how to return isValid = false when test(E) for both regex
let arr =[A_MANAGER,B_MANAGER,C_MANAGER,A_EMPLOYEE,C_EMPLOYEE]
arr.map((item)=>{
let isValid = ( new RegExp("MANAGER").test(item) && !new RegExp("EMPLOYEE").test(item))
console.log(isValid)
})
problem statement
let M= MANAGER, E=EMPLOYEE
M M = true
M E = true
E M = true
E E =false
One approach would be to parse the input array into two separate arrays of employees and managers. Then, starting with a collection of all people, detect whether or not a person be a manager and employee at the same time.
var arr = ["A_MANAGER", "B_MANAGER", "C_MANAGER", "A_EMPLOYEE", "C_EMPLOYEE"];
var employees = [];
var managers = [];
var all = ["A", "B", "C"];
for (var i=0; i < arr.length; ++i) {
if (/_MANAGER$/.test(arr[i])) {
managers.push(arr[i].replace(/_.*$/, ""));
}
else {
employees.push(arr[i].replace(/_.*$/, ""));
}
}
for (var i=0; i < all.length; ++i) {
if (managers.includes(all[i]) && employees.includes(all[i])) {
console.log("Person " + all[i] + " is both a manager and an employee.");
}
else {
var type = managers.includes(all[i]) ? "manager" : "employee";
console.log("Person " + all[i] + " is a " + type + " only.");
}
}
Note that ideally you should also maintain a separate list/array of all people.
You can simply do this with regex but also passing the actual array to compare values. Currently you are just matching the current item in both times and not the current and succeeding one.
You can do something like
let arr =['A_MANAGER','B_MANAGER','C_MANAGER','A_EMPLOYEE','C_EMPLOYEE']
arr.map((total,key,arr)=>{
if(arr[key+1] != undefined){
return !(new RegExp(/EMPLOYEE/).test(total) && new RegExp(/EMPLOYEE/).test(arr[key+1]));
}
//Output : True,True,True,False
You can check the working example here
https://www.w3schools.com/code/tryit.asp?filename=GS103YAMF97P
Related
I have an array which looks like this:
["1,8", "4,6,8", "8,9", "6,9"]
1/ I would like to turn it in to this
[1,8,4,6,8,8,9,6,9]
2/ I would then like to find matching values, by looking for the most number:
[8]
This first has been solved with this:
var carArray = ["1,8", "4,6,8,7,7,7,7", "8,9", "6,9"];
//1) create single array
var arr = carArray.join().split(',');
//2) find most occurring
var counts = {}; //object to hold count for each occurence
var max = 0, maxOccurring;
arr.forEach(function(el){
var cnt = (counts[el] || 0); //previous count
counts[el] = ++cnt;
if(cnt > max && cnt > 1){ //only register if more than once (cnt>1)
max=cnt;
maxOccurring = el;
}
});
if(maxOccurring){
//there was an element more than once, maxOccuring contains that element
setResult('Most occuring: ' + maxOccurring + ' (' + max + ' times)');
}
else{
//3)/4) ???
setResult('sorting?');
}
//below is only for test display purposes
function setResult(res){
console.log(res);
}
3/ If the are no matching values like this
[1,8,4,6,5,7]
4/ Then I need to compare this array to another array, such as this
[6,7,4,1,2,8,9,5]
If the first number in <4> array above appears in <3> array, then get that number, ie in the above example I need to get 6. The <4> array will be static values and not change. The numbers is <3> will be dynamic.
