The emphasis here is on the word exactly. This needs to work for any number of permutations, so hopefully my example is clear enough.
Given a string of random letters, is it possible (using RegEx) to match an exact number of letters within the given string?
So if I have a string (str1) containing letters ABZBABJDCDAZ and I wanted to match the letters JDBBAA (str2), my function should return true because str1 contains all the right letters enough times. If however str1 were to be changed to ABAJDCDA, then the function would return false as str2 requires that str1 have at least 2 instances of the letter B.
This is what I have so far using a range:
const findLetters = (str1, str2) => {
const regex = new RegExp(`[${str2}]`, 'g')
const result = (str1.match(regex))
console.log(result)
}
findLetters('ABZBABJDCDAZ', 'JDBBAA')
As you can see it matches the right letters, but it matches all instances of them. Is there any way to do what I'm trying to do using RegEx? The reason I'm focusing on RegEx here is because I need this code to be highly optimised, and so far my other functions using Array.every() and indexOf() are just too slow.
Note: My function only requires to return a true/false value.
Try (here we sort letters of both strings and then create regexp like A.*A.*B.*B.*D.*J)
const findLetters = (str1, str2) => {
const regex = new RegExp([...str2].sort().join`.*`)
return regex.test([...str1].sort().join``)
}
console.log( findLetters('ABZBABJDCDAZ', 'JDBBAA') );
console.log( findLetters('ABAJDCDA', 'JDBBAA') );
I dont know if regex is the right way for this as this can also get very expensive. Regex is fast, but not always the fastest.
const findLetters2 = (strSearchIn, strSearchFor) => {
var strSearchInSorted = strSearchIn.split('').sort(function(a, b) {
return a.localeCompare(b);
});
var strSearchForSorted = strSearchFor.split('').sort(function(a, b) {
return a.localeCompare(b);
});
return hasAllChars(strSearchInSorted, strSearchForSorted);
}
const hasAllChars = (searchInCharList, searchCharList) => {
var counter = 0;
for (i = 0; i < searchCharList.length; i++) {
var found = false;
for (counter; counter < searchInCharList.length;) {
counter++;
if (searchCharList[i] == searchInCharList[counter - 1]) {
found = true;
break;
}
}
if (found == false) return false;
}
return true;
}
// No-Regex solution
console.log('true: ' + findLetters2('abcABC', 'abcABC'));
console.log('true: ' + findLetters2('abcABC', 'acbACB'));
console.log('true: ' + findLetters2('abcABCx', 'acbACB'));
console.log('false: ' + findLetters2('abcABC', 'acbACBx'));
console.log('true: ' + findLetters2('ahfffmbbbertwcAtzrBCasdf', 'acbACB'));
console.log('false: ' + findLetters2('abcABC', 'acbAACB'));
Feel free to test it's speed and to optimize it as I'm no js expert. This solution should iterate each string once after sorting. Sorting is thanks to https://stackoverflow.com/a/51169/9338645.
Related
I have a string like "this/ is an example abc/def/fgh/uio to give you an example"
I'd like to target the longest word and replace on this substring any "/" by a "+".
I manage to identify the longest word and I would know how to replace ALL "/" by a "+" BUT I don't know how to replace the "/" only in the longest word.
Here's what I've got so far
//identify longest word in string
function longestWord(str) {
var words = str.split(' ');
return words.reduce(longer);
}
function longer(champ, contender) {
return (contender.length > champ.length) ? contender: champ;
}
//purely given an exemple, some strigns won't be exactly like this
var text2 = "this/ is an example abc/def/fgh/uio to give you an example"
if (longestWord(text2) > 30 ) {
text2.replace(/\//g, ' / ');
}
The problem is this will also replace the "/" on the substring "this/", and I don't want that.
How to achieve this?
Your longestWord function returns the longest word in the string, so you can pass that string alone (not a regular expression) as the first argument to .replace, and replace with (the second argument) the /\//g called on that longest word:
function getLongestWord(str) {
var words = str.split(' ');
return words.reduce(longer);
}
function longer(champ, contender) {
return (contender.length > champ.length) ? contender: champ;
}
var text2 = "this/ is an example abc/def/fgh/uio to give you an example"
const longestWord = getLongestWord(text2);
const output = text2.replace(longestWord, longestWord.replace(/\//g, '+'));
console.log(output);
#CertainPermance's solution is far more elegant (and I think performant) than this, but as I'd written the answer I thought I may as well put it in.
