I am new to JavaScript and i will appreciate some help . I try to search array for element but i cant find the right solution . First i tried this , but no success.
var find = function(string, array) {
for(i=0;i>=array.length-1;i++){
if(array[i]==string){
return true;
}
else{
return false;
}
}
};
Then i tried this
var find = function(string, array) {
if(array.indexOf(string)>-1){
return true;}
else{
return false;
}
};
but it doesn't work with numbers
This are my tests
Test.assertEquals(find("hello", ["bye bye","hello"]), true);
Test.assertEquals(find("2", ["bye bye","2"]), true);
Test.assertEquals(find("2", ["bye bye",2]), false);
You are returning false the first time an element is found that doesn't match what you are looking for. You should only return false once the entire array has been processed. Your loop is also incorrect, if i is 0, it will never be greater than or equal to the array length unless the array is empty:
var find = function(string, array) {
for(i=0; i < array.length; i++) {
if(array[i]==string) {
return true;
}
}
return false;
};
You should also focus on naming conventions. Your function is called find, but it doesn't actually return the found element. I would name the function contains.
This is a good learning exercise, but don't reinvent the wheel:
[1,2,3,4,5,6].indexOf(foo) > -1
This will return true if foo is in the array.
Assuming you have a Test.assetEquals Function just use the build in array function:
Test.assertEquals((["bye bye","hello"].indexOf('hello') > -1),true);
With type conversion:
var a = [1, 2, 3];
if (a.some(function (x) { return x == "2" })) {
...
}
Strict comparison:
if (a.indexOf("2") >= 0) {
...
}
Related
I made a function to search words in a string given. The function called str_find receives the word and then the parameters to search. These parameters can be an array with strings or only a string.
For example, if I call the function like: str_find("123456789&", ["", "123456789&"]); it should give me true, but for a rare reason sometimes it gives me false.
Can you help me please?
My code:
function str_find(word, find){
if(!Array.isArray(find)){
if (word.indexOf(find) > 0) {
return true;
} else {
return false;
}
} else {
var flag = false;
for (var i = 0, length = find.length; i < length; i++) {
if (word.indexOf(find[i]) > 0) {
flag = true;
break;
}
}
return flag;
}
}
You can take advantage of both String and Array having an includes() prototype method
function str_find(needle, haystack) {
return haystack.includes(needle);
}
console.log(str_find("123456789&", ["", "123456789&"]))
console.log(str_find("foo", 'foobar'))
MDN points out the reason in an obscure way:
Return value
The value of the first element in the array that satisfies the
provided testing function. Otherwise, undefined is returned.
When the string is found at the very first position of the searched string, it returns a found location index of 0 - which is a falsy value:
Thus your logical test if (word.indexOf(find) > 0) returns false when this occurs and gives the false indication that the string was not found, when in fact it was found at the very beginning of the search string.
You can change this behavior by testing for -1 rather than > 0.
Alternatively, you can choose to use other Array methods that are less obscure.
const data = ["", "123456789&"];
const str = "123456789&";
const tests = ["4", "", "123456789&", ";", ["456", "123456789&"]];
function str_find(toFind, toSearch) {
let found = false;
if (!Array.isArray(toFind)) {
if (toSearch.indexOf(toFind) !== -1) {
found = true;
}
} else {
for (var i = 0; i < toFind.length; i++) {
found = str_find(toFind[i],toSearch);
}
}
return found;
}
tests.forEach(t=>console.log("test: ", t, str_find(t, data)));
this is my first question on SO. I have an assignment where I'm supposed to create a function that has two parameters--the first being an array of strings, and the second being a string to possibly match in the array.
I have two variations of the function: one uses a "for" loop, the other uses the .forEach() method. The for loop accurately returns "true"/"false" depending on whether or not the second parameter exists within the array. The forEach always returns "false".
