I'm working on a lab assignment for a web applications class and am stuck on implementing a word counter for a basic HTML webpage. The setup of the tests and HTML are already done for us. I simply need to write a function called countWords that takes a string and returns the number of words. It works differently from your traditional word counter though. A word is defined as anything A-Z. Everything else is considered not part of a word. So, if the string is just "234##$^" then the word count is 0. So, I'm not just counting white space like most word counters. All the answers I've found on StackOverflow to similar questions try to just count white space and don't work for my situation. Hence why I made a new question.
My idea was to have a return statement that matches any grouping of a-z using a regular expression and return the length. Then, have a conditional to check for the empty string or string with no letters a-z.
function countWords(s) {
if(s === "" || s === "%$#^23#") {
return 0
}
return s.match(/[^a-z]/gi).length
}
Right now the if statement is just matching the two test cases so that I can pass my tests. I'm not sure how to go about writing another match regular expression to check for no letters in the string or the empty string. Any help is appreciated! Been stuck for a while.
const str1 = '%$#^23#';
const str2 = 'String with ___ special characters and #$&# white spaces !!!';
const str3 = 'Special &$%# characters --> and %$#^5# connected,words but our <++##||++> function,still_works!';
const wordCount = (str) => str.replace(/[\W_\d]/g,' ').split(' ').filter(Boolean).length;
console.log(wordCount(str1)); // 0
console.log(wordCount(str2)); // 7
console.log(wordCount(str3)); // 11
use "regex" to replace all special characters, underscores, numbers, and extra white spaces with an empty space
--> replace(/[\W_\d]/g,' ')
convert the string into an array
--> .split(' ')
use filter to remove all empty string(s) in the array
--> .filter(Boolean)
then, get the word count with "length"
--> .length
You first need to filter the string, remove all the special characters and numbers:
var filtered_test = my_text.replace(/[^a-zA-Z ]/g, '');
then do a normal split and count:
var words = filtered_test.split(" ");
console.log(words.length); //prints out the count of words
You can use a functional replace method to chunk all of the "words" into an array, then simply return the array length. This has the added benefit of providing a 0 count:
explanatory version:
function countWords(str, words = []) {
str.replace(/[A-Z]+/gi, (m) => words.push(m));
return words.length;
}
minimal version:
let countWords = (str, words = []) =>
( str.replace(/[A-Z]+/gi, (m) => words.push(m)), words.length );
let countWords = (str, words = []) => (str.replace(/[A-Z]+/gi, (m) => words.push(m)), words.length);
console.log( "##asdfadf###asfadf: " + countWords("##asdfadf###asfadf") )
console.log("##13424#$#$#$%: " + countWords("##13424#$#$#$%"));
How about this regular expression: /.*?[a-z]+.*?(\s|$)/gi
Use return s.match(/.*?[a-z]+.*?(\s|$)/gi).length
Anything with at least 1 letter in it is counted. Then the phrase O##ne two $#!+ ##%Three four^&&$ five would count as 5 words.
Edit: If you want to be evil to pass your test cases when there are 0 matches use (input.match(/.*?[a-z]+.*?(\s|$)/gi) || "").length
Related
I'm trying to create a regex that will select the numbers/numbers with commas(if easier, can trim commas later) that do not have a parentheses after and not the numbers inside the parentheses should not be selected either.
Used with the JavaScript's String.match method
Example strings
9(296,178),5,3(123),10
10,9(296,178),2,5,3(123),3(124,125)
10,7,5(296,293,444,1255),3(218),2,4
What i have so far:
/((^\d+[^\(])|(,\d+,)|(,*\d+$))/gm
I tried this in regex101 and underlined the numbers i would like to match and x on the one that should not.
You could start with a substitution to remove all the unwanted parts:
/\d*\(.*?\),?//gm
Demo
This leaves you with
5,10
10,2,5,
10,7,2,4
which makes the matching pretty straight forward:
/(\d+)/gm
If you want it as a single match expression you could use a negative lookbehind:
/(?<!\([\d,]*)(\d+)(?:,|$)/gm
Demo - and here's the same matching expression as a runnable javascript (skeleton code borrowed from Wiktor's answer):
const text = `9(296,178),5,3(123),10
10,9(296,178),2,5,3(123),3(124,125)
10,7,5(296,293,444,1255),3(218),2,4`;
const matches = Array.from(text.matchAll(/(?<!\([\d,]*)(\d+)(?:,|$)/gm), x=>x[1])
console.log(matches);
Here, I'd recommend the so-called "best regex trick ever": just match what you do not need (negative contexts) and then match and capture what you need, and grab the captured items only.
