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I've a little problem.
I'm using NodeJS as backend. Now, an user has a field "biography", where the user can write something about himself.
Suppose that this field has 220 maxlength, and suppose this as input:
👶🏻👦🏻👧🏻👨🏻👩🏻👱🏻♀️👱🏻👴🏻👵🏻👲🏻👳🏻♀️👳🏻👮🏻♀️👮🏻👷🏻♀️👷🏻💂🏻♀️💂🏻🕵🏻♀️👩🏻⚕️👨🏻⚕️👩🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾
As you can see there aren't 220 emojis (there are 37 emojis), but if I do in my nodejs server
console.log(bio.length)
where bio is the input text, I got 221. How could I "parse" the string input to get the correct length? Is it a problem about unicode?
SOLVED
I used this library: https://github.com/orling/grapheme-splitter
I tried that:
var Grapheme = require('grapheme-splitter');
var splitter = new Grapheme();
console.log(splitter.splitGraphemes(bio).length);
and the length is 37. It works very well!
str.length gives the count of UTF-16 units.
Unicode-proof way to get string length in codepoints (in characters) is [...str].length as iterable protocol splits the string to codepoints.
If we need the length in graphemes (grapheme clusters), we have these native ways:
a. Unicode property escapes in RegExp. See for example: Unicode-aware version of \w or Matching emoji.
b. Intl.Segmenter — coming soon, probably in ES2021. Can be tested with a flag in the last V8 versions (realization was synced with the last spec in V8 86). Unflagged (shipped) in V8 87.
See also:
The Absolute Minimum Every Software Developer Absolutely, Positively Must Know About Unicode and Character Sets (No Excuses!)
What every JavaScript developer should know about Unicode
JavaScript has a Unicode problem
Unicode-aware regular expressions in ES2015
ES6 Strings (and Unicode, ❤) in Depth
JavaScript for impatient programmers. Unicode – a brief introduction
TL;DR there are solutions, but they don’t work in every case. Unicode can feel like a dark art.
There seems to be limitations in various solutions I have seen presented, with the issue going beyond emojis and covering other characters in the Unicode range. Consider é can be stored as é or e + ‘, if using combing characters. This can even lead to two strings that look the same not being equal. Also note, in certain cases a single emoji can be 11 characters when stored and as a result 22 bytes, assuming UTF16.
The way this is handled and how characters are combined, or displayed, can even vary between browsers and operating systems. So, while you may think you cracked it, there is a risk another environment breaks this. Be sure to test where it matters.
Now, there is the front-end vs back-end problem: you solved the character count problem so it works well for human users, now your single emoji blows right past the allocated field size in the database. Less of an issue with databases such as mongo, but can be one with SQL databases, where field allocation was conservative. This means how you solve your problem will depend where the hardest limitation comes in.
Note, that a basic solution does involve converting a string to an array and getting the length, accepting limitations:
Array.from(str)
This will fall apart when characters are combined and dealing with astral planes.
A few high level approaches, that take into account limitations:
use approaches that solve the front-end issue, as best as possible, and then ensure storage issues are resolved
be more conservative with the advertised front-end limits, if the database or other storage can’t be adjusted
limit the character types that can be entered
clearly indicate limitations of the length calculation
Additionally, given the complexity of the issue it may be worth seeing if there is a popular JS library that already deals with this? I did not find one at the time of writing. Hopefully this is something that would become core to Javascript at some point.
Other pages to read:
https://blog.jonnew.com/posts/poo-dot-length-equals-two
https://mathiasbynens.be/notes/javascript-unicode
https://www.contentful.com/blog/2016/12/06/unicode-javascript-and-the-emoji-family/
https://dmitripavlutin.com/what-every-javascript-developer-should-know-about-unicode/
I answered to a similar question here
But basically, here it is :
'👍'.match(/./gu).length == 1
As :
'👍'.length == 2
More precision in my original post
function fancyCount2(str){
const joiner = "\u{200D}";
const split = str.split(joiner);
let count = 0;
for(const s of split){
//removing the variation selectors
const num = Array.from(s.split(/[\ufe00-\ufe0f]/).join("")).length;
count += num;
}
//assuming the joiners are used appropriately
return count / split.length;
}
With a regex that can parse emojis, this can be done easily and without the use of external libraries. Please see the code snippets for examples. 👷🏻♀️
Note that grapheme-splitter as suggested in the question will overcount and split apart compound emojis that contain other emojis, such as this one: 🧑🏻🤝🧑🏻. This is reported as three distinct "graphemes", 🧑🏻 and 🤝 and 🧑🏻
Here we are using the 'compact', literal version so it'll fit, but there's a safe, long version that uses Unicode escapes as well.
For more info on the regex see also this answer.