Can someone explain why? Code below:
.forEach() version:
function insideArray (array, word) {
var value;
array.forEach(function(each) {
if(each === word) {
value = "true";
return value;
}
else {
value = "false";
}
});
return value;
}
for loop version:
function insideArray (array, word) {
var value;
for(var i = 0; i < array.length; i++) {
if(array[i] === word) {
value = "true";
return value;
}
else {
value = "false";
}
}
return value;
}
A sample array:
var heroArray = [ "spiderman", "wolverine", "batman", "greenArrow", "boosterGold" ];
Testing .forEach():
insideArray(heroArray, "spiderman");
"false"
Testing for loop:
insideArray(heroArray, "spiderman");
"true"
Thanks for any help in advance!
This happens because you are returning from callback function of forEach and not from insideArray() so in the end insideArray() will always return false except in the case you match the last element in the array. you can fix the code by initializing value with false and removing else condition like this:
function insideArray (array, word) {
var value = "false";
array.forEach(function(each) {
if(each === word) {
value = "true";
}
});
return value;
}
Also note that you can write more simple code for this by using indexOf:
function insideArray (array, word) {
return array.indexOf(word) >= 0 ? "true" : "false";
}
The .forEach method doesn't return anything, so I'd modify your code to the following:
function insideArray (array, word) {
var value = "false";
array.forEach(function(each) {
if(each === word) {
value = "true";
}
});
return value
};
You can't break/return out of a Array#forEach loop, though you could throw and catch from it (but don't). Array#some would probably be a better choice, if you intend to test like that. A better, though slightly less well supported, method would be to use Array#includes (polyfills are available, or transpile with babel). Another solution would be Array#indexof
function insideArray(array, word) {
return String(array.some(function(each) {
return each === word;
}));
}
var heroArray = ['spiderman', 'wolverine', 'batman', 'greenArrow', 'boosterGold'];
console.log(insideArray(heroArray, 'supergirl'));
console.log(insideArray(heroArray, 'batman'));
const insideArray = (array, word) => String(array.includes(word));
const heroArray = ['spiderman', 'wolverine', 'batman', 'greenArrow', 'boosterGold'];
console.log(insideArray(heroArray, 'supergirl'));
console.log(insideArray(heroArray, 'batman'));
I found a solution to this homework question, but I dont feel its the most efficient way to tackle the problem. Interested in other solutions I should explore.
Question:
Write a function named allEqual that returns true if every character in the string is the same
Example:
If you pass "aaa" it should return true
If you pass "aba" it should return false
*/
My Code
var stringAE = "aba";
function allEqual(string) {
var stringAENew = "";
for (var i = 0; i < string.length; i++) {
if (string[0] === string[i]) {
stringAENew += string[i];
console.log(stringAENew)
}
}
return stringAENew === string;
}
allEqual(stringAE)
Simple solution using .every().
function allEqual(input) {
return input.split('').every(char => char === input[0]);
}
console.log(allEqual('aba')); // false
console.log(allEqual('aaa')); // true
console.log(allEqual('')); // true
You can return false immediately once you find a character that doesn't match the first character. If you make it through the whole loop, return true because all the characters must have matched.
function allEqual(string) {
for (var i = 1; i < string.length; i++) {
if (string[i] != string[0]) {
return false;
}
}
return true;
}
You can also start your loop at i = 1, since the first character is obviously equal to itself, so there's no need to test it.
Can be done with regex too
function allEqual(str) {
return /^(.)\1*$/.test(str);
}
Although probably not so effective.