If you want to match integer numbers that are not matched with \d+\([^()]*\) pattern (a number followed with a parenthetical substring), you can match this pattern or match and capture the \d+, one or more digit matching pattern, and then simply grab Group 1 values from matches:
const text = `9(296,178),5,3(123),10
10,9(296,178),2,5,3(123),3(124,125)
10,7,5(296,293,444,1255),3(218),2,4`;
const matches = Array.from(text.matchAll(/\d+\([^()]*\)|(\d+)/g), x=> x[1] ?? "").filter(Boolean)
console.log(matches);
Details:
text.matchAll(/\d+\([^()]*\)|(\d+)/g) - matches one or more digits (\d+) + ( (with \() + any zero or more chars other than ( and ) (with [^()]*) + \) (see \)), or (|) one or more digits captured into Group 1 ((\d+))
Array.from(..., x=> x[1] ?? "") - gets Group 1 value, or, if not assigned, just adds an empty string
.filter(Boolean) - removes empty strings.
Using several replacement regexes
var textA = `9(296,178),5,3(123),10
10,9(296,178),2,5,3(123),3(124,125)
10,7,5(296,293,444,1255),3(218),2,4
`
console.log('A', textA)
var textB = textA.replace(/\(.*?\),?/g, ';')
console.log('B', textB)
var textC = textB.replace(/^\d+|\d+$|\d*;\d*/gm, '')
console.log('C', textC)
var textD = textC.replace(/,+/g, ' ').trim(',')
console.log('D', textD)
With a loop
Here is a solution which splits the lines on comma and loops over the pieces:
var inside = false;
var result = [];
`9(296,178),5,3(123),10
10,9(296,178),2,5,3(123),3(124,125)
10,7,5(296,293,444,1255),3(218),2,4
`.split("\n").map(line => {
let pieceArray = line.split(",")
pieceArray.forEach((piece, k) => {
if (piece.includes('(')) {
inside = true
} else if (piece.includes(')')) {
inside = false
} else if (!inside && k > 0 && k < pieceArray.length-1 && !pieceArray[k-1].includes(')')) {
result.push(piece)
}
})
})
console.log(result)
It does print the expected result: ["5", "7"]
I'm fairly new to JavaScript (and development in general). I wanted to try a challenge from Codewars. The challenge was to process a string through a function that would flip any words that were over 5 characters and return the original string with those flipped words. Here's the code I came up with (It did work!).
//this function turns each word into an array that will get flipped.
let wordFlipper = (word) => {
var splitWord = word.split(''); //convert word to array
var reversedWord = splitWord.reverse(); //flips the indexes for the array
var joinReversedWord = reversedWord.join('').toString(); //turns array back to a string.
return joinReversedWord;
}
function spinWords(phrase){
let finalArray = [];
let wordsToArray = phrase.split(' ');
const processFlipWords = wordsToArray.forEach(word => {
if (word.toString().length > 4) {
var flippedWord = wordFlipper(word); //here's where we call the function wordFlipper()
finalArray.push(flippedWord);
}
else {
finalArray.push(word);
}
});
return finalArray.join(' ');
}
How would you experts suggest writing this? I'm sure I'm not being too efficient at writing this code.
Thank you!
Here's what it looks like inside codewars!
I'd use a regular expression to match 5 or more word characters (\w{5,}), and have a replacer function (String.replace()) return the reversed (reverse()) word:
const spinWords = phrase => phrase.replace(
/\w{5,}/g,
word => [...word].reverse().join('')
);
console.log(spinWords('foo barbar more words go here'));
\w matches a word character - something from A to Z, case-insensitive, or a digit, or an underscore.
The brackets indicates the number of times to repeat the previous token. {5,} starts with a 5 (so, "at least 5") and has nothing after the comma ("up to any number").
Then /g, the global flag, matches and replaces every substring that matches this pattern, not just the first.