It's fairly similar, in truth, though in this instance we get the index of the word and use that to perform the replace, which at the time of writing I thought was necessary. Now looking at the better solution, I realise such a check is not needed, as the longest word in a string will not feature in any other words, so it's easy and safe to simply perform a replace on it.
const data = "this/ is an example abc/def/fgh/uio to give you an example";
const getLongestWordIndex = stringIn => stringIn
.split(' ')
.reduce(
(prev, curr, i) => curr.length > prev.length ? {
index: i,
length: curr.length
} : prev,
{
length: -1,
index: -1
}
).index
const replaceLongestWord = (sentence, replacer) => {
const longestWordIndex = getLongestWordIndex(sentence);
const words = data.split(' ');
return Object.values({
...words,
[longestWordIndex]: replacer(words[longestWordIndex])
}).join(' ')
}
const wordReplaceFunction = word => word.replace(/\//g, '+')
const result = replaceLongestWord(data, wordReplaceFunction);
console.dir(result)
I have written a code that removes all consonants before a vowel from a string and replaces it with an 'r' and in the case, the string starts with a vowel it should return the word without doing anything to it. Now, I want to add two things I came up with to it but unfortunately, I have not been able to:
1. When the string input is all consonants then it should do nothing and just return the string.
2. If user types in space like so ' ' then it should be trimmed.
How do I place this logic in the code below without affecting what is already working?
const scoobyDoo = str => {
if(typeof str !== 'string'){
return 'This function accepts strings only';
}
let newStr = str.toLowerCase().split('');
let arrWord = newStr.length;
let regex = /[aeiou]/gi;
if (newStr[0].match(regex)){
let nothing = newStr.join('');
return nothing;
}
else {
for (let i = 0; i < arrWord; i++){
let vowelIndex = newStr.indexOf(str.match(regex)[i].toLowerCase());
newStr.splice(0, vowelIndex, 'r');
return newStr.join('');
}
}
}
console.log(scoobyDoo('scooby'));//works as expected returns 'rooby'
console.log(scoobyDoo('ethane'));//works as expected returns 'ethane'
console.log(scoobyDoo('why'));// should return 'why'
console.log(scoobyDoo(' '));// should return trimmed space and a
text telling the user only spaces were entered.
I realise this doesn't really answer your question, but your existing logic is very complicated and you could achieve the same result with String.trim, .toLowerCase and .replace:
console.log('scooby'.trim().toLowerCase().replace(/^(?=.*?[aeiou])[^aeiou]+/, 'r'))
rooby
console.log('ethane'.trim().toLowerCase().replace(/^(?=.*?[aeiou])[^aeiou]+/, 'r'))
ethane
console.log('why'.trim().toLowerCase().replace(/^(?=.*?[aeiou])[^aeiou]+/, 'r'))
why
console.log('*' + ' '.trim().toLowerCase().replace(/^(?=.*?[aeiou])[^aeiou]+/, 'r') + '*')
**
The regexp uses a positive lookahead to ensure that there is a vowel in the string, and if so replaces all leading consonants with an r.
To do something more in line with your existing function, you could try this. It still makes extensive use of regex functions though.
const scoobyDoo = str => {
if(typeof str !== 'string'){
return 'This function accepts strings only';
}
// is it a blank string?
if (str.match(/^\s+$/)) {
return '';
}
// does it start with a vowel? if so, nothing to do
if (str.match(/^[aeiou]/i)) {
return str;
}
// does it only contain consonants?
if (!str.match(/[aeiou]/i)) {
return str;
}
// must not start with a vowel but still include one
return str.replace(/^[^aeiou]+/i, 'r');
}
I'm aware of the CSS attribute text-transform: capitalize but can anyone help me with replicating this using Javascript?
I would like to pass an argument to my function which will return the string with the first letter of each word capitalized.
I've got this far but I'm stuck trying to break my array of strings in to chunks:
function upper(x){
x = x.split(" ");
// this function should return chunks but when called I'm getting undefined
Array.prototype.chunk = function ( n ) {
return [ this.slice( 0, n ) ].concat( this.slice(n).chunk(n) );
};
x = x.chunk;
}
upper("chimpanzees like cigars")
after the chunk I'm guessing I need to again split each chunk in to the first character and the remaining characters, use .toUpperCase() on the first character, join it back up with the remaining and then join up the chunks again in to a string?