EDIT Not the most elegant of answers, but I do have something working now. I didn't compare the original array directly with the second array, instead used simple if/else statements to do what I needed:
var carArray = ["1,5", "4", "8,2", "3,9,1,1,1"];
//1) create single array
var arr = carArray.join().split(',');
//2) find most occurring
var counts = {}; //object to hold count for each occurence
var max = 0, maxOccurring;
arr.forEach(function(el){
var cnt = (counts[el] || 0); //previous count
counts[el] = ++cnt;
if(cnt > max && cnt > 1){ //only register if more than once (cnt>1)
max=cnt;
maxOccurring = el;
}
});
if(maxOccurring){
//there was an element more than once, maxOccuring contains that element
console.log('Most occuring: ' + maxOccurring + ' (' + max + ' times)');
console.log(maxOccurring);
}
else {
// If not occuring, match from a list
if(jQuery.inArray("6", arr) !== -1) { console.log('6'); }
else if(jQuery.inArray("9", arr) !== -1) { console.log('9'); }
else if(jQuery.inArray("7", arr) !== -1) { console.log('7'); }
else if(jQuery.inArray("5", arr) !== -1) { console.log('5'); }
else if(jQuery.inArray("4", arr) !== -1) { console.log('4'); }
else if(jQuery.inArray("1", arr) !== -1) { console.log('1'); }
else { console.log('not found'); }
}
Example Fiddle
Step 1 is fairly easy by using javascript's join and split methods respectively:
var arr = carArray .join().split(',');
For step 2, several methods can be used, the most common one using an object and using the elements themselves as properties. Since you only need to get the most occurring value if there is a reoccurring value, it can be used in the same loop:
var counts = {}; //object to hold count for each occurence
var max = 0, maxOccurring;
arr.forEach(function(el){
var cnt = (counts[el] || 0); //previous count
counts[el] = ++cnt;
if(cnt > max && cnt > 1){ //only register if more than once (cnt>1)
max=cnt;
maxOccurring = el;
}
});
After the above, the variable maxOccurring will contain the reoccurring value (if any) and max will contain the times it occured
For step 4 the easiest way is to loop through the compare array and get the element that occurs in the input array:
var cmpArr = ['6','7','4','1','2','8','9','5'];
//find the first occurrence inside the cmpArr
res = function(){ for(var i= 0 ; i < cmpArr.length; i++){ if(arr.indexOf(cmpArr[i]) !== -1)return cmpArr[i];}}();
The above uses an in place function which is called immediately to be able to use return. You could also just use a loop and assign res when found, then break from the loop.
Last update, an alternate fiddle where the above is converted to a single function: http://jsfiddle.net/v9hhsdny/5/
Well first of all the following code results in four matching answers since the jQuery selectors are the same.
var questionAnswer1 = $(this).find('input[name=questionText]').val();
var questionAnswer2 = $(this).find('input[name=questionText]').val();
var questionAnswer3 = $(this).find('input[name=questionText]').val();
var questionAnswer4 = $(this).find('input[name=questionText]').val();
var carArray = [questionAnswer1, questionAnswer2, questionAnswer3, questionAnswer4];
You could use the eq(index) method of jQuery to select the appropriate element. However having 4 inputs with the same name is a bad practice.
Well lets say that the carArray has 4 different values which all consist out of comma separated numbers. You could then do the following:
var newArr = [];
carArray.forEach(function(e) {
e.split(",").forEach(function(n) {
newArr.push(n);
});
});
Well then we got to find the most occurring number. JavaScript doesn't have any functions for that so we will have to find an algorithm for that. I found the following algorithm on this stackoverflow page
var count = function(ary, classifier) {
return ary.reduce(function(counter, item) {
var p = (classifier || String)(item);
counter[p] = counter.hasOwnProperty(p) ? counter[p] + 1 : 1;
return counter;
}, {})
}
var occurances = count(newArr);
It isn't clear to me what you're trying to do in step 3 and 4, so can't answer those at the moment.