The callback function runs for every matched substring, where the argument is the matched substring, and what is returned gets replaced at that point in the original string.
I'm working with a string where I need to extract the first n characters up to where numbers begin. What would be the best way to do this as sometimes the string starts with a number: 7EUSA8889er898 I would need to extract 7EUSA But other string examples would be SWFX74849948, I would need to extract SWFX from that string.
Not sure how to do this with regex my limited knowledge is blocking me at this point:
^(\w{4}) that just gets me the first four characters but I don't really have a stopping point as sometimes the string could be somelongstring292894830982 which would require me to get somelongstring
Using \w will match a word character which includes characters and digits and an underscore.
You could match an optional digit [0-9]? from the start of the string ^and then match 1+ times A-Za-z
^[0-9]?[A-Za-z]+
Regex demo
const regex = /^[0-9]?[A-Za-z]+/;
[
"7EUSA8889er898",
"somelongstring292894830982",
"SWFX74849948"
].forEach(s => console.log(s.match(regex)[0]));
Can use this regex code:
(^\d+?[a-zA-Z]+)|(^\d+|[a-zA-Z]+)
I try with exmaple and good worked:
1- somelongstring292894830982 -> somelongstring
2- 7sdfsdf5456 -> 7sdfsdf
3- 875werwer54556 -> 875werwer
If you want to create function where the RegExp is parametrized by n parameter, this would be
function getStr(str,n) {
var pattern = "\\d?\\w{0,"+n+"}";
var reg = new RegExp(pattern);
var result = reg.exec(str);
if(result[0]) return result[0].substr(0,n);
}
There are answers to this but here is another way to do it.
var string1 = '7EUSA8889er898';
var string2 = 'SWFX74849948';
var Extract = function (args) {
var C = args.split(''); // Split string in array
var NI = []; // Store indexes of all numbers
// Loop through list -> if char is a number add its index
C.map(function (I) { return /^\d+$/.test(I) === true ? NI.push(C.indexOf(I)) : ''; });
// Get the items between the first and second occurence of a number
return C.slice(NI[0] === 0 ? NI[0] + 1 : 0, NI[1]).join('');
};
console.log(Extract(string1));
console.log(Extract(string2));
Output
EUSA
SWFX7
Since it's hard to tell what you are trying to match, I'd go with a general regex
^\d?\D+(?=\d)
I am trying to remove some spaces from a few dynamically generated strings. Which space I remove depends on the length of the string. The strings change all the time so in order to know how many spaces there are, I iterate over the string and increment a variable every time the iteration encounters a space. I can already remove all of a specific type of character with str.replace(' ',''); where 'str' is the name of my string, but I only need to remove a specific occurrence of a space, not all the spaces. So let's say my string is
var str = "Hello, this is a test.";
How can I remove ONLY the space after the word "is"? (Assuming that the next string will be different so I can't just write str.replace('is ','is'); because the word "is" might not be in the next string).
I checked documentation on .replace, but there are no other parameters that it accepts so I can't tell it just to replace the nth instance of a space.
If you want to go by indexes of the spaces:
var str = 'Hello, this is a test.';
function replace(str, indexes){
return str.split(' ').reduce(function(prev, curr, i){
var separator = ~indexes.indexOf(i) ? '' : ' ';
return prev + separator + curr;
});
}
console.log(replace(str, [2,3]));
http://jsfiddle.net/96Lvpcew/1/
As it is easy for you to get the index of the space (as you are iterating over the string) , you can create a new string without the space by doing:
str = str.substr(0, index)+ str.substr(index);
where index is the index of the space you want to remove.
I came up with this for unknown indices
function removeNthSpace(str, n) {
var spacelessArray = str.split(' ');
return spacelessArray
.slice(0, n - 1) // left prefix part may be '', saves spaces
.concat([spacelessArray.slice(n - 1, n + 1).join('')]) // middle part: the one without the space
.concat(spacelessArray.slice(n + 1)).join(' '); // right part, saves spaces
}
Do you know which space you want to remove because of word count or chars count?
