rest.push(word);
return ret;
}
function even(list) {
var d = list.reduce(function(d, w) { d[w] = !d[w]; return d; }, {});
return Object.keys(d).filter(function(w) { return !d[w]; });
}
console.log(even(["hello", "hi", "hello", "elephant", "hi"]));
console.log(even(["hello", "yo", "yo", "hi", "hello", "yo", "elephant", "hi"]));
Explanation: Use the array .reduce() method to create an object (d) with a property for each word (w) with a boolean value indicating whether the word has an odd number of occurrences. Then .filter() the keys to get all the ones that are not odd.
If you previously sort the array you can filter it as required in just a code line like this :
var even = (str) => str.sort().filter((element, index, arr) => index+1 === arr.lastIndexOf(element));
console.log(even(["hello", "hello", "hi", "elephant", "hi", "hi"])); //[ 'hello', 'hi' ]
I've been trying to find a reasonably concise way to set the dimensions of an empty multidimensional JavaScript array, but with no success so far.
First, I tried to initialize an empty 10x10x10 array using var theArray = new Array(10, 10 10), but instead, it only created a 1-dimensional array with 3 elements.
I've figured out how to initialize an empty 10x10x10 array using nested for-loops, but it's extremely tedious to write the array initializer this way. Initializing multidimensional arrays using nested for-loops can be quite tedious: is there a more concise way to set the dimensions of empty multidimensional arrays in JavaScript (with arbitrarily many dimensions)?
//Initializing an empty 10x10x10 array:
var theArray = new Array();
for(var a = 0; a < 10; a++){
theArray[a] = new Array();
for(var b = 0; b < 10; b++){
theArray[a][b] = new Array();
for(var c = 0; c < 10; c++){
theArray[a][b][c] = 10
}
}
}
console.log(JSON.stringify(theArray));
Adapted from this answer:
function createArray(length) {
var arr = new Array(length || 0),
i = length;
if (arguments.length > 1) {
var args = Array.prototype.slice.call(arguments, 1);
while(i--) arr[i] = createArray.apply(this, args);
}
return arr;
}
Simply call with an argument for the length of each dimension.
Usage examples:
var multiArray = createArray(10,10,10); Gives a 3-dimensional array of equal length.
var weirdArray = createArray(34,6,42,2); Gives a 4-dimensional array of unequal lengths.
function multiDimArrayInit(dimensions, leafValue) {
if (!dimensions.length) {
return leafValue;
}
var arr = [];
var subDimensions = dimensions.slice(1);
for (var i = 0; i < dimensions[0]; i++) {
arr.push(multiDimArrayInit(subDimensions, leafValue));
}
return arr;
}
console.log(multiDimArrayInit([2,8], "hi")); // counting the nested "hi"'s yields 16 of them
demo http://jsfiddle.net/WPrs3/
Here is my take on the problem: nArray utility function
function nArray() {
var arr = new Array();
var args = Array.prototype.slice.call(arguments, 1);
for(var i=0;i<arguments[0];i++) {
arr[i] = (arguments.length > 1 && nArray.apply(this, args)) || undefined;
}
return arr;
}
Usage example:
var arr = nArray(3, 3, 3);
Results in 3x3x3 array of undefined values.
Running code with some tests also available as a Fiddle here: http://jsfiddle.net/EqT3r/7/
The more dimension you have, the more you have interest in using one single flat array and a getter /setter function for your array.
Because for a [d1 X d2 X d3 X .. X dn] you'll be creating d2*d3*...*dn arrays instead of one, and when accessing, you'll make n indirection instead of 1.
The interface would look like :
var myNArray = new NArray(10,20,10);
var oneValue = myNArray.get(5,8,3);
myNArray.set(8,3,2, 'the value of (8,3,2)');
the implementation depends on your preference for a fixed-size
n-dimensionnal array or an array able to push/pop and the like.
A more succinct version of #chris code:
function multiDim (dims, leaf) {
dims = Array.isArray (dims) ? dims.slice () : [dims];
return Array.apply (null, Array (dims.shift ())).map (function (v, i) {
return dims.length
? multiDim (dims, typeof leaf == 'string' ? leaf.replace ('%i', i + ' %i') : leaf)
: typeof leaf == 'string' ? leaf.replace ('%i', i) : leaf;
});
}
console.log (JSON.stringify (multiDim ([2,2], "hi %i"), null, ' '));
Produces :
[
[
"hi 0 0",
"hi 0 1"
],
[
"hi 1 0",
"hi 1 1"
]
]
In this version you can pass the first argument as a number for single dimension array.