/*the pattern (compact version)*/
var emojiPattern = 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🐜|🐝|🪲|🐞|🦗|🪳|🕷|🕸|🦂|🦟|🪰|🪱|🦠|💐|🌸|💮|🏵|🌹|🥀|🌺|🌻|🌼|🌷|🌱|🪴|🌲|🌳|🌴|🌵|🌾|🌿|🍀|🍁|🍂|🍃|🍇|🍈|🍉|🍊|🍋|🍌|🍍|🥭|🍎|🍏|🍐|🍑|🍒|🍓|🫐|🥝|🍅|🫒|🥥|🥑|🍆|🥔|🥕|🌽|🌶|🫑|🥒|🥬|🥦|🧄|🧅|🍄|🥜|🌰|🍞|🥐|🥖|🫓|🥨|🥯|🥞|🧇|🧀|🍖|🍗|🥩|🥓|🍔|🍟|🍕|🌭|🥪|🌮|🌯|🫔|🥙|🧆|🥚|🍳|🥘|🍲|🫕|🥣|🥗|🍿|🧈|🧂|🥫|🍱|🍘|🍙|🍚|🍛|🍜|🍝|🍠|🍢|🍣|🍤|🍥|🥮|🍡|🥟|🥠|🥡|🦀|🦞|🦐|🦑|🦪|🍦|🍧|🍨|🍩|🍪|🎂|🍰|🧁|🥧|🍫|🍬|🍭|🍮|🍯|🍼|🥛|🫖|🍵|🍶|🍾|🍷|🍸|🍹|🍺|🍻|🥂|🥃|🥤|🧋|🧃|🧉|🧊|🥢|🍽|🍴|🥄|🔪|🏺|🌍|🌎|🌏|🌐|🗺|🗾|🧭|🏔|🌋|🗻|🏕|🏖|🏜|🏝|🏞|🏟|🏛|🏗|🧱|🪨|🪵|🛖|🏘|🏚|🏠|🏡|🏢|🏣|🏤|🏥|🏦|🏨|🏩|🏪|🏫|🏬|🏭|🏯|🏰|💒|🗼|🗽|🕌|🛕|🕍|🕋|🌁|🌃|🏙|🌄|🌅|🌆|🌇|🌉|🎠|🎡|🎢|💈|🎪|🚂|🚃|🚄|🚅|🚆|🚇|🚈|🚉|🚊|🚝|🚞|🚋|🚌|🚍|🚎|🚐|🚑|🚒|🚓|🚔|🚕|🚖|🚗|🚘|🚙|🛻|🚚|🚛|🚜|🏎|🏍|🛵|🦽|🦼|🛺|🚲|🛴|🛹|🛼|🚏|🛣|🛤|🛢|🚨|🚥|🚦|🛑|🚧|🛶|🚤|🛳|🛥|🚢|🛩|🛫|🛬|🪂|💺|🚁|🚟|🚠|🚡|🛰|🚀|🛸|🛎|🧳|🕰|🕛|🕧|🕐|🕜|🕑|🕝|🕒|🕞|🕓|🕟|🕔|🕠|🕕|🕡|🕖|🕢|🕗|🕣|🕘|🕤|🕙|🕥|🕚|🕦|🌑|🌒|🌓|🌔|🌕|🌖|🌗|🌘|🌙|🌚|🌛|🌜|🌡|🌝|🌞|🪐|🌟|🌠|🌌|🌤|🌥|🌦|🌧|🌨|🌩|🌪|🌫|🌬|🌀|🌈|🌂|🔥|💧|🌊|🎃|🎄|🎆|🎇|🧨|🎈|🎉|🎊|🎋|🎍|🎎|🎏|🎐|🎑|🧧|🎀|🎁|🎗|🎟|🎫|🎖|🏆|🏅|🥇|🥈|🥉|🥎|🏀|🏐|🏈|🏉|🎾|🥏|🎳|🏏|🏑|🏒|🥍|🏓|🏸|🥊|🥋|🥅|🎣|🤿|🎽|🎿|🛷|🥌|🎯|🪀|🪁|🎱|🔮|🪄|🧿|🎮|🕹|🎰|🎲|🧩|🧸|🪅|🪆|🃏|🀄|🎴|🎭|🖼|🎨|🧵|🪡|🧶|🪢|👓|🕶|🥽|🥼|🦺|👔|👕|👖|🧣|🧤|🧥|🧦|👗|👘|🥻|🩱|🩲|🩳|👙|👚|👛|👜|👝|🛍|🎒|🩴|👞|👟|🥾|🥿|👠|👡|🩰|👢|👑|👒|🎩|🎓|🧢|🪖|📿|💄|💍|💎|🔇|🔈|🔉|🔊|📢|📣|📯|🔔|🔕|🎼|🎵|🎶|🎙|🎚|🎛|🎤|🎧|📻|🎷|🪗|🎸|🎹|🎺|🎻|🪕|🥁|🪘|📱|📲|📞|📟|📠|🔋|🔌|💻|🖥|🖨|🖱|🖲|💽|💾|💿|📀|🧮|🎥|🎞|📽|🎬|📺|📷|📸|📹|📼|🔍|🔎|🕯|💡|🔦|🏮|🪔|📔|📕|📖|📗|📘|📙|📚|📓|📒|📃|📜|📄|📰|🗞|📑|🔖|🏷|💰|🪙|💴|💵|💶|💷|💸|💳|🧾|💹|📧|📨|📩|📤|📥|📦|📫|📪|📬|📭|📮|🗳|🖋|🖊|🖌|🖍|📝|💼|📁|📂|🗂|📅|📆|🗒|🗓|📇|📈|📉|📊|📋|📌|📍|📎|🖇|📏|📐|🗃|🗄|🗑|🔒|🔓|🔏|🔐|🔑|🗝|🔨|🪓|🛠|🗡|🔫|🪃|🏹|🛡|🪚|🔧|🪛|🔩|🗜|🦯|🔗|🪝|🧰|🧲|🪜|🧪|🧫|🧬|🔬|🔭|📡|💉|🩸|💊|🩹|🩺|🚪|🛗|🪞|🪟|🛏|🛋|🪑|🚽|🪠|🚿|🛁|🪤|🪒|🧴|🧷|🧹|🧺|🧻|🪣|🧼|🪥|🧽|🧯|🛒|🚬|🪦|🗿|🪧|🏧|🚮|🚰|🚹|🚺|🚻|🚼|🚾|🛂|🛃|🛄|🛅|🚸|🚫|🚳|🚭|🚯|🚱|🚷|📵|🔞|🔃|🔄|🔙|🔚|🔛|🔜|🔝|🛐|🕉|🕎|🔯|🔀|🔁|🔂|🔼|🔽|🎦|🔅|🔆|📶|📳|📴|💱|💲|🔱|📛|🔰|🔟|🔠|🔡|🔢|🔣|🔤|🅰|🆎|🅱|🆑|🆒|🆓|🆔|🆕|🆖|🅾|🆗|🅿|🆘|🆙|🆚|🈁|🈂|🈷|🈶|🈯|🉐|🈹|🈚|🈲|🉑|🈸|🈴|🈳|🈺|🈵|🔴|🟠|🟡|🟢|🔵|🟣|🟤|🟥|🟧|🟨|🟩|🟦|🟪|🟫|🔶|🔷|🔸|🔹|🔺|🔻|💠|🔘|🔳|🔲|🏁|🚩|🎌|🏴|🏳|🏻|🏼|🏽|🏾|🏿|☺|☹|☠|❣|❤|✋|✌|☝|✊|✍|⛷|⛹|☘|☕|⛰|⛪|⛩|⛲|⛺|♨|⛽|⚓|⛵|⛴|✈|⌛|⏳|⌚|⏰|⏱|⏲|☀|⭐|☁|⛅|⛈|☂|☔|⛱|⚡|❄|☃|⛄|☄|✨|⚽|⚾|⛳|⛸|♠|♥|♦|♣|♟|⛑|☎|⌨|✉|✏|✒|✂|⛏|⚒|⚔|⚙|⚖|⛓|⚗|⚰|⚱|♿|⚠|⛔|☢|☣|⬆|↗|➡|↘|⬇|↙|⬅|↖|↕|↔|↩|↪|⤴|⤵|⚛|✡|☸|☯|✝|☦|☪|☮|♈|♉|♊|♋|♌|♍|♎|♏|♐|♑|♒|♓|⛎|▶|⏩|⏭|⏯|◀|⏪|⏮|⏫|⏬|⏸|⏹|⏺|⏏|♀|♂|⚧|✖|➕|➖|➗|♾|‼|⁉|❓|❔|❕|❗|〰|⚕|♻|⚜|⭕|✅|☑|✔|❌|❎|➰|➿|〽|✳|✴|❇|©|®|™|ℹ|Ⓜ|㊗|㊙|⚫|⚪|⬛|⬜|◼|◻|◾|◽|▪|▫)`
/*compile the pattern string into a regex*/
let emoRegex = new RegExp(emojiPattern, "g");
/*count of emojis*/
let emoCount = [..."👶🏻👦🏻👧🏻👨🏻👩🏻👱🏻♀️👱🏻👴🏻👵🏻👲🏻👳🏻♀️👳🏻👮🏻♀️👮🏻👷🏻♀️👷🏻💂🏻♀️💂🏻🕵🏻♀️👩🏻⚕️👨🏻⚕️👩🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾".matchAll(emoRegex)].length
console.log(emoCount) //37
/*modifying the pattern to count other characters too*/
let generalCounter = new RegExp(emojiPattern+"|.", "g") //emoji or regular character
let allCount = [..."$%^ other stuff equalling 28👶👦🏻👧🏻👨🏻👩🏻👱🏻♀️👱🏻👴🏻👵🏻👲🏻👳🏻♀️👳🏻👮🏻♀️👮🏻👷🏻♀️👷🏻💂🏻♀️💂🏻🕵🏻♀️👩🏻⚕️👨🏻⚕️👩🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾".matchAll(generalCounter)].length
console.log(allCount) //28+37 = 65
As you can see from the below example, this is to do with unicode encoding,
There's some great resources such as the one I took this example from.
https://blog.jonnew.com/posts/poo-dot-length-equals-two
