I have a variable consisting of multiple sentences separated with dots named var a. Then I want to check if Result value is equal to one of the sentences on var a:
Here is the code:
var a = "my name is jack. your name is sara. her name is joe. his name is mike";
var nameSplit = a.split(".");
console.log(nameSplit.length);
console.log(nameSplit);
var Result = "his name is mike";
var i;
for (i = 0; i < nameSplit.length; i++) {
if (nameSplit[i].includes(Result)) {
console.log("Result is one of the sentences on var a");
}
}
The code works fine .
The problem is I have no idea if this code works fast and efficient if I have 1000 sentences or 10000 sentences on var a?
Is there any modification that makes the code faster to execute?
Is there any better solution to do this?
You can use Array.prototype.includes to check if an array includes a string. or String.prototype.includes to match a substring.
var a = "my name is jack. your name is sara. her name is joe. his name is mike";
var nameSplit = a.split(". ");
var Result = "his name is mike";
nameSplit.includes(Result) // true
or
var a = "my name is jack. your name is sara. her name is joe. his name is mike";
var Result = "his name is mike";
a.includes(Result) // true
You need to ask yourself how many substring you want to find in a, if just one, then you dont need to split as split itself takes time and space. (If interested , you can further check KMP algorithm which can efficiently match word W in a string S.)
If you have a huge amount of substrings to be matched, you then can use split, which costs more space to trade off time.
I think the fastest way should be with
str.indexOf('his name is mike') !== -1
but for 1k words i dont think its relevant
Your question says you "want to check if Result value is equal to one of the sentences" but your code doesn't check for equality. It allows partial matches — for example if Result is i it matches the i in my name is jack..
If you want to match the whole sentences, you could split the sentences into a Set and then checking for matches is a constant time operation. But they'll need to be exact matches, including whitespace.
var a = "my name is jack. your name is sara. her name is joe. his name is mike";
var nameSplit = new Set(a.split(/\.\s+/g))
var Result = "his name is mike";
if (nameSplit.has(Result)) {
console.log("Result is one of the sentences on var a");
}
Result = "his name is";
if (nameSplit.has(Result)) {
console.log("Result is one of the sentences on var a");
} else {
console.log("not in sentences")
}
Related
The code the presence of a single word in a sentence and it's working fine.
var str ="My best food is beans and plantain. Yam is also good but I prefer yam porrage"
if(str.match(/(^|\W)food($|\W)/)) {
alert('Word Match');
//alert(' The matched word is' +matched_word);
}else {
alert('Word not found');
}
Here is my issue: I need to check presence of multiple words in a sentence (eg: food,beans,plantains etc) and then also alert the matched word.
something like //alert(' The matched word is' +matched_word);
I guess I have to passed the searched words in an array as per below:
var words_checked = ["food", "beans", "plantain"];
You can construct a regular expression by joining the array of words by |, then surround it with word boundaries \b:
var words_checked = ['foo', 'bar', 'baz']
const pattern = new RegExp(String.raw`\b(?:${words_checked.join('|')})\b`);
var str = 'fooNotAStandaloneWord baz something';
console.log('Match:', str.match(pattern)[0]);
Here's a way to solve this. Simply loop through the list of words to check, build the regex as you go and check to see if there is a match. You can read up on how to build Regexp objects here
var str ="My best food is beans and plantain. Yam is also good but I prefer
yam porrage"
var words = [
"food",
"beans",
"plantain",
"potato"
]
for (let word of words) {
let regex = new RegExp(`(^|\\W)${word}($|\\W)`)
if (str.match(regex)) {
console.log(`The matched word is ${word}`);
} else {
console.log('Word not found');
}
}
var text = "I am happy, We all are happy";
var count = countOccurences(text, "happy");
// count will return 2
//I am passing the whole line and also the word for which i want to find the number of occurences
// The code splits the string and find the length of the word
function countOccurences(string, word){
string.split(word).length - 1;
}
for example i have this two strings:
string1:
"hi sir may name is Jone"
string2
"hi may name is Jone"
i have this this regex:
var regex = XRegExp('hi(?:(?!hi|Jone).)*?Jone', 'gs');
will match both of them but i want to modify the regex to match only in limited length of the whole string
i want to match the string two "hi may name is Jone" as had less words length how to do it..
If you want to get the string with the least amount of words that also matches your regex, you could split and use a whitespace as a separator and check the length of the returned array.
As an example with an array of strings, you could create var longestString = ""; which will at the end of the loop contain the shortest matched string.