Including %i in the leaf value will provide index values in the leaf values.
Play with it at : http://jsfiddle.net/jstoolsmith/r3eMR/
Very simple function, generate an array with any number of dimensions. Specify length of each dimension and the content which for me is '' usually
function arrayGen(content,dims,dim1Len,dim2Len,dim3Len...) {
var args = arguments;
function loop(dim) {
var array = [];
for (var a = 0; a < args[dim + 1]; a++) {
if (dims > dim) {
array[a] = loop(dim + 1);
} else if (dims == dim) {
array[a] = content;
}
}
return array;
}
var thisArray = loop(1);
return thisArray;
};
I use this function very often, it saves a lot of time
I have 2 array:
var array1 = [[5,10],[6,10],[7,10],[8,10],[9,10]];
var array2 = [[1,10],[2,10],[3,10],[4,10],[5,40],[6,40]];
Want to get 1 merged array with the sum of corresponding keys;
var array1 = [[1,10],[2,10],[3,10],[4,10],[5,50],[6,50],[7,10],[8,10],[9,10]];
Both arrays have unique keys, but the corresponding keys needs to be summed.
I tried loops, concat, etc but can't get the result i need.
anybody done this before?
You can use .reduce() to pass along an object that tracks the found sets, and does the addition.
DEMO: http://jsfiddle.net/aUXLV/
var array1 = [[5,10],[6,10],[7,10],[8,10],[9,10]];
var array2 = [[1,10],[2,10],[3,10],[4,10],[5,40],[6,40]];
var result =
array1.concat(array2)
.reduce(function(ob, ar) {
if (!(ar[0] in ob.nums)) {
ob.nums[ar[0]] = ar
ob.result.push(ar)
} else
ob.nums[ar[0]][1] += ar[1]
return ob
}, {nums:{}, result:[]}).result
If you need the result to be sorted, then add this to the end:
.sort(function(a,b) {
return a[0] - b[0];
})
This is one way to do it:
var sums = {}; // will keep a map of number => sum
// for each input array (insert as many as you like)
[array1, array2].forEach(function(array) {
//for each pair in that array
array.forEach(function(pair) {
// increase the appropriate sum
sums[pair[0]] = pair[1] + (sums[pair[0]] || 0);
});
});
// now transform the object sums back into an array of pairs
var results = [];
for(var key in sums) {
results.push([key, sums[key]]);
}
See it in action.
a short routine can be coded using [].map()
var array1 = [[5,10],[6,10],[7,10],[8,10],[9,10]];
var array2 = [[1,10],[2,10],[3,10],[4,10],[5,40],[6,40]];
array1=array2.concat(array1).map(function(a){
var v=this[a[0]]=this[a[0]]||[a[0]];
v[1]=(v[1]||0)+a[1];
return this;
},[])[0].slice(1);
alert(JSON.stringify(array1));
//shows: [[1,10],[2,10],[3,10],[4,10],[5,50],[6,50],[7,10],[8,10],[9,10]]
i like how it's just 3 line of code, doesn't need any internal function calls like push() or sort() or even an if() statement.
Try this:
var array1 = [[5,10],[6,10],[7,10],[8,10],[9,10]];
var array2 = [[1,10],[2,10],[3,10],[4,10],[5,40],[6,40]];
var res = [];
someReasonableName(array1, res);
someReasonableName(array2, res);
function someReasonableName(arr, res) {
var arrLen = arr.length
, i = 0
;
for(i; i < arrLen; i++) {
var ar = arr[i]
, index = ar[0]
, value = ar[1]
;
if(!res[index]) {
res[index] = [index, 0];
}
res[index][1] += value;
}
}
console.log(JSON.stringify(res, null, 2));
So, the result may have holes. Just like 0th index. Use the below function if you want to ensure there are no holes.
function compact(arr) {
var i = 0
, arrLen = arr.length
, res = []
;
for(i; i < arrLen; i++) {
var v = arr[i]
;
if(v) {
res[res.length] = v;
}
}
return res;
}
So, you can do:
var holesRemoved = compact(res);
And finally if you don't want the 0th elem of res. Do res.shift();
Disclaimer: I am not good with giving reasonable names.
The simple solution is like this.
function sumArrays(...arrays) {
const n = arrays.reduce((max, xs) => Math.max(max, xs.length), 0);
const result = Array.from({ length: n });
return result.map((_, i) => arrays.map(xs => xs[i] || 0).reduce((sum, x) => sum + x, 0));
}
console.log(...sumArrays([0, 1, 2], [1, 2, 3, 4], [1, 2])); // 2 5 5 4