Is there a simpler method for doing this?
I came up with a solution for both a single word and also for an array of words. It will also ensure that all other letters are lowercase for good measure. I used the Airbnb style guide as well. I hope this helps!
const mixedArr = ['foo', 'bAr', 'Bas', 'toTESmaGoaTs'];
const word = 'taMpa';
function capitalizeOne(str) {
return str.charAt(0).toUpperCase().concat(str.slice(1).toLowerCase());
}
function capitalizeMany(args) {
return args.map(e => {
return e.charAt(0).toUpperCase().concat(e.slice(1).toLowerCase());
});
};
const cappedSingle = capitalizeOne(word);
const cappedMany = capitalizeMany(mixedArr);
console.log(cappedSingle);
console.log(cappedMany);
The map function is perfect for this.
w[0].toUpperCase() : Use this to capitalize the first letter of each word
w.slice(1): Return the string from the second character on
EDGE Case
If the user doesn't enter a string, the map function will not work and an error will be raised. This can be guarded against by checking if the user actually entered something.
var userInput = prompt("Enter a string");
var capitalizedString = userInput == "" ? "Invalid String" :
userInput.split(/\s+/).map(w => w[0].toUpperCase() + w.slice(1)).join(' ');
console.log(capitalizedString);
You can use the following solution which doesn't use regex.
function capitalize(str=''){
return str.trim().split('')
.map((char,i) => i === 0 ? char.toUpperCase() : char )
.reduce((final,char)=> final += char, '' )
}
capitalize(' hello') // Hello
"abcd efg ijk lmn".replace(/\b(.)/g, (m => m.toUpperCase())) // Abcd Efg Ijk Lmn
You may want to try a regex approach:
function upperCaseFirst(value) {
var regex = /(\b[a-z](?!\s))/g;
return value ? value.replace(regex, function (v) {
return v.toUpperCase();
}) : '';
}
This will grab the first letter of every word on a sentence and capitalize it, but if you only want the first letter of the sentence, you can just remove the g modifier at the end of the regex declaration.
or you could just iterate the string and do the job:
function capitalize(lowerStr){
var result = "";
var isSpacePrevious = false;
for (var i=0; i<lowerStr.length; i++){
if (i== 0 || isSpacePrevious){
result += lowerStr[i].toUpperCase();
isSpacePrevious = false;
continue;
}
if (lowerStr[i] === ' '){
isSpacePrevious = true;
}
result += lowerStr[i];
}
return result;
}
What is the best way to bold a part of string in Javascript?
I have an array of objects. Each object has a name. There is also an input parameter.
If, for example, you write "sa" in input, it automatically searches in array looking for objects with names that contain "sa" string.
When I print all the names, I want to bold the part of the name that coincide with the input text.
For example, if I search for "Ma":
Maria
Amaria
etc...
I need a solution that doesn't use jQuery. Help is appreciated.
PD: The final strings are in the tag. I create a list using angular ng-repeat.
This is the code:
$scope.users = data;
for (var i = data.length - 1; i >= 0; i--) {
data[i].name=data[i].name.replace($scope.modelCiudad,"<b>"+$scope.modelCiudad+"</b>");
};
ModelCiudad is the input text content var. And data is the array of objects.
In this code if for example ModelCiudad is "ma" the result of each is:
<b>Ma</b>ria
not Maria
You can use Javascript's str.replace() method, where str is equal to all of the text you want to search through.
var str = "Hello";
var substr = "el";
str.replace(substr, '<b>' + substr + '</b>');
The above will only replace the first instance of substr. If you want to handle replacing multiple substrings within a string, you have to use a regular expression with the g modifier.
function boldString(str, substr) {
var strRegExp = new RegExp(substr, 'g');
return str.replace(strRegExp, '<b>'+substr+'</b>');
}
In practice calling boldString would looks something like:
boldString("Hello, can you help me?", "el");
// Returns: H<b>el</b>lo can you h<b>el</b>p me?
Which when rendered by the browser will look something like: Hello can you help me?
Here is a JSFiddle with an example: https://jsfiddle.net/1rennp8r/3/
A concise ES6 solution could look something like this:
const boldString = (str, substr) => str.replace(RegExp(substr, 'g'), `<b>${substr}</b>`);
Where str is the string you want to modify, and substr is the substring to bold.