var ary = ["1,8", "4,6,8", "8,9", "6,9"];
var splitted = ary.reduce(function(acc, item) {
return acc.concat(item.split(','));
}, []);
var occurences = splitted.reduce(function(acc, item) {
if (!acc.hasOwnProperty(item)) acc[item] = 0;
acc[item] += 1;
return acc;
},{}),
biggest = Object.keys(occurences).reduce(function (acc, key) {
if (occurences[key] > acc.occurences) {
acc.name = key;
acc.occurences = occurences[key];
}
return acc;
},{'name':'none','occurences':0}).name;
var vals=["1,8", "4,6,8", "8,9", "6,9"];
// 1) turn into number array
var arrNew=[];
for(var i=0; i<vals.length; i++)
{
arrLine=vals[i].split(",");
for (var j=0;j<arrLine.length;j++) { arrNew.push (parseInt(arrLine[j])) }
}
//result:
alert(arrNew.join(";");
// 2) find most common
var found=[];
for(var i=0; i<arrNew.length; i++) {
// make an array of the number of occurrances of each value
if (found["num"+newArray[i]]) {
found["num"+newArray[i]] ++ ;
} else {
found["num"+newArray[i]]=1;
}
}
var mostCommon={count:0,val:"ROGUE"};
for (x in found) {
if (found[x] > mostCommon.count) {
mostCommon.count=found[x].count;
mostCommon.val=x;
}
}
// result :
alert(mostCommon.val);
//3) not quite sure what you meant there
// 4) unique values:
// at this point the 'found' list contains unique vals
var arrUnique=[];
for (x in found) {
arrUnique.push[x];
}
// result :
alert(arrUnique.join(";"))
//sort:
arrUnique.sort(function(a, b){return a-b});
(This won't work in most browsers) but on a side note, when ES6 becomes widely supported, your solution could look like this:
var arr1 = ["1,8", "4,6,8", "8,9", "6,9"];
var arr2 = arr1.join().split(',');
var s = Array.from(new Set(arr2)); //Array populated by unique values, ["1", "8", "4", "6", "9"]
Thought you might like to see a glimpse of the future!
1.
var orgArray = ['1,8', '4,6,8', '8,9', '6,9'];
var newArray = [];
for (var i in orgArray) {
var tmpArray = orgArray[i].split(',');
for (var j in tmpArray) {
newArray.push(Number(tmpArray[j]));
}
}
2.
var counts = {};
var most = null;
for (var i in newArray) {
var num = newArray[i];
if (typeof counts[num] === 'undefined') {
counts[num] = 1;
} else {
++(counts[num]);
}
if (most == null || counts[num] > counts[most]) {
most = num;
} else if (most != null && counts[num] === counts[most]) {
most = null;
}
}
I don't understand the question 3 and 4 (what "unique order" means) so I can't answer those questions.
I got this string:
var longText="This is a superuser test, super user is is super important!";
I want to know how many times the string "su" is in longText and the position of each "su".
I was trying with:
var nr4 = longText.replace("su", "").length;
And the difference of lenght between the main text and the nr4 divided by "su" lenght beeing 2 is resulting a number of repetitions but i bet there is a better way of doing it.
For example
var parts=longText.split("su");
alert(parts.length-1); // length will be two if there is one "su"
More details using exec
FIDDLE
var re =/su/g, pos=[];
while ((result = re.exec(longText)) !== null) {
pos.push(result.index);
}
if (pos.length>0) alert(pos.length+" found at "+pos.join(","));
Use exec. Example amended from the MDN code. len contains the number of times su appears.
var myRe = /su/g;
var str = "This is a superuser test, super user is is super important!";
var myArray, len = 0;
while ((myArray = myRe.exec(str)) !== null) {
len++;
var msg = "Found " + myArray[0] + ". ";
msg += "Next match starts at " + myRe.lastIndex;
console.log(msg, len);
}
// "Found su. Next match starts at 12" 1
// "Found su. Next match starts at 28" 2
// "Found su. Next match starts at 45" 3
DEMO
Could do :
var indexesOf = function(baseString, strToMatch){
var baseStr = new String(baseString);
var wordLen = strToMatch.length;
var listSu = [];
// Number of strToMatch occurences
var nb = baseStr.split(strToMatch).length - 1;
for (var i = 0, len = nb; i < len; i++){
var ioF = baseStr.indexOf(strToMatch);
baseStr = baseStr.slice(ioF + wordLen, baseStr.length);
if (i > 0){
ioF = ioF + listSu[i-1] + wordLen;
}
listSu.push(ioF);
}
return listSu;
}
indexesOf("This is a superuser test, super user is is super important!","su");
return [10, 26, 43]
var longText="This is a superuser test, super user is is super important!";
var count = 0;
while(longText.indexOf("su") != -1) { // NB the indexOf() method is case sensitive!
longText = longText.replace("su",""); //replace first occurence of 'su' with a void string
count++;
}
I asked a pretty detailed question earlier but it got flagged as a duplicate even though it wasn't.