This ES6 solution also works for strings with Unicode code points in other than the first plane, i.e. with codes outside of the 16 bit range:
function allEqual(string) {
return [...string].every( (x, _, a) => x === a[0]);
}
console.log(allEqual('aaaa')); // true
console.log(allEqual('aaaba')); // false
// Next one fails in solutions that don't support multi-plane unicode:
console.log(allEqual('𝌆𝌆𝌆')); // true
console.log(allEqual('')); // true
There's no reason to construct a result string. Just go over all the characters and compare them to the first one (as you've been doing). If you found a different character, the result is false. If you've gone over all the characters and haven't found a different one, the answer is true (note that this includes the edge cases of an empty string and a single character string):
function allEqual(string) {
for (var i = 1; i < string.length; i++) {
if (string[0] !== string[i]) {
return false;
}
}
return true;
}
I'm a little late for the party, but as I needed to do this on a project, I came up with another approach:
function allEqual(input) {
return input === '' || new Set(input).size === 1;
}
console.log(['', 'aaa', '11', '####', 'aba', '12', '###%', null, undefined].map(item => ({
item,
allEqual: allEqual(item),
})));
Possible Duplicate:
Easiest way to find duplicate values in a javascript array
How do I check if an array has duplicate values?
If some elements in the array are the same, then return true. Otherwise, return false.
['hello','goodbye','hey'] //return false because no duplicates exist
['hello','goodbye','hello'] // return true because duplicates exist
Notice I don't care about finding the duplication, only want Boolean result whether arrays contains duplications.
If you have an ES2015 environment (as of this writing: io.js, IE11, Chrome, Firefox, WebKit nightly), then the following will work, and will be fast (viz. O(n)):
function hasDuplicates(array) {
return (new Set(array)).size !== array.length;
}
If you only need string values in the array, the following will work:
function hasDuplicates(array) {
var valuesSoFar = Object.create(null);
for (var i = 0; i < array.length; ++i) {
var value = array[i];
if (value in valuesSoFar) {
return true;
}
valuesSoFar[value] = true;
}
return false;
}
We use a "hash table" valuesSoFar whose keys are the values we've seen in the array so far. We do a lookup using in to see if that value has been spotted already; if so, we bail out of the loop and return true.
If you need a function that works for more than just string values, the following will work, but isn't as performant; it's O(n2) instead of O(n).
function hasDuplicates(array) {
var valuesSoFar = [];
for (var i = 0; i < array.length; ++i) {
var value = array[i];
if (valuesSoFar.indexOf(value) !== -1) {
return true;
}
valuesSoFar.push(value);
}
return false;
}
The difference is simply that we use an array instead of a hash table for valuesSoFar, since JavaScript "hash tables" (i.e. objects) only have string keys. This means we lose the O(1) lookup time of in, instead getting an O(n) lookup time of indexOf.
You could use SET to remove duplicates and compare, If you copy the array into a set it will remove any duplicates. Then simply compare the length of the array to the size of the set.
function hasDuplicates(a) {
const noDups = new Set(a);
return a.length !== noDups.size;
}
One line solutions with ES6
const arr1 = ['hello','goodbye','hey']
const arr2 = ['hello','goodbye','hello']
const hasDuplicates = (arr) => arr.length !== new Set(arr).size;
console.log(hasDuplicates(arr1)) //return false because no duplicates exist
console.log(hasDuplicates(arr2)) //return true because duplicates exist
const s1 = ['hello','goodbye','hey'].some((e, i, arr) => arr.indexOf(e) !== i)
const s2 = ['hello','goodbye','hello'].some((e, i, arr) => arr.indexOf(e) !== i);
console.log(s1) //return false because no duplicates exist
console.log(s2) //return true because duplicates exist
Another approach (also for object/array elements within the array1) could be2:
function chkDuplicates(arr,justCheck){
var len = arr.length, tmp = {}, arrtmp = arr.slice(), dupes = [];
arrtmp.sort();
while(len--){
var val = arrtmp[len];
if (/nul|nan|infini/i.test(String(val))){
val = String(val);
}
if (tmp[JSON.stringify(val)]){
if (justCheck) {return true;}
dupes.push(val);
}
tmp[JSON.stringify(val)] = true;
}
return justCheck ? false : dupes.length ? dupes : null;
}
//usages
chkDuplicates([1,2,3,4,5],true); //=> false
chkDuplicates([1,2,3,4,5,9,10,5,1,2],true); //=> true
chkDuplicates([{a:1,b:2},1,2,3,4,{a:1,b:2},[1,2,3]],true); //=> true
chkDuplicates([null,1,2,3,4,{a:1,b:2},NaN],true); //=> false
chkDuplicates([1,2,3,4,5,1,2]); //=> [1,2]
chkDuplicates([1,2,3,4,5]); //=> null
See also...