In the loop, first check if there is a match and if longestString is an empty string. If that is the case then set the variable so you have a match to compare against future possible matches.
var strings = [
"test",
"hi sir may name is Jone",
"hi may name is Jone",
"hi Jone",
"hi may name is Jone test",
"hi i am Jone",
"may name is Jone test",
"hi may name is Jane test test 2"
];
var regex = /hi(?:(?!hi|Jone).)*?Jone/;
var longestString = "";
strings.forEach((str) => {
var match = XRegExp.match(str, regex);
if (match && longestString === "") {
longestString = str;
return;
}
if (match && str.split(" ").length < longestString.split(" ").length) {
longestString = str;
}
});
console.log(longestString);
<script src="https://unpkg.com/xregexp/xregexp-all.js"></script>
If you want to match a limited length of the whole string using only regex, I think you can do something like this:
var index = 15;
var testString1 = "hi may name is Jone";
var testString2 = "hi sir may name is Jone";
testString1.match("^.{1," + index + "}Jone"); // This will match
testString2.match("^.{1," + index + "}Jone"); // This string is longer and will not match
Explanation of the regex ^.{1, n}Jone.
^ : should match the start of the string.
.{1, n}Jone : matches everything between 1 to n until the pattern is fully matched.
In this case we define the n as index so this "limit" can be dynamic.
Hope this helps!
I'd like to know if it is possible to replace every matching pattern in the string with not one but different values each time.
Let's say I found 5 matches in a text and I want to replace first match with a string, second match with another string, third match with another and so on... is it achievable?
var synonyms = ["extremely", "exceedingly", "exceptionally", "especially", "tremendously"];
"I'm very upset, very distress, very agitated, very annoyed and very pissed".replace(/very/g, function() {
//replace 5 matches of the keyword every with 5 synonyms in the array
});
You may try to replace the matches inside a replace callback function:
var synonyms = ["extremely", "exceedingly", "exceptionally", "especially", "tremendously"];
var cnt = 0;
console.log("I'm very upset, very distress, very agitated, very annoyed and very pissed (and very anxious)".replace(/very/g, function($0) {
if (cnt === synonyms.length) cnt = 0;
return synonyms[cnt++]; //replace 5 matches of the keyword every with 5 synonyms in the array
}));
If you have more matches than there are items in the array, the cnt will make sure the array items will be used from the first one again.
A simple recursive approach. Be sure your synonyms array has enough elements to cover all matches in your string.
let synonyms = ["extremely", "exceedingly", "exceptionally"]
let yourString = "I'm very happy, very joyful, and very handsome."
let rex = /very/
function r (s, i) {
let newStr = s.replace(rex, synonyms[i])
if (newStr === s)
return s
return r(newStr, i+1)
}
r(yourString, 0)
I would caution that if your replacement would also match your regex, you need to add an additional check.
function replaceExpressionWithSynonymsInText(text, regX, synonymList) {
var
list = [];
function getSynonym() {
if (list.length <= 0) {
list = Array.from(synonymList);
}
return list.shift();
}
return text.replace(regX, getSynonym);
}
var
synonymList = ["extremely", "exceedingly", "exceptionally", "especially", "tremendously"],
textSource = "I'm very upset, very distress, very agitated, very annoyed and very pissed",
finalText = replaceExpressionWithSynonymsInText(textSource, (/very/g), synonymList);
console.log("synonymList : ", synonymList);
console.log("textSource : ", textSource);
console.log("finalText : ", finalText);
The advantages of the above approach are, firstly one does not alter the list of synonyms,
secondly working internally with an ever new copy of the provided list and shifting it,
makes additional counters obsolete and also provides the opportunity of being able to
shuffle the new copy (once it has been emptied), thus achieving a more random replacement.
Using the example you've provided, here's what I would do.
First I would set up some variables
var text = "I'm very upset, very distress, very agitated, very annoyed and very pissed";
var regex = /very/;
var synonyms = ["extremely", "exceedingly", "exceptionally", "especially", "tremendously"];
Then count the number of matches
var count = text.match(/very/g).length;
Then I would run a loop to replace the matches with the values from the array
for(var x = 0; x < count; x++) {
text = text.replace(regex, synonyms[x]);
}
You can do it with the use of Replace() function, where you use 'g' option for global matching (finds all occurrences of searched expression). For the second argument you can use a function which returns values from your predefined array.
Here is a little fiddle where you can try it out.
var str = "test test test";
var rep = ["one", "two", "three"];
var ix = 0;
var res = str.replace(/test/g, function() {
if (ix == rep.length)
ix = 0;
return rep[ix++];
});
$("#result").text(res);
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<p id="result">
Result...
</p>
Yes it is achievable. There may be a more efficient answer than this, but the brute force way is to double the length of your regex. i.e. Instead of searching just A, search (/A){optionalText}(/A) and then replace /1 /2 as needed. If you need help with the regex itself, provide some code for what you're searching for and someone with more rep than me can probably comment the actual regexp.
I have a text which goes like this...
var string = '~a=123~b=234~c=345~b=456'
I need to extract the string such that it splits into
['~a=123~b=234~c=345','']
That is, I need to split the string with /b=.*/ pattern but it should match the last found pattern. How to achieve this using RegEx?
Note: The numbers present after the equal is randomly generated.
Edit:
The above one was just an example. I did not make the question clear I guess.
Generalized String being...