console.log("👩❤️💋👩".length === 11);
For anyone interested, I had a similar problem where I wanted to count the length of an emoji at the end of a string.
This is the solution I came up with:
var emoji = new RegExp('(\\p{Extended_Pictographic})((\u200D\\p{Extended_Pictographic})*)$', 'u');
var testStrings = ['👨👩👧', '😂', '🌲'];
for(var string = 0; string < testStrings.length; string++){
var match = testStrings[string].match(emoji);
var chars = match == null ? 0 : match[0].length;
console.log(testStrings[string] + ': ' + chars);
}
Explanation: \\p{Extended_Pictographic} matches an emoji like 😂, consisting of two characters. Emojis like 👨👩👧 consists of 4 emojis (👨, 👩, 👧,👦) combined by a zero width joiner (\u200D).
The regex matches any emoji at the end ($). If there is a match the length is counted. I am sure it could be adopted for your use-case by matching all emojis in a given string and then subtracting the surplus. It's not a complete implementation for your particular question but I hope this gets you on the right track.
use lodash toArray method
console.log(_.toArray("👨👩👧").length); // 1
console.log(_.toArray("👨👩👧🧍♂️👩👧👧").length); // 3
Check here for Codesandbox
I suggest using the runes package to accomplish correct multi-byte string conversions cause else you will get more issues if using reducers and more to reverse strings for example.
Take a look at this great small package: runes
Is it possible? With no library's or memory of the changes as they're made?
The example I'm showing below is probably more edited than the text I will be evaluating (although much shorter). For my use case, I will be editing a transcription that is usually very accurate and I need to be able to know if a new word is added & where so I can approximate a timecode for the new word, as well as shift the existing timecodes forwards to combat the offset that the new word has created.
originalString = "Hello, this sum example txt. Hopefully this is possible!";
editedString = "Hello, this is some example text. Maybe this is not impossible.";
//The edited words are: "sum/some", "txt/text", "Hopefully/Maybe", "possible/impossible"
//It would be useful to get the edited words also - but not essential
//the new words are: "is", "not"
//Output would be
newWordIdxs = [2, 9];
First post here and have only been coding for 3 or 4 months so any tips on how to ask better questions are very welcome.
How to estimate an word was edited? According to the example you posted, the word 'some' may be added, and 'sum' is removed. It's more logical to find which word or character is removed or added by excluding the same words or characters in sequence. This javascript library may help , https://www.npmjs.com/package/diff
I've a little problem.
I'm using NodeJS as backend. Now, an user has a field "biography", where the user can write something about himself.