ES12 introduces a new string method str.replaceAll() which obviates the need for regex if replacing all occurrences at once. It's usage in this case would look something like this:
const boldString = (str, substr) => str.replaceAll(substr, `<b>${substr}</b>`);
I should mention that in order for these latter approaches to work, your environment must support ES6/ES12 (or use a tool like Babel to transpile).
Another important note is that all of these approaches are case sensitive.
Here's a pure JS solution that preserves the original case (ignoring the case of the query thus):
const boldQuery = (str, query) => {
const n = str.toUpperCase();
const q = query.toUpperCase();
const x = n.indexOf(q);
if (!q || x === -1) {
return str; // bail early
}
const l = q.length;
return str.substr(0, x) + '<b>' + str.substr(x, l) + '</b>' + str.substr(x + l);
}
Test:
boldQuery('Maria', 'mar'); // "<b>Mar</b>ia"
boldQuery('Almaria', 'Mar'); // "Al<b>mar</b>ia"
I ran into a similar problem today - except I wanted to match whole words and not substrings. so if const text = 'The quick brown foxes jumped' and const word = 'foxes' than I want the result to be 'The quick brown <strong>foxes</strong> jumped'; however if const word = 'fox', than I expect no change.
I ended up doing something similar to the following:
const pattern = `(\\s|\\b)(${word})(\\s|\\b)`;
const regexp = new RegExp(pattern, 'ig'); // ignore case (optional) and match all
const replaceMask = `$1<strong>$2</strong>$3`;
return text.replace(regexp, replaceMask);
First I get the exact word which is either before/after some whitespace or a word boundary, and then I replace it with the same whitespace (if any) and word, except the word is wrapped in a <strong> tag.
Here is a version I came up with if you want to style words or individual characters at their index in react/javascript.
replaceAt( yourArrayOfIndexes, yourString/orArrayOfStrings )
Working example: https://codesandbox.io/s/ov7zxp9mjq
function replaceAt(indexArray, [...string]) {
const replaceValue = i => string[i] = <b>{string[i]}</b>;
indexArray.forEach(replaceValue);
return string;
}
And here is another alternate method
function replaceAt(indexArray, [...string]) {
const startTag = '<b>';
const endTag = '</b>';
const tagLetter = i => string.splice(i, 1, startTag + string[i] + endTag);
indexArray.forEach(tagLetter);
return string.join('');
}
And another...
function replaceAt(indexArray, [...string]) {
for (let i = 0; i < indexArray.length; i++) {
string = Object.assign(string, {
[indexArray[i]]: <b>{string[indexArray[i]]}</b>
});
}
return string;
}
Above solutions are great, but are limited! Imagine a test scenerio where you want to match case insensitive query in a string and they could be multiple matches.
For example
Query: ma
String: The Amazing Spiderman
Expected Result: The Amazing Spiderman
For above scenerio, use this:
const boldMatchText = (text,searchInput) => {
let str = text.toLowerCase();
const query = searchInput.toLowerCase();
let result = "";
let queryLoc = str.indexOf(query);
if (queryLoc === -1) {
result += text;
} else
do {
result += ` ${text.substr(0, queryLoc)}
<b>${text.substr(queryLoc, query.length)}</b>`;
str = str.substr(queryLoc + query.length, str.length);
text = text.substr(queryLoc + query.length, str.length);
queryLoc = str.indexOf(query);
} while (text.length > 0 && queryLoc !== -1);
return result + text;
};
Given a string
'1.2.3.4.5'
I would like to get this output
'1.2345'
(In case there are no dots in the string, the string should be returned unchanged.)
I wrote this
function process( input ) {
var index = input.indexOf( '.' );
if ( index > -1 ) {
input = input.substr( 0, index + 1 ) +
input.slice( index ).replace( /\./g, '' );
}
return input;
}
Live demo: http://jsfiddle.net/EDTNK/1/
It works but I was hoping for a slightly more elegant solution...
There is a pretty short solution (assuming input is your string):
var output = input.split('.');
output = output.shift() + '.' + output.join('');
If input is "1.2.3.4", then output will be equal to "1.234".
See this jsfiddle for a proof. Of course you can enclose it in a function, if you find it necessary.