1 needs a browser that supports JSON, or a JSON library if not.
2 edit: function can now be used for simple check or to return an array of duplicate values
You can take benefit of indexOf and lastIndexOf. if both indexes are not same, you have duplicate.
function containsDuplicates(a) {
for (let i = 0; i < a.length; i++) {
if (a.indexOf(a[i]) !== a.lastIndexOf(a[i])) {
return true
}
}
return false
}
If you are dealing with simple values, you can use array.some() and indexOf()
for example let's say vals is ["b", "a", "a", "c"]
const allUnique = !vals.some((v, i) => vals.indexOf(v) < i);
some() will return true if any expression returns true. Here we'll iterate values (from the index 0) and call the indexOf() that will return the index of the first occurrence of given item (or -1 if not in the array). If its id is smaller that the current one, there must be at least one same value before it. thus iteration 3 will return true as "a" (at index 2) is first found at index 1.
is just simple, you can use the Array.prototype.every function
function isUnique(arr) {
const isAllUniqueItems = input.every((value, index, arr) => {
return arr.indexOf(value) === index; //check if any duplicate value is in other index
});
return isAllUniqueItems;
}
One nice thing about solutions that use Set is O(1) performance on looking up existing items in a list, rather than having to loop back over it.
One nice thing about solutions that use Some is short-circuiting when the duplicate is found early, so you don't have to continue evaluating the rest of the array when the condition is already met.
One solution that combines both is to incrementally build a set, early terminate if the current element exists in the set, otherwise add it and move on to the next element.
const hasDuplicates = (arr) => {
let set = new Set()
return arr.some(el => {
if (set.has(el)) return true
set.add(el)
})
}
hasDuplicates(["a","b","b"]) // true
hasDuplicates(["a","b","c"]) // false
According to JSBench.me, should preform pretty well for the varried use cases. The set size approach is fastest with no dupes, and checking some + indexOf is fatest with a very early dupe, but this solution performs well in both scenarios, making it a good all-around implementation.
function hasAllUniqueChars( s ){
for(let c=0; c<s.length; c++){
for(let d=c+1; d<s.length; d++){
if((s[c]==s[d])){
return false;
}
}
}
return true;
}
I need to check if array contains at least one empty elements. If any of the one element is empty then it will return false.
Example:
var my_arr = new Array();
my_arr[0] = "";
my_arr[1] = " hi ";
my_arr[2] = "";
The 0th and 2nd array elements are "empty".
You can check by looping through the array with a simple for, like this:
function NoneEmpty(arr) {
for(var i=0; i<arr.length; i++) {
if(arr[i] === "") return false;
}
return true;
}
You can give it a try here, the reason we're not using .indexOf() here is lack of support in IE, otherwise it'd be even simpler like this:
function NoneEmpty(arr) {
return arr.indexOf("") === -1;
}
But alas, IE doesn't support this function on arrays, at least not yet.
You have to check in through loop.
function checkArray(my_arr){
for(var i=0;i<my_arr.length;i++){
if(my_arr[i] === "")
return false;
}
return true;
}
You can try jQuery.inArray() function:
return jQuery.inArray("", my_arr)
Using a "higher order function" like filter instead of looping can sometimes make for faster, safer, and more readable code. Here, you could filter the array to remove items that are not the empty string, then check the length of the resultant array.