<word1>=<random_alphanumeric_word>~<word2>=<random_alphanumeric_word>..~..~..<word2>=<random_alphanumeric_word>
All have random length and all wordi are alphabets, the whole string length is not fixed. the only text known would be <word2>. Hence I needed RegEx for it and pattern being /<word2>=.*/
This doesn't sound like a job for regexen considering that you want to extract a specific piece. Instead, you can just use lastIndexOf to split the string in two:
var lio = str.lastIndexOf('b=');
var arr = [];
var arr[0] = str.substr(0, lio);
var arr[1] = str.substr(lio);
http://jsfiddle.net/NJn6j/
I don't think I'd personally use a regex for this type of problem, but you can extract the last option pair with a regex like this:
var str = '~a=123~b=234~c=345~b=456';
var matches = str.match(/^(.*)~([^=]+=[^=]+)$/);
// matches[1] = "~a=123~b=234~c=345"
// matches[2] = "b=456"
Demo: http://jsfiddle.net/jfriend00/SGMRC/
Assuming the format is (~, alphanumeric name, =, and numbers) repeated arbitrary number of times. The most important assumption here is that ~ appear once for each name-value pair, and it doesn't appear in the name.
You can remove the last token by a simple replacement:
str.replace(/(.*)~.*/, '$1')
This works by using the greedy property of * to force it to match the last ~ in the input.
This can also be achieved with lastIndexOf, since you only need to know the index of the last ~:
str.substring(0, (str.lastIndexOf('~') + 1 || str.length() + 1) - 1)
(Well, I don't know if the code above is good JS or not... I would rather write in a few lines. The above is just for showing one-liner solution).
A RegExp that will give a result that you may could use is:
string.match(/[a-z]*?=(.*?((?=~)|$))/gi);
// ["a=123", "b=234", "c=345", "b=456"]
But in your case the simplest solution is to split the string before extract the content:
var results = string.split('~'); // ["", "a=123", "b=234", "c=345", "b=456"]
Now will be easy to extract the key and result to add to an object:
var myObj = {};
results.forEach(function (item) {
if(item) {
var r = item.split('=');
if (!myObj[r[0]]) {
myObj[r[0]] = [r[1]];
} else {
myObj[r[0]].push(r[1]);
}
}
});
console.log(myObj);
Object:
a: ["123"]
b: ["234", "456"]
c: ["345"]
(?=.*(~b=[^~]*))\1
will get it done in one match, but if there are duplicate entries it will go to the first. Performance also isn't great and if you string.replace it will destroy all duplicates. It would pass your example, but against '~a=123~b=234~c=345~b=234' it would go to the first 'b=234'.
.*(~b=[^~]*)
will run a lot faster, but it requires another step because the match comes out in a group:
var re = /.*(~b=[^~]*)/.exec(string);
var result = re[1]; //~b=234
var array = string.split(re[1]);
This method will also have the with exact duplicates. Another option is:
var regex = /.*(~b=[^~]*)/g;
var re = regex.exec(string);
var result = re[1];
// if you want an array from either side of the string:
var array = [string.slice(0, regex.lastIndex - re[1].length - 1), string.slice(regex.lastIndex, string.length)];
This actually finds the exact location of the last match and removes it regex.lastIndex - re[1].length - 1 is my guess for the index to remove the ellipsis from the leading side, but I didn't test it so it might be off by 1.
From server I get data as such:
"07.00 PROGRAM DESCRIPTION"
"07.20 PROGRAM DESCRIPTION 2"
I want to split them into a 2 indexed array such as: ["07.00", "PROGRAM DESCRIPTION 2"]. Regular split( " " ) would not work me as the description part contains severaral " " spaces.
I will be grateful for any suggestion.
Regards
You could use:
var parts = str.split(' '),
time = parts.shift(),
description = parts.join(' ');
or, to get your array:
var parts = str.split(' ');
parts[1] = parts.slice(1).join(' ');
;)
You need somekind of a pattern, which is reliable. If it's always the case that you need to split just between the first whitespace character to you can do:
var blub = "07.00 PROGRAM DESCRIPTION",
pos = blub.indexOf(" "),
arr = [];
arr[0] = blub.slice(0, pos);
arr[1] = blub.slice(pos + 1);
or you might just want to use regular expression. Since I don't pretend to be a genius on that field here is my little suggestion:
var blub = "07.00 PROGRAM DESCRIPTION",
arr = /(\d+\.\d+)\s(.*)/.exec(blub);
var pattern = /([0-9]{2}\.[0-9]{2})\s(.+)/;
var data = "07.00 PROGRAM DESCRIPTION";
var parsed = pattern.exec(data);
console.log(parsed); // (Array) ["07.00 PROGRAM DESCRIPTION", "07.00", "PROGRAM DESCRIPTION"]
this is flexible and easier to adapt in case the format changes (just change the pattern and use that var anywhere else in your code)
The substring method:
var time = row.substring(0,4);
var description = row.substring(5);
Or, with the split method:
row = row.split(" ",1);
The second parameter is the maximum number of splits... so it'll only split at the first space. Edit: this won't work. Use the first method instead.