EDIT:
Taking into account your additional requirement (to not modify the output if there is no dot found), the solution could look like this:
var output = input.split('.');
output = output.shift() + (output.length ? '.' + output.join('') : '');
which will leave eg. "1234" (no dot found) unchanged. See this jsfiddle for updated code.
It would be a lot easier with reg exp if browsers supported look behinds.
One way with a regular expression:
function process( str ) {
return str.replace( /^([^.]*\.)(.*)$/, function ( a, b, c ) {
return b + c.replace( /\./g, '' );
});
}
You can try something like this:
str = str.replace(/\./,"#").replace(/\./g,"").replace(/#/,".");
But you have to be sure that the character # is not used in the string; or replace it accordingly.
Or this, without the above limitation:
str = str.replace(/^(.*?\.)(.*)$/, function($0, $1, $2) {
return $1 + $2.replace(/\./g,"");
});
You could also do something like this, i also don't know if this is "simpler", but it uses just indexOf, replace and substr.
var str = "7.8.9.2.3";
var strBak = str;
var firstDot = str.indexOf(".");
str = str.replace(/\./g,"");
str = str.substr(0,firstDot)+"."+str.substr(1,str.length-1);
document.write(str);
Shai.
Here is another approach:
function process(input) {
var n = 0;
return input.replace(/\./g, function() { return n++ > 0 ? '' : '.'; });
}
But one could say that this is based on side effects and therefore not really elegant.
This isn't necessarily more elegant, but it's another way to skin the cat:
var process = function (input) {
var output = input;
if (typeof input === 'string' && input !== '') {
input = input.split('.');
if (input.length > 1) {
output = [input.shift(), input.join('')].join('.');
}
}
return output;
};
Not sure what is supposed to happen if "." is the first character, I'd check for -1 in indexOf, also if you use substr once might as well use it twice.
if ( index != -1 ) {
input = input.substr( 0, index + 1 ) + input.substr(index + 1).replace( /\./g, '' );
}
var i = s.indexOf(".");
var result = s.substr(0, i+1) + s.substr(i+1).replace(/\./g, "");
Somewhat tricky. Works using the fact that indexOf returns -1 if the item is not found.
Trying to keep this as short and readable as possible, you can do the following:
JavaScript
var match = string.match(/^[^.]*\.|[^.]+/g);
string = match ? match.join('') : string;
Requires a second line of code, because if match() returns null, we'll get an exception trying to call join() on null. (Improvements welcome.)
Objective-J / Cappuccino (superset of JavaScript)
string = [string.match(/^[^.]*\.|[^.]+/g) componentsJoinedByString:''] || string;
Can do it in a single line, because its selectors (such as componentsJoinedByString:) simply return null when sent to a null value, rather than throwing an exception.
As for the regular expression, I'm matching all substrings consisting of either (a) the start of the string + any potential number of non-dot characters + a dot, or (b) any existing number of non-dot characters. When we join all matches back together, we have essentially removed any dot except the first.
var input = '14.1.2';
reversed = input.split("").reverse().join("");
reversed = reversed.replace(\.(?=.*\.), '' );
input = reversed.split("").reverse().join("");
Based on #Tadek's answer above. This function takes other locales into consideration.
For example, some locales will use a comma for the decimal separator and a period for the thousand separator (e.g. -451.161,432e-12).
First we convert anything other than 1) numbers; 2) negative sign; 3) exponent sign into a period ("-451.161.432e-12").
Next we split by period (["-451", "161", "432e-12"]) and pop out the right-most value ("432e-12"), then join with the rest ("-451161.432e-12")
(Note that I'm tossing out the thousand separators, but those could easily be added in the join step (.join(','))
var ensureDecimalSeparatorIsPeriod = function (value) {
var numericString = value.toString();
var splitByDecimal = numericString.replace(/[^\d.e-]/g, '.').split('.');
if (splitByDecimal.length < 2) {
return numericString;
}
var rightOfDecimalPlace = splitByDecimal.pop();
return splitByDecimal.join('') + '.' + rightOfDecimalPlace;
};
let str = "12.1223....1322311..";
let finStr = str.replace(/(\d*.)(.*)/, '$1') + str.replace(/(\d*.)(.*)/, '$2').replace(/\./g,'');
console.log(finStr)
const [integer, ...decimals] = '233.423.3.32.23.244.14...23'.split('.');
const result = [integer, decimals.join('')].join('.')
Same solution offered but using the spread operator.
It's a matter of opinion but I think it improves readability.