So I will try and explain what I am trying to do as simply as I can. I want to search a text string for words that begin with specific letters such as "mak", "mind" and "mass" (which are in an array) and end with either nothing extra or "e" or "er". That would be in this instance "mak", "make", "maker", "mind", "minde", "minder", "mass", "masse", "masser".
The code I am using only finds the first match for each word in the array, for instance "mak", "mind" and "mass" in the example.
derPro = ['mak','mind', 'mass', ;
for(i = 0; i < derPro.length; i++){
searchTerm = new RegExp(
"\\b" + derPro[i] + "\\b|" +
"\\b" + derPro[i] + "e\\b|" +
"\\b" + derPro[i] + "er\\b,'gi'");
word = testText.match(searchTerm, "gi");
This should work:
var derPro = ['mak','mind', 'mass'];
var searchTerm = new RegExp('\\b((?:' + derPro.join('|') + ')(?:er?)?)\\b', "gi");
//=> /\b((?:mak|mind|mass)(?:er?)?)\b/gi
// now match the regex in a while loop
var matches=[]
while (m = searchTerm.exec('mass maker minde')) matches.push(m[1]);
console.log(matches);
//=> ["mass", "maker", "minde"]
Get the matched group from index 1.
/(\b(mak|mind|mass)(e|er)?\b)/gi
Online demo
This works, i know you asked for a regex, but I thought this would be useful.
var keys = [
'mak',
'mind',
'mass'
];
var test_words = [
'mak',
'make',
'maker',
'mind',
'mass'
];
var matches = [];
for(var i = 0; i < keys.length; i++) {
var key = keys[i];
for(var ii = 0; ii < test_words.length; ii++) {
var test_word = test_words[ii];
if(test_word.substr(0, key.length) == key) {
if( test_word.substr(-1) == 'e' || test_word.substr(-2) == 'er') {
matches.push(test_word);
}
}
}
}
console.log(matches);
// [make, maker, minde]
jsfiddle
I'm currently implementing a substring search. From the algorithm, I get array of substrings occurence positions where each element is in the form of [startPos, endPos].
For example (in javascript array):
[[1,3], [8,10], [15,18]]
And the string to highlight is:
ACGATCGATCGGATCGAGCGATCGAGCGATCGAT
I want to highlight (in HTML using <b>) the original string, so it will highlight or bold the string from position 1 to 3, then 8 to 10, then 15 to 18, etc (0-indexed).
A<b>CGA</b>TCGA<b>TCG</b>GATC<b>GAGC</b>GATCGAGCGATCGAT
This is what I have tried (JavaScript):
function hilightAtPositions(text, posArray) {
var startPos, endPos;
var startTag = "<b>";
var endTag = "</b>";
var hilightedText = "";
for (var i = 0; i < posArray.length; i++) {
startPos = posArray[i][0];
endPos = posArray[i][1];
hilightedText = [text.slice(0, startPos), startTag, text.slice(startPos, endPos), endTag, text.slice(endPos)].join('');
}
return hilightedText;
}
But it highlights just a range from the posArray (and I know it is still incorrect yet). So, how can I highlight a string given multiple occurrences position?
Looking at this question, and following John3136's suggestion of going from tail to head, you could do:
String.prototype.splice = function( idx, rem, s ) {
return (this.slice(0,idx) + s + this.slice(idx + Math.abs(rem)));
};
function hilightAtPositions(text, posArray) {
var startPos, endPos;
posArray = posArray.sort(function(a,b){ return a[0] - b[0];});
for (var i = posArray.length-1; i >= 0; i--) {
startPos = posArray[i][0];
endPos = posArray[i][1];
text= text.splice(endPos, 0, "</b>");
text= text.splice(startPos, 0, "<b>");
}
return text;
}
Note that in your code, you are overwriting hilightedText with each iteration, losing your changes.
