I am trying to loop over an array argument and return the first n elements of the passed array without using standard javascript functions such as slice, concat, push, pop etc...
var n = 0;
var anyArray = Array;
var SR = {};
SR.first = function(anyArray,n){
var isArray = (Object.prototype.toString.apply(anyArray) === '[object Array]');
var specification = (typeof n === "number");
if(isArray && specification){
for(i = 0; i < n; i++){
return Array(anyArray[i]);
}
}
else if (isArray || !specification){
return anyArray[0];
}
}
I do not want to build the return array "anyArray" by using +=. So, how would I proceed to have it return some thing like this [1,2,3,4] when "SR.first([1,2,3,4,5,6,7], 4);" is called?
var newArr = Array.apply(null, anyArray); // new Array using original content
newArr.length = n; // truncate the length of the new Array
return newArr; // return it
One small edge case will be when anyArray has only one member, which is a number. You'll need to guard against that scenario.
I don't understand why you would not want to use Array operations like push and slice, but this would work:
if ( isArray && specification ) {
var result = [];
for ( var i = 0; i < n; i++ ) {
result[i] = anyArray[i];
}
return result;
}
else ...
If for some reason you really don't want to use native javascript functions, you can assign each element to your return array one by one.
var returnArray = [];
if(isArray && specification) {
for(i = 0; i < n; i++) {
returnArray[i] = anyArray[i];
}
}
return returnArray;
SR.first = function(anyArray,n){
var newArray = [];
for(i = 0; i < n; i++){
newArray[newArray.length] = anyArray[i];
}
return newArray;
}
First of all - global variable it's really bad practice! You don't need declare anyArray and n, becouse its a function arguments, and its declared on function call.
Second problem - that you can put number of elements bigger than array length - you must check this situation.
var SR = {};
SR.first = function(anyArray,n){
var isArray = (anyArray instanceof Array),
specification = (typeof n === 'number'),
tmp = new Array;
console.log(isArray, specification);
if(isArray && specification){
for(i = 0, l = anyArray.length; i < n && i < l; i++){
tmp[i] = anyArray[i];
}
return tmp;
} else if (isArray || !specification){
return anyArray[0];
}
}
Related
I want to count how often a number in an Array occurs. For example, in Python I can use Collections.Counter to create a dictionary of how frequently an item occurs in a list.
This is as far as I've gotten in JavaScript:
var array = [1,4,4,5,5,7];
var obj = {};
for (var i=0; i < array.length; i++) {
/* obj[array[i]] = +=1 */ <= pseudo code
}
How can I create this frequency counter object?
Close but you can't increment undefined so you need to set initial value if it doesn't exist
var array = [1,4,4,5,5,7];
var obj = {};
for (var i=0; i < array.length; i++) {
obj[array[i]] = (obj[array[i]] || 0) +1 ;
}
You were almost there. See below code:
var array = [1,4,4,5,5,7];
var obj = {};
for (var i=0; i < array.length; i++) {
obj[array[i]] = (obj[array[i]] || 0 ) +1;
}
console.log(obj);
Create an object and check if that specific key exist.If exist then increase it's value by 1
var array = [1, 4, 4, 5, 5, 7];
var obj = {};
for (var i = 0; i < array.length; i++) {
if (obj.hasOwnProperty(array[i])) {
obj[array[i]] += 1;
} else {
obj[array[i]] = 1;
}
}
console.log(obj)
You can use the ? : ternary operator to set initial value as 1 and then increment it on subsequent matches.
var array = [1,4,4,5,5,7];
var obj = {};
for (var i=0; i < array.length; i++) {
obj[array[i]] = obj[array[i]]?obj[array[i]]+1:1;
}
console.log(obj);
If the array is always going to be same, and you are going to check frequency of multiple items in the same array without it it being modified, #JohanP's answer is good.
But if you are only going to check frequency of only one item, or the array can change, creating the object is nothing but extra overhead.
In that case, you can do something like this:
const getItemFrequency = function(array, item) {
return array.filter(i => i === item).length;
}
var array = [1,4,4,5,5,7];
console.log(getItemFrequency(array, 4));
Concise logic written as proper function:
function countArrayItemFrequecy(array) {
const length = array.length;
const map = {};
for ( let i = 0; i < length; i++ ) {
let currentItem = array[i];
if (typeof map[currentItem] !== 'undefined' ) {
map[currentItem]++
} else {
map[currentItem] = 1
}
}
return map;
}
You need to make sure to assign default value to your frequency object for the first occurrence of the item. As a shortcut you can use ternary operator
var array = [1,4,4,5,5,7];
var obj = {};
for (var i=0; i < array.length; i++) {
obj[array[i]] = obj[array[i]] ? obj[array[i]]++ : 1;
}
which is the same as:
var array = [1,4,4,5,5,7];
var obj = {};
for (var i=0; i < array.length; i++) {
if (obj[array[i]]) {
obj[array[i]]++;
} else {
obj[array[i]] = 1;
}
}
You can use Object.assign: below clones map and then increments/adds the counter. These are pure (no side effects/param reassignment), single-purpose functions.