Suppose that this field has 220 maxlength, and suppose this as input:
👶🏻👦🏻👧🏻👨🏻👩🏻👱🏻♀️👱🏻👴🏻👵🏻👲🏻👳🏻♀️👳🏻👮🏻♀️👮🏻👷🏻♀️👷🏻💂🏻♀️💂🏻🕵🏻♀️👩🏻⚕️👨🏻⚕️👩🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾
As you can see there aren't 220 emojis (there are 37 emojis), but if I do in my nodejs server
console.log(bio.length)
where bio is the input text, I got 221. How could I "parse" the string input to get the correct length? Is it a problem about unicode?
SOLVED
I used this library: https://github.com/orling/grapheme-splitter
I tried that:
var Grapheme = require('grapheme-splitter');
var splitter = new Grapheme();
console.log(splitter.splitGraphemes(bio).length);
and the length is 37. It works very well!
str.length gives the count of UTF-16 units.
Unicode-proof way to get string length in codepoints (in characters) is [...str].length as iterable protocol splits the string to codepoints.
If we need the length in graphemes (grapheme clusters), we have these native ways:
a. Unicode property escapes in RegExp. See for example: Unicode-aware version of \w or Matching emoji.
b. Intl.Segmenter — coming soon, probably in ES2021. Can be tested with a flag in the last V8 versions (realization was synced with the last spec in V8 86). Unflagged (shipped) in V8 87.
See also:
The Absolute Minimum Every Software Developer Absolutely, Positively Must Know About Unicode and Character Sets (No Excuses!)
What every JavaScript developer should know about Unicode
JavaScript has a Unicode problem
Unicode-aware regular expressions in ES2015
ES6 Strings (and Unicode, ❤) in Depth
JavaScript for impatient programmers. Unicode – a brief introduction
TL;DR there are solutions, but they don’t work in every case. Unicode can feel like a dark art.
There seems to be limitations in various solutions I have seen presented, with the issue going beyond emojis and covering other characters in the Unicode range. Consider é can be stored as é or e + ‘, if using combing characters. This can even lead to two strings that look the same not being equal. Also note, in certain cases a single emoji can be 11 characters when stored and as a result 22 bytes, assuming UTF16.
The way this is handled and how characters are combined, or displayed, can even vary between browsers and operating systems. So, while you may think you cracked it, there is a risk another environment breaks this. Be sure to test where it matters.
Now, there is the front-end vs back-end problem: you solved the character count problem so it works well for human users, now your single emoji blows right past the allocated field size in the database. Less of an issue with databases such as mongo, but can be one with SQL databases, where field allocation was conservative. This means how you solve your problem will depend where the hardest limitation comes in.
Note, that a basic solution does involve converting a string to an array and getting the length, accepting limitations:
Array.from(str)
This will fall apart when characters are combined and dealing with astral planes.
A few high level approaches, that take into account limitations:
use approaches that solve the front-end issue, as best as possible, and then ensure storage issues are resolved
be more conservative with the advertised front-end limits, if the database or other storage can’t be adjusted
limit the character types that can be entered
clearly indicate limitations of the length calculation
Additionally, given the complexity of the issue it may be worth seeing if there is a popular JS library that already deals with this? I did not find one at the time of writing. Hopefully this is something that would become core to Javascript at some point.
Other pages to read:
https://blog.jonnew.com/posts/poo-dot-length-equals-two
https://mathiasbynens.be/notes/javascript-unicode
https://www.contentful.com/blog/2016/12/06/unicode-javascript-and-the-emoji-family/
https://dmitripavlutin.com/what-every-javascript-developer-should-know-about-unicode/
I answered to a similar question here
But basically, here it is :
'👍'.match(/./gu).length == 1
As :
'👍'.length == 2
More precision in my original post
function fancyCount2(str){
const joiner = "\u{200D}";
const split = str.split(joiner);
let count = 0;
for(const s of split){
//removing the variation selectors
const num = Array.from(s.split(/[\ufe00-\ufe0f]/).join("")).length;
count += num;
}
//assuming the joiners are used appropriately
return count / split.length;
}
With a regex that can parse emojis, this can be done easily and without the use of external libraries. Please see the code snippets for examples. 👷🏻♀️
Note that grapheme-splitter as suggested in the question will overcount and split apart compound emojis that contain other emojis, such as this one: 🧑🏻🤝🧑🏻. This is reported as three distinct "graphemes", 🧑🏻 and 🤝 and 🧑🏻
Here we are using the 'compact', literal version so it'll fit, but there's a safe, long version that uses Unicode escapes as well.
For more info on the regex see also this answer.