Try this:
var stringToHighlight = "ACGATCGATCGGATCGAGCGATCGAGCGATCGAT";
var highlightPositions = [[1,3], [8,10], [15,18]];
var lengthDelta = 0;
for (var highlight in highlightPositions) {
var start = highlightPositions[highlight][0] + lengthDelta;
var end = highlightPositions[highlight][1] + lengthDelta;
var first = stringToHighlight.substring(0, start);
var second = stringToHighlight.substring(start, end + 1);
var third = stringToHighlight.substring(end + 1);
stringToHighlight = first + "<b>" + second + "</b>" + third;
lengthDelta += ("<b></b>").length;
}
alert(stringToHighlight);
Demo: http://jsfiddle.net/kPkk3/
Assuming that you're trying to highlight search terms or something like that. Why not replace the term with the bolding?
example:
term: abc
var text = 'abcdefgabcqq';
var term = 'abc';
text.replace(term, '<b>' + term + '</b>');
This would allow you to avoid worrying about positions, assuming that you are trying to highlight a specific string.
Assuming your list of segments is ordered from lowest start to highest, try doing your array from last to first.
That way you are not changing parts of the string you haven't reached yet.
Just change the loop to:
for (var i = posArray.length-1; i >=0; i--) {
If you want to check for multiple string matches and highlight them, this code snippet works.
function highlightMatch(text, matchString) {
let textArr = text.split(' ');
let returnArr = [];
for(let i=0; i<textArr.length; i++) {
let subStrMatch = textArr[i].toLowerCase().indexOf(matchString.toLowerCase());
if(subStrMatch !== -1) {
let subStr = textArr[i].split('');
let subStrReturn = [];
for(let j=0 ;j<subStr.length; j++) {
if(j === subStrMatch) {
subStrReturn.push('<strong>' + subStr[j]);
} else if (j === subStrMatch + (matchString.length-1)){
subStrReturn.push(subStr[j] + '<strong>');
} else {
subStrReturn.push(subStr[j]);
}
}
returnArr.push(subStrReturn.join(''));
} else {
returnArr.push(textArr[i]);
}
}
return returnArr;
}
highlightMatch('Multi Test returns multiple results', 'multi');
=> (5) ['<strong>Multi<strong>', 'Test', 'returns', '<strong>multi<strong>ple', 'results']
I have a string that looks something like the following 'test:1;hello:five;just:23'. With this string I need to be able to do the following.
....
var test = MergeTokens('test:1;hello:five;just:23', 'yes:23;test:567');
...
The end result should be 'test:567;hello:five;just:23;yes:23' (note the exact order of the tokens is not that important).
Just wondering if anyone has any smart ideas of how to go about this. I was thinking a regex replace on each of the tokens on right and if a replace didn't occur because there was not match just append it. But maybe there is better way.
Cheers
Anthony
Edit: The right side should override the left. The left being what was originally there and the right side being the new content. Another way of looking at it, is that you only keep the tokens on the left if they don't exist on the right and you keep all the tokens on the right.
#Ferdinand
Thanks for the reply. The problem is the efficiency with which the solution you proposed. I was initially thinking down similar lines but discounted it due to the O(n*z) complexity of the merge (where n and z is the number tokens on the left and right respectively) let alone the splitting and joining.
Hence why I was trying to look down the path of a regex. Maybe behind the scenes, regex is just as bad or worse, but having a regex which removes any token from the left string that exists on the right (O(n) for the total amount of token on the right) and then just add the 2 string together (i.e. vat test = test1 + test2) seems more efficient. thanks
I would use join() and split() to create some utility functions to pack and unpack your token data to an object:
// Unpacks a token string into an object.
function splitTokens(str) {
var data = {}, pairs = str.split(';');
for (var i = 0; i < pairs.length; ++i) {
var pair = pairs[i].split(':');
data[pair[0]] = pair[1];
}
return data;
}
// Packs an object into a token string.
function joinTokens(data) {
var pairs = [];
for (var key in data) {
pairs.push(key + ":" + data[key]);
}
return pairs.join(';');
}
Using these, merging is easy:
// Merges all token strings (