addToMap does the same thing as { ...map, map[e]: [e]: (map[e] || 0) + 1 }, but that requires babel.
const addToMap = (map, e) => Object.assign({}, map, { [e]: (map[e] || 0) + 1 });
const buildMap = a => a.reduce(addToMap, {});
Using Array.reduce:
arr.reduce(function (acc, item) {
acc[item] = (acc[item] || 0) + 1;
return acc;
}, {});
Example:
var arr = [1,1,2,4,1,4];
var counts = arr.reduce(function (acc, item) {
acc[item] = (acc[item] || 0) + 1;
return acc;
}, {});
console.log(counts);
hello I would like to build an array that every element will be a pair of objects , Something like this
var Shelves = new arr[][]
var books = new Books[] ;
Shelves[book[i],book[j=i+1]],[book[i+1],book[j=i+1]] and so on......;
I mean that I understand how to go with a for loop and to get the elements 'but how to push them in pairs array? arr.push doesn't work :(
build1ArrPairs(1Arr) {
if (1Arr != undefined || 1Arr!=null) {
for (var i = 0; i < 1Arr.length; i = i + 1) {
for (var j = i + 1; j <= 1Arr.length; j++) {
this.1ArrPair.push(1Arr[i] 1Arr[j]);
break;
}
}
}
}
Thanks :)
Alternatively, you can use array#reduce to group your array.
var names = ['a', 'b','c','d','e'];
var result = names.reduce((r,w,i) => {
let index = Math.floor(i/2);
if(!Array.isArray(r[index]))
r[index] = [];
r[index].push(w);
return r;
},[]);
console.log(result);
First of all, variables names can't start with a number.
Second, initialize an array, then add your elements - two at a time - and return the array:
var ans = [];
if (Arr != undefined || Arr != null) {
for (var i=0; i<(Arr.length-1); i+=2) { // Note the loop stops 2 elements before the last one
ans.push([Arr[i], Arr[i+1]]);
// ^ Note the comma - it's missing in your code
}
}
return ans;
Doing some jasmine testing with JS. So I have the testing function which seems reasonable(only running the first test currently)
describe('Anagram', function() {
it('no matches',function() {
var subject = new Anagram('diaper');
var matches = subject.matches([ 'hello', 'world', 'zombies', 'pants']);
expect(matches).toEqual([]);
});
Then I have my simple function
var Anagram = function(string){
this.word = string;
};
Anagram.prototype.matches = function(array){
var answer = [];
var splitWord = this.word.split('').sort();
for(var i = 0; i < array.length; i++){
var isAnagram = true;
var splitItem = array[i].split('').sort();
for(var j = 0; j < splitWord.length; j++){
if(splitWord[j] !== splitItem[j]){
isAnagram = false;
}
}
if(isAnagram === true){
answer.push(array[i]);
}
}
return answer;
};
module.export = Anagram;
My function is supposed to take a string and then look at an array of strings and return the anagram strings. I keep getting TypeError: Anagram is not a function. Tried looking for answers and most of it seemed to relate to semi colons which I think are not an issue since I use one after variable declaration and method declaration. I'd really just like to know what that typeError means if it's a common one to know or what the most likely reasons to get it are.
To check whether two strings are anagram or not, one can use this function. If the strings are anagram, then it will return true.
function isAnagram(str1,str2){
// str to alphabet array
let arr1 = Array.from(str1);
let arr2 = Array.from(str2);
//checking length
let n1 = arr1.length;
let n2 = arr2.length;
if (n1 !== n2){
return false;
}
//sorting array
arr1.sort();
arr2.sort();
for (let i=0; i<n1; i++ ){
if ( arr1[i] !== arr2[i]){
return false;
}
}
return true;
}
// Example: checking "listen and silent
console.log(isAnagram("listen", "silent"));
Let's say I'm writing a function like so:
function longestString (someArray) {
// code
}
If someArray = ['word','longer phrase',['a','b','c'],1234567891011121314151617], I would want the function to only return the longest string in the array and ignore the integers and other arrays that may also lie within it. I tried this:
function longestString (someArray) {
return someArray.sort(function (a, b) { return b.length - a.length; })[0];
}
It didn't work, and I am now stuck. :/
Filter only string element and do it
function longestString (someArray) {
return someArray
.filter(function(a){ return typeof(a)=='string' })
.sort(function (a, b) { return b.length - a.length; })[0];
}
Try this:
function longestString(someArray) {
var result = "";
for (var i = 0; i < someArray.length; i++) {
if ((typeof someArray[i] === "string") && (someArray[i].length > result.length)) {
result = someArray[i];
}
}
return result;
}
try to do :
function longestString (someArray) {
$longest_string = '';
foreach ($someArray as $value)
{
$current_length = strlen($longest_string);
if(strlen($value) > $current_length ) $longest_string = $value;
}
return $longest_string;
}
Try this.