Basic JavaScript
var my_arr = ["", "hi", ""]
// only keep items that are the empty string
new_arr = my_arr.filter(function(item) {
return item === ""
})
// if filtered array is not empty, there are empty strings
console.log(new_arr);
console.log(new_arr.length === 0);
Modern Javascript: One-liner
var my_arr = ["", "hi", ""]
var result = my_arr.filter(item => item === "").length === 0
console.log(result);
A note about performance
Looping is likely faster in this case, since you can stop looping as soon as you find an empty string. I might still choose to use filter for code succinctness and readability, but either strategy is defensible.
If you needed to loop over all the elements in the array, however-- perhaps to check if every item is the empty string-- filter would likely be much faster than a for loop!
Nowadays we can use Array.includes
my_arr.includes("")
Returns a Boolean
You could do a simple help method for this:
function hasEmptyValues(ary) {
var l = ary.length,
i = 0;
for (i = 0; i < l; i += 1) {
if (!ary[i]) {
return false;
}
}
return true;
}
//check for empty
var isEmpty = hasEmptyValues(myArray);
EDIT: This checks for false, undefined, NaN, null, "" and 0.
EDIT2: Misread the true/false expectation.
..fredrik
function containsEmpty(a) {
return [].concat(a).sort().reverse().pop() === "";
}
alert(containsEmpty(['1','','qwerty','100'])); // true
alert(containsEmpty(['1','2','qwerty','100'])); // false
my_arr.includes("")
This returned undefined instead of a boolean value so here's an alternative.
function checkEmptyString(item){
if (item.trim().length > 0) return false;
else return true;
};
function checkIfArrayContainsEmptyString(array) {
const containsEmptyString = array.some(checkEmptyString);
return containsEmptyString;
};
console.log(checkIfArrayContainsEmptyString(["","hey","","this","is","my","solution"]))
// *returns true*
console.log(checkIfArrayContainsEmptyString(["yay","it","works"]))
// *returns false*
yourArray.join('').length > 0
Join your array without any space in between and check for its length. If the length, turns out to be greater than zero that means array was not empty. If length is less than or equal to zero, then array was empty.
I see in your comments beneath the question that the code example you give is PHP, so I was wondering if you were actually going for the PHP one? In PHP it would be:
function hasEmpty($array)
{
foreach($array as $bit)
{
if(empty($bit)) return true;
}
return false;
}
Otherwise if you actually did need JavaScript, I refer to Nick Craver's answer
Just do a len(my_arr[i]) == 0; inside a loop to check if string is empty or not.
var containsEmpty = !my_arr.some(function(e){return (!e || 0 === e.length);});
This checks for 0, false, undefined, "" and NaN.
It's also a one liner and works for IE 9 and greater.
One line solution to check if string have empty element
let emptyStrings = strArray.filter(str => str.trim().length <= 0);
let strArray = ['str1', '', 'str2', ' ', 'str3', ' ']
let emptyStrings = strArray.filter(str => str.trim().length <= 0);
console.log(emptyStrings)
One line solution to get non-empty strings from an array
let nonEmptyStrings = strArray.filter(str => str.trim().length > 0);
let strArray = ['str1', '', 'str2', ' ', 'str3', ' ']
let nonEmptyStrings = strArray.filter(str => str.trim().length > 0);
console.log(nonEmptyStrings)
If you only care about empty strings then this will do it:
const arr = ["hi","hello","","jj"]
('' in arr) //returns false
the last line checks if an empty string was found in the array.
I don't know if this is the most performant way, but here's a one liner in ES2015+:
// true if not empty strings
// false if there are empty strings
my_arr.filter(x => x).length === my_arr.length
The .filter(x => x) will return all the elements of the array that are not empty nor undefined. You then compare the length of the original array. If they are different, that means that the array contains empty strings.
You have to check in through the array of some functions.
if isEmptyValue is true that means the array has an empty string otherwise not.
const arr=['A','B','','D'];
const isEmptyValue = arr.some(item => item.trim() === '');
console.log(isEmptyValue)
array.includes("") works just fine.
Let a = ["content1", "" , "content2"];
console.log(a.includes(""));
//Output in console
true