If char count, you can Rafaels Cardoso's answer,
If word count you can split them with space and join however you want:
var wordArray = str.split(" ");
var newStr = "";
wordIndex = 3; // or whatever you want
for (i; i<wordArray.length; i++) {
newStr+=wordArray[i];
if (i!=wordIndex) {
newStr+=' ';
}
}
I think your best bet is to split the string into an array based on placement of spaces in the string, splice off the space you don't want, and rejoin the array into a string.
Check this out:
var x = "Hello, this is a test.";
var n = 3; // we want to remove the third space
var arr = x.split(/([ ])/); // copy to an array based on space placement
// arr: ["Hello,"," ","this"," ","is"," ","a"," ","test."]
arr.splice(n*2-1,1); // Remove the third space
x = arr.join("");
alert(x); // "Hello, this isa test."
Further Notes
The first thing to note is that str.replace(' ',''); will actually only replace the first instance of a space character. String.replace() also accepts a regular expression as the first parameter, which you'll want to use for more complex replacements.
To actually replace all spaces in the string, you could do str.replace(/ /g,""); and to replace all whitespace (including spaces, tabs, and newlines), you could do str.replace(/\s/g,"");
To fiddle around with different regular expressions and see what they mean, I recommend using http://www.regexr.com
A lot of the functions on the JavaScript String object that seem to take strings as parameters can also take regular expressions, including .split() and .search().
I have searched for solution but did not find yet.
I have the following string.
1. hello
2. HELLO
3. hello_world
4. HELLO_WORLD
5. Hello World
I want to convert them to following:
1. Hello
2. Hello
3. HelloWorld
4. HelloWorld
5. HelloWorld
If there is No space and underscore in string just uppercase first and all others to lowercase. If words are separated by underscore or space then Uppercase first letter of each word and remove space and underscore. How can I do this in JavaScript.
Thanks
Here is a regex solution:
First lowercase the string:
str = str.toLowerCase();
Replace all _ and spaces and first characters in a word with upper case character:
str = str.replace(/(?:_| |\b)(\w)/g, function(str, p1) { return p1.toUpperCase()})
DEMO
Update: Less steps ;)
Explanation:
/ // start of regex
(?: // starts a non capturing group
_| |\b // match underscore, space, or any other word boundary character
// (which in the end is only the beginning of the string ^)
) // end of group
( // start capturing group
\w // match word character
) // end of group
/g // and of regex and search the whole string
The value of the capturing group is available as p1 in the function, and the whole expression is replaced by the return value of the function.
You could do something like this:
function toPascalCase(str) {
var arr = str.split(/\s|_/);
for(var i=0,l=arr.length; i<l; i++) {
arr[i] = arr[i].substr(0,1).toUpperCase() +
(arr[i].length > 1 ? arr[i].substr(1).toLowerCase() : "");
}
return arr.join("");
}
You can test it out here, the approach is pretty simple, .split() the string into an array when finding either whitespace or an underscore. Then loop through the array, upper-casing the first letter, lower-casing the rest...then take that array of title-case words and .join() it together into one string again.
function foo(str) {
return $(str.split(/\s|_/)).map(function() {
return this.charAt(0).toUpperCase() + this.slice(1).toLowerCase();
}).get().join("");
}
Working demo: http://jsfiddle.net/KSJe3/3/
(I used Nicks regular expression in the demo)
Edit: Another version of the code - I replaced map() with $.map():
function foo(str) {
return $.map(str.split(/\s|_/), function(word) {
return word.charAt(0).toUpperCase() + word.slice(1).toLowerCase();
}).join("");
}
Working demo: http://jsfiddle.net/KSJe3/4/
An ES6 / functional update of #NickCraver's answer. As with #NickCraver's answer this function will handle multiple spaces / underscores properly by filtering them out.
const pascalWord = x => x[0].toUpperCase() + x.slice(1).toLowerCase();
const toPascalCase2 = (str) => (
str.split(/\s|_/)
.filter(x => x)
.map(pascalWord)
.join('')
);
const tests = [
'hello',
'HELLO',
'hello_world',
'HELLO_WORLD',
'Hello World',
'HELLO__WORLD__',
'Hello World_',
].map(toPascalCase2).join('<br>');
document.write(tests);
var city = city.replace(/\s+/g,' ') //replace all spaceses to singele speace
city = city.replace(/\b\w/g,city => city .toUpperCase()) //after speace letter convert capital