/*the pattern (compact version)*/
var emojiPattern = 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🐜|🐝|🪲|🐞|🦗|🪳|🕷|🕸|🦂|🦟|🪰|🪱|🦠|💐|🌸|💮|🏵|🌹|🥀|🌺|🌻|🌼|🌷|🌱|🪴|🌲|🌳|🌴|🌵|🌾|🌿|🍀|🍁|🍂|🍃|🍇|🍈|🍉|🍊|🍋|🍌|🍍|🥭|🍎|🍏|🍐|🍑|🍒|🍓|🫐|🥝|🍅|🫒|🥥|🥑|🍆|🥔|🥕|🌽|🌶|🫑|🥒|🥬|🥦|🧄|🧅|🍄|🥜|🌰|🍞|🥐|🥖|🫓|🥨|🥯|🥞|🧇|🧀|🍖|🍗|🥩|🥓|🍔|🍟|🍕|🌭|🥪|🌮|🌯|🫔|🥙|🧆|🥚|🍳|🥘|🍲|🫕|🥣|🥗|🍿|🧈|🧂|🥫|🍱|🍘|🍙|🍚|🍛|🍜|🍝|🍠|🍢|🍣|🍤|🍥|🥮|🍡|🥟|🥠|🥡|🦀|🦞|🦐|🦑|🦪|🍦|🍧|🍨|🍩|🍪|🎂|🍰|🧁|🥧|🍫|🍬|🍭|🍮|🍯|🍼|🥛|🫖|🍵|🍶|🍾|🍷|🍸|🍹|🍺|🍻|🥂|🥃|🥤|🧋|🧃|🧉|🧊|🥢|🍽|🍴|🥄|🔪|🏺|🌍|🌎|🌏|🌐|🗺|🗾|🧭|🏔|🌋|🗻|🏕|🏖|🏜|🏝|🏞|🏟|🏛|🏗|🧱|🪨|🪵|🛖|🏘|🏚|🏠|🏡|🏢|🏣|🏤|🏥|🏦|🏨|🏩|🏪|🏫|🏬|🏭|🏯|🏰|💒|🗼|🗽|🕌|🛕|🕍|🕋|🌁|🌃|🏙|🌄|🌅|🌆|🌇|🌉|🎠|🎡|🎢|💈|🎪|🚂|🚃|🚄|🚅|🚆|🚇|🚈|🚉|🚊|🚝|🚞|🚋|🚌|🚍|🚎|🚐|🚑|🚒|🚓|🚔|🚕|🚖|🚗|🚘|🚙|🛻|🚚|🚛|🚜|🏎|🏍|🛵|🦽|🦼|🛺|🚲|🛴|🛹|🛼|🚏|🛣|🛤|🛢|🚨|🚥|🚦|🛑|🚧|🛶|🚤|🛳|🛥|🚢|🛩|🛫|🛬|🪂|💺|🚁|🚟|🚠|🚡|🛰|🚀|🛸|🛎|🧳|🕰|🕛|🕧|🕐|🕜|🕑|🕝|🕒|🕞|🕓|🕟|🕔|🕠|🕕|🕡|🕖|🕢|🕗|🕣|🕘|🕤|🕙|🕥|🕚|🕦|🌑|🌒|🌓|🌔|🌕|🌖|🌗|🌘|🌙|🌚|🌛|🌜|🌡|🌝|🌞|🪐|🌟|🌠|🌌|🌤|🌥|🌦|🌧|🌨|🌩|🌪|🌫|🌬|🌀|🌈|🌂|🔥|💧|🌊|🎃|🎄|🎆|🎇|🧨|🎈|🎉|🎊|🎋|🎍|🎎|🎏|🎐|🎑|🧧|🎀|🎁|🎗|🎟|🎫|🎖|🏆|🏅|🥇|🥈|🥉|🥎|🏀|🏐|🏈|🏉|🎾|🥏|🎳|🏏|🏑|🏒|🥍|🏓|🏸|🥊|🥋|🥅|🎣|🤿|🎽|🎿|🛷|🥌|🎯|🪀|🪁|🎱|🔮|🪄|🧿|🎮|🕹|🎰|🎲|🧩|🧸|🪅|🪆|🃏|🀄|🎴|🎭|🖼|🎨|🧵|🪡|🧶|🪢|👓|🕶|🥽|🥼|🦺|👔|👕|👖|🧣|🧤|🧥|🧦|👗|👘|🥻|🩱|🩲|🩳|👙|👚|👛|👜|👝|🛍|🎒|🩴|👞|👟|🥾|🥿|👠|👡|🩰|👢|👑|👒|🎩|🎓|🧢|🪖|📿|💄|💍|💎|🔇|🔈|🔉|🔊|📢|📣|📯|🔔|🔕|🎼|🎵|🎶|🎙|🎚|🎛|🎤|🎧|📻|🎷|🪗|🎸|🎹|🎺|🎻|🪕|🥁|🪘|📱|📲|📞|📟|📠|🔋|🔌|💻|🖥|🖨|🖱|🖲|💽|💾|💿|📀|🧮|🎥|🎞|📽|🎬|📺|📷|📸|📹|📼|🔍|🔎|🕯|💡|🔦|🏮|🪔|📔|📕|📖|📗|📘|📙|📚|📓|📒|📃|📜|📄|📰|🗞|📑|🔖|🏷|💰|🪙|💴|💵|💶|💷|💸|💳|🧾|💹|📧|📨|📩|📤|📥|📦|📫|📪|📬|📭|📮|🗳|🖋|🖊|🖌|🖍|📝|💼|📁|📂|🗂|📅|📆|🗒|🗓|📇|📈|📉|📊|📋|📌|📍|📎|🖇|📏|📐|🗃|🗄|🗑|🔒|🔓|🔏|🔐|🔑|🗝|🔨|🪓|🛠|🗡|🔫|🪃|🏹|🛡|🪚|🔧|🪛|🔩|🗜|🦯|🔗|🪝|🧰|🧲|🪜|🧪|🧫|🧬|🔬|🔭|📡|💉|🩸|💊|🩹|🩺|🚪|🛗|🪞|🪟|🛏|🛋|🪑|🚽|🪠|🚿|🛁|🪤|🪒|🧴|🧷|🧹|🧺|🧻|🪣|🧼|🪥|🧽|🧯|🛒|🚬|🪦|🗿|🪧|🏧|🚮|🚰|🚹|🚺|🚻|🚼|🚾|🛂|🛃|🛄|🛅|🚸|🚫|🚳|🚭|🚯|🚱|🚷|📵|🔞|🔃|🔄|🔙|🔚|🔛|🔜|🔝|🛐|🕉|🕎|🔯|🔀|🔁|🔂|🔼|🔽|🎦|🔅|🔆|📶|📳|📴|💱|💲|🔱|📛|🔰|🔟|🔠|🔡|🔢|🔣|🔤|🅰|🆎|🅱|🆑|🆒|🆓|🆔|🆕|🆖|🅾|🆗|🅿|🆘|🆙|🆚|🈁|🈂|🈷|🈶|🈯|🉐|🈹|🈚|🈲|🉑|🈸|🈴|🈳|🈺|🈵|🔴|🟠|🟡|🟢|🔵|🟣|🟤|🟥|🟧|🟨|🟩|🟦|🟪|🟫|🔶|🔷|🔸|🔹|🔺|🔻|💠|🔘|🔳|🔲|🏁|🚩|🎌|🏴|🏳|🏻|🏼|🏽|🏾|🏿|☺|☹|☠|❣|❤|✋|✌|☝|✊|✍|⛷|⛹|☘|☕|⛰|⛪|⛩|⛲|⛺|♨|⛽|⚓|⛵|⛴|✈|⌛|⏳|⌚|⏰|⏱|⏲|☀|⭐|☁|⛅|⛈|☂|☔|⛱|⚡|❄|☃|⛄|☄|✨|⚽|⚾|⛳|⛸|♠|♥|♦|♣|♟|⛑|☎|⌨|✉|✏|✒|✂|⛏|⚒|⚔|⚙|⚖|⛓|⚗|⚰|⚱|♿|⚠|⛔|☢|☣|⬆|↗|➡|↘|⬇|↙|⬅|↖|↕|↔|↩|↪|⤴|⤵|⚛|✡|☸|☯|✝|☦|☪|☮|♈|♉|♊|♋|♌|♍|♎|♏|♐|♑|♒|♓|⛎|▶|⏩|⏭|⏯|◀|⏪|⏮|⏫|⏬|⏸|⏹|⏺|⏏|♀|♂|⚧|✖|➕|➖|➗|♾|‼|⁉|❓|❔|❕|❗|〰|⚕|♻|⚜|⭕|✅|☑|✔|❌|❎|➰|➿|〽|✳|✴|❇|©|®|™|ℹ|Ⓜ|㊗|㊙|⚫|⚪|⬛|⬜|◼|◻|◾|◽|▪|▫)`
/*compile the pattern string into a regex*/
let emoRegex = new RegExp(emojiPattern, "g");
/*count of emojis*/
let emoCount = [..."👶🏻👦🏻👧🏻👨🏻👩🏻👱🏻♀️👱🏻👴🏻👵🏻👲🏻👳🏻♀️👳🏻👮🏻♀️👮🏻👷🏻♀️👷🏻💂🏻♀️💂🏻🕵🏻♀️👩🏻⚕️👨🏻⚕️👩🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾".matchAll(emoRegex)].length
console.log(emoCount) //37
/*modifying the pattern to count other characters too*/
let generalCounter = new RegExp(emojiPattern+"|.", "g") //emoji or regular character
let allCount = [..."$%^ other stuff equalling 28👶👦🏻👧🏻👨🏻👩🏻👱🏻♀️👱🏻👴🏻👵🏻👲🏻👳🏻♀️👳🏻👮🏻♀️👮🏻👷🏻♀️👷🏻💂🏻♀️💂🏻🕵🏻♀️👩🏻⚕️👨🏻⚕️👩🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾👨🏻🌾".matchAll(generalCounter)].length
console.log(allCount) //28+37 = 65
As you can see from the below example, this is to do with unicode encoding,
There's some great resources such as the one I took this example from.
https://blog.jonnew.com/posts/poo-dot-length-equals-two
console.log("👩❤️💋👩".length === 11);
For anyone interested, I had a similar problem where I wanted to count the length of an emoji at the end of a string.
This is the solution I came up with:
var emoji = new RegExp('(\\p{Extended_Pictographic})((\u200D\\p{Extended_Pictographic})*)$', 'u');
var testStrings = ['👨👩👧', '😂', '🌲'];
for(var string = 0; string < testStrings.length; string++){
var match = testStrings[string].match(emoji);
var chars = match == null ? 0 : match[0].length;
console.log(testStrings[string] + ': ' + chars);
}
Explanation: \\p{Extended_Pictographic} matches an emoji like 😂, consisting of two characters. Emojis like 👨👩👧 consists of 4 emojis (👨, 👩, 👧,👦) combined by a zero width joiner (\u200D).
The regex matches any emoji at the end ($). If there is a match the length is counted. I am sure it could be adopted for your use-case by matching all emojis in a given string and then subtracting the surplus. It's not a complete implementation for your particular question but I hope this gets you on the right track.
use lodash toArray method
console.log(_.toArray("👨👩👧").length); // 1
console.log(_.toArray("👨👩👧🧍♂️👩👧👧").length); // 3
Check here for Codesandbox
I suggest using the runes package to accomplish correct multi-byte string conversions cause else you will get more issues if using reducers and more to reverse strings for example.
Take a look at this great small package: runes
I have a string, eg:
Lenovo K6 Power (Silver, 32 GB)(4 GB RAM), smart phone
I want to remove all parentheses and contents within the parentheses, commas etc and then extract the first five words only so that I get the result as
Lenovo K6 Power smart phone
Is there any method to apply regex to get this result?
Here's one way of doing it:
var str = 'Lenovo K6 Power (Silver, 32 GB)(4 GB RAM)';
document.write(str.match(/\w+/g).slice(0,5).join(' '));
It gets all words into an array (match(/\w+/g)), then gets the first five (slice(0,5)), to join then back to a string separated by space (join(' ')).
(And... Considering the question is tagged with regex, I believe a word could be defined as consisting of regex word characters, i.e. \w.)
Edit
The question has changed so the answer isn't correct anymore. Here's an update snippet that works with the new criteria:
var str = 'Lenovo K6 Power (Silver, 32 GB)(4 GB RAM), smart phone';
document.write(str.split(/(?:\W*\([^)]*\))*\W+/).slice(0,5).join(' '));
This one split's the string instead, using the regex (?:\W*\([^)]*\))*\W+ which will match everything but word characters (\W), unless they're inside parentheses (everything inside parentheses is matched).
spliting on that will give an array with only the desired words. Therefrom the logic is the same.
var s1 = "Lenovo K6 Power (Silver, 32 GB)(4 GB RAM), smart phone";
var s2 = s1.replace(/\([^)]*\)|, /g,'')
console.log(s2) //Output : "Lenovo K6 Power smart phone"
var myString = "Lenovo K6 Power (Silver, 32 GB)(4 GB RAM), smart phone";
while (/\(.*\)/.test(myString)) {
myString = myString.replace(/\(.*?\)/.exec(myString)[0],'');
}
console.log(myString.match(/\w+/g));
The first snippet matches all parentheses pairs as long as there are some and removes them, them it matches all remaining words.