var a = ['word', 'longer phrase', ['a', 'b', 'c'], 1234567891011121314151617];
returnLongestString(a);
function returnLongestString(arr) {
var longestString = '';
for (var i = 0; i < arr.length; i++) {
if (typeof arr[i] == "string" && arr[i].length > longestString.length) {
longestString = arr[i];
}
};
return longestString;
}
When I saw this question, it had 0 answers, when I finished writing and testing my function, there were five. But this is my approach:
The key here is to first filter only the strings. Then, list all the lengths on an array. Then using that array you can get a match on the bigger one.
function longestString (someArray) {
//we only need strings, so first we will filter all the data
var stringsOnly = [];
for (var i = 0; i < someArray.length; i++) {
if(typeof(someArray[i]) === 'string'){
stringsOnly.push(someArray[i]);
}
};
//Now with an array of just strings, we can get their indivial lenghts
var stringLengths = [];
for (var i = 0; i < stringsOnly.length; i++) {
var currentString = stringsOnly[i];
stringLengths.push(currentString.length);
};
//Get the max length
var maxLength = Math.max.apply(Math,stringLengths);
//get a string wich length equals to maxLength
for (var i = 0; i < stringsOnly.length; i++) {
var theString = stringsOnly[i];
if(theString.length === maxLength){return theString};
};
}
This function will return the largest string. If more than one string have the same length, It will return the first one. However, if you want to get various strings, you could make some little modificationson the function above:
function multipleLongestString (someArray) {
//we only need strings, so first we will filter all the data
var stringsOnly = [];
for (var i = 0; i < someArray.length; i++) {
if(typeof(someArray[i]) === 'string'){
stringsOnly.push(someArray[i]);
}
};
//Now with an array of just strings, we can get their indivial lenghts
var stringLengths = [];
for (var i = 0; i < stringsOnly.length; i++) {
var currentString = stringsOnly[i];
stringLengths.push(currentString.length);
};
//Get the max length
var maxLength = Math.max.apply(Math,stringLengths);
//modification here
longests = [];
//get a string wich length equals to maxLength
for (var i = 0; i < stringsOnly.length; i++) {
var theString = stringsOnly[i];
if(theString.length === maxLength){longests.push(theString)};
};
return longests;
}
That is an extra that may help you later. But if you just want the largest one, use the first function. I hope that my answer is relevant.
:)
Heres my approach at it. Really all you want to know is if the phrase is longer than the previous and if its a string.
jsFiddle here.
var someArray = ['word','longer phrase',['a','b','c'],1234567891011121314151617]
var longestString = function(arr) {
var longest = "";
for (var i = 0; i < arr.length; i++) {
var value = arr[i];
if (typeof value === "string") {
longest = arr[i];
}
}
alert(longest);
}
longestString(someArray);
Here is one to get your brain going! I've not had any luck with it.
[1,2,1,1,2,1,1,1,2,2]
[1,2,1,1,2,1]
I would like to use the second array to find the values in the first, but they must be in the same order.
Once for I would like it to return the next key up from the last key in the second array.
So in this example it would use the first six digits in the first array and then return 6 as the key after the final one in the second array.
var a2 = [1,2,1,1,2,1,1,1,2,2]
var a1 = [1,2,1,1,0,1]
function find(arr1, arr2) {
var len = 1
var result = 0;
var s2 = arr2.toString();
for (len=1;len <= a1.length; len++)
{
var aa1 = arr1.slice(0, len)
var s1 = aa1.toString();
if(s2.indexOf(s1)>=0){
result = aa1.length;
}
else {
break;
}
}
return result;
}
alert(find(a1, a2));
var find = function(haystack, needle) {
var doesMatch = function(offset) {
for (var i = 0; i < needle.length; i++) {
if (haystack[i+offset] !== needle[i]) {
return false;
}
}
return true;
};
for (var j=0; j < haystack.length - needle.length; j++) {
if (doesMatch(j)) {
return j;
}
}
return -1;
};
This is quick, this is dirty, and this is correct only if your data doesn't include any comma.
var needle = [1,2,1,1,2,1];
var haystack = [1,2,1,1,2,1,1,1,2,2];
if ( needle.length <= 0 ) return 0;
var fromStr = ','+haystack.toString()+','
var findStr = ','+needle.toString()+','
// Find ',1,2,1,1,2,1,' in ',1,2,1,1,2,1,1,1,2,2,'
var pos = fromStr.indexOf(findStr);
// Count the end position requested
return pos >= 0 ? fromStr.slice(0,pos+1).match(/,/g).length + needle.length - 1 : -1;
Note: The comma at head and tail is to make sure [22,12] doesn't match [2,1].