Output: Obj... ["Lenovo", "K6", "Power", "smart", "phone"]
This is a general solution, to always only get the first 5 Elements change the console log to
var obj = myString.match(/\w+/g);
for (var i = 0; i < 5; i ++)
{
console.log(obj[i]);
}
Your question is extremely trivial and can be answered with the most basic JavaScript skills. I strongly suggest you go back and review your tutorials and intros, and try solving your problem yourself.
To remove something, you simply do
string.replace(WHAT, '')
In other words, you replace something with nothing (the empty string ''). In the case you mentioned, this, a simple Google search for something like "javascript remove regexp" will give you plenty of pointers. In this case, one of the first results actually is about removing parentheses.
In your case, I guess you finally decided you want to remove parentheses and what's inside them. In about the first five minutes of learning regexp, you should have learned to write
/\(.*?\)/g
^^ an actual left paren
^^^ any number of characters
^^ an actual right paren
^ match this over and over again
If you need help with this, try an online regexp tester such as regex101.com. It will also give you a readable version of your regexp.
The only thing moderately advanced about this is the .*?, where the ? means "non-greedy"--in other words, take characters up only to the next right paren.
I'm sure you already learned why you have to write \( to match a left parenthesis, right? The \ escapes the parentheses, because by itself the parentheses would have a special meaning to regexp. You know the g flag too, right? That means replace all the matches.
To find the first five tokens, you first need to split your string into tokens. I'm sure you recall from your studies the basic Array methods, including--drum roll--split! Split your string with string.split(' '). That will split on single spaces. If you want to split on any whitespace, you could try string.split(/\s+/).
Now go back and read the documentation for split real carefully, although I know you already have. Look carefully at the second argument, called limit. It does exactly what you want. It splits into segments, but no more than specified by limit.
The solution to your problem, which you could easily have come up with if you had spent about five minutes studying the documentation and experimenting, is
input.replace(/\(.*?\)/g, '').split(/\s+/, 5)
Unfortunately, your approach of posting on Stack Overflow is not going to scale well at all. You can't post here every time there is some minor problem you cannot figure out yourself. To be perfectly frank, if you cannot learn how to learn, then you better give up on being a programmer and try brick-laying instead. You need to learn how to figure out things yourself. Before anything else, you need to learn how to read (and digest) the documentation. Very early in your career, you're also going to need to learn how to debug your programs, since Stack Overflow is no better a way to get your programs debugged than it is to get them written for you in the first place. If you simply cannot bring yourself to read documents or learn by yourself, and can only work by asking other people how to do every little thing, then find a chat room or forum where there are people with nothing better to do than answer such questions. That is not what Stack Overflow is.
console.log('Lenovo K6 Power (Silver, 32 GB)(4 GB RAM), smart phone'.replace(/\(.*?\)/g, '').split(/\s+/, 5));
Fixing this code so as to remove the comma is left as an exercise for you to use your new-found learning powers on. Hint: you may want to use the regexp feature called "alternation", which is represented by the vertical bar or pipe |. You may also find yourself needing to use character classes, which is another thing you should have learned about very early in your regexp studies.
None of this has anything to do with TypeScript, or Angular, as you seem to have thought when you initially posted the question. It's a little concerning that you seem to think that doing basic regexp or string or array manipulation would somehow be a TypeScript or Angular issue. TypeScript is merely a typing layer on top of JavaScript. Angular is a framework for building web apps. Neither replaces JavaScript, or provides any new basic language capability. In fact, to use either effectively, you must know JavaScript well.
Readability aside, are there any discernable differences (performance perhaps) between using
str.indexOf("src")
and
str.match(/src/)
I personally prefer match (and regexp) but colleagues seem to go the other way. We were wondering if it mattered ...?
EDIT:
I should have said at the outset that this is for functions that will be doing partial plain-string matching (to pick up identifiers in class attributes for JQuery) rather than full regexp searches with wildcards etc.
class='redBorder DisablesGuiClass-2345-2d73-83hf-8293'
So it's the difference between:
string.indexOf('DisablesGuiClass-');
and
string.match(/DisablesGuiClass-/)
RegExp is indeed slower than indexOf (you can see it here), though normally this shouldn't be an issue. With RegExp, you also have to make sure the string is properly escaped, which is an extra thing to think about.
Both of those issues aside, if two tools do exactly what you need them to, why not choose the simpler one?
Your comparison may not be entirely fair. indexOf is used with plain strings and is therefore very fast; match takes a regular expression - of course it may be slower in comparison, but if you want to do a regex match, you won't get far with indexOf. On the other hand, regular expression engines can be optimized, and have been improving in performance in the last years.
In your case, where you're looking for a verbatim string, indexOf should be sufficient. There is still one application for regexes, though: If you need to match entire words and want to avoid matching substrings, then regular expressions give you "word boundary anchors". For example:
indexOf('bar')
will find bar three times in bar, fubar, barmy, whereas
match(/\bbar\b/)
will only match bar when it is not part of a longer word.
As you can see in the comments, some comparisons have been done that show that a regex may be faster than indexOf - if it's performance-critical, you may need to profile your code.
Here all possible ways (relatively) to search for string
// 1. includes (introduced in ES6)
var string = "string to search for substring",
substring = "sea";
string.includes(substring);
// 2. string.indexOf
var string = "string to search for substring",
substring = "sea";
string.indexOf(substring) !== -1;
// 3. RegExp: test
var string = "string to search for substring",
expr = /sea/; // no quotes here
expr.test(string);
// 4. string.match
var string = "string to search for substring",
expr = "/sea/";
string.match(expr);
//5. string.search
var string = "string to search for substring",
expr = "/sea/";
string.search(expr);
Here a src: https://koukia.ca/top-6-ways-to-search-for-a-string-in-javascript-and-performance-benchmarks-ce3e9b81ad31
Benchmarks seem to be twisted specially for es6 includes , read the comments.
In resume:
if you don't need the matches.
=> Either you need regex and so use test. Otherwise es6 includes or indexOf. Still test vs indexOf are close.
And for includes vs indexOf:
They seem to be the same : https://jsperf.com/array-indexof-vs-includes/4 (if it was different it would be wierd, they mostly perform the same except for the differences that they expose check this)
And for my own benchmark test. here it is http://jsben.ch/fFnA0
You can test it (it's browser dependent) [test multiple time]
here how it performed (multiple run indexOf and includes one beat the other, and they are close). So they are the same. [here using the same test platform as the article above].
And here for the a long text version (8 times longer)
http://jsben.ch/wSBA2
Tested both chrome and firefox, same thing.
Notice jsben.ch doesn't handle memory overflow (or there limits correctly. It doesn't show any message) so result can get wrong if you add more then 8 text duplication (8 work well). But the conclusion is for very big text all three perform the same way. Otherwise for short indexOf and includes are the same and test a little bit slower. or Can be the same as it seemed in chrome (firefox 60 it is slower).
Notice with jsben.ch: don't freak out if you get inconsistant result. Try different time and see if it's consistent or not. Change browser, sometimes they just run totally wrong. Bug or bad handling of memory. Or something.
ex:
Here too my benchmark on jsperf (better details, and handle graphs for multiple browsers)
(top is chrome)
normal text
https://jsperf.com/indexof-vs-includes-vs-test-2019
resume: includes and indexOf have same perofrmance. test slower.
(seem all three perform the same in chrom)
Long text (12 time longer then normal)
https://jsperf.com/indexof-vs-includes-vs-test-2019-long-text-str/
resume: All the three perform the same. (chrome and firefox)
very short string
https://jsperf.com/indexof-vs-includes-vs-test-2019-too-short-string/
resume: includes and indexOf perform the same and test slower.
Note: about the benchmark above. For the very short string version (jsperf) had an big error for chrome. Seeing by my eyes. around 60 sample was run for both indexOf and includes same way (repeated a lot of time). And test a little bit less and so slower.
don't be fooled with the wrong graph. It's clear wrong. Same test work ok for firefox, surely it's a bug.
Here the illustration: (the first image was the test on firefox)
waaaa. Suddenly indexOf became superman. But as i said i did the test, and looked at the number of samples it was around 60. Both indexOf and includes and they performed the same. A bug on jspref. Except for this one (maybe because of a memory restriction related problem) all the rest was consistent, it give more details. And you see how many simple happen in real time.
Final resume
indexOf vs includes => Same performance
test => can be slower for short strings or text. And the same for long texts. And it make sense for the overhead that the regex engine add. In chrome it seemed it doesn't matter at all.
If you're trying to search for substring occurrences case-insensitively then match seems to be faster than a combination of indexOf and toLowerCase()
Check here - http://jsperf.com/regexp-vs-indexof/152
You ask whether str.indexOf('target') or str.match(/target/) should be preferred. As other posters have suggested, the use cases and return types of these methods are different. The first asks "where in str can I first find 'target'?" The second asks "does str match the regex and, if so, what are all of the matches for any associated capture groups?"
The issue is that neither one technically is designed to ask the simpler question "does the string contain the substring?" There is something that is explicitly designed to do so:
var doesStringContainTarget = /target/.test(str);
There are several advantages to using regex.test(string):
It returns a boolean, which is what you care about
It is more performant than str.match(/target/) (and rivals str.indexOf('target'))
If for some reason, str is undefined or null, you'll get false (the desired result) instead of throwing a TypeError
Using indexOf should, in theory, be faster than a regex when you're just searching for some plain text, but you should do some comparative benchmarks yourself if you're concerned about performance.
If you prefer match and it's fast enough for your needs then go for it.
For what it's worth, I agree with your colleagues on this: I'd use indexOf when searching for a plain string, and use match etc only when I need the extra functionality provided by regular expressions.
Performance wise indexOf will at the very least be slightly faster than match. It all comes down to the specific implementation. When deciding which to use ask yourself the following question:
Will an integer index suffice or do I
need the functionality of a RegExp
match result?
The return values are different
Aside from the performance implications, which are addressed by other answers, it is important to note that the return values for each method are different; so the methods cannot merely be substituted without also changing your logic.
Return value of .indexOf: integer
The index within the calling String object of the first occurrence of the specified value, starting the search at fromIndex.Returns -1 if the value is not found.
Return value of .match: array
An Array containing the entire match result and any parentheses-captured matched results.Returns null if there were no matches.
Because .indexOf returns 0 if the calling string begins with the specified value, a simple truthy test will fail.
For example:
Given this class…
class='DisablesGuiClass-2345-2d73-83hf-8293 redBorder'
…the return values for each would differ:
// returns `0`, evaluates to `false`
if (string.indexOf('DisablesGuiClass-')) {
… // this block is skipped.
}
vs.
// returns `["DisablesGuiClass-"]`, evaluates to `true`
if (string.match(/DisablesGuiClass-/)) {
… // this block is run.
}
The correct way to run a truthy test with the return from .indexOf is to test against -1:
if (string.indexOf('DisablesGuiClass-') !== -1) {
// ^returns `0` ^evaluates to `true`
… // this block is run.
}
remember Internet Explorer 8 doesnt understand indexOf.
But if nobody of your users uses ie8 (google analytics would tell you) than omit this answer.
possible solution to fix ie8:
How to fix Array indexOf() in JavaScript for Internet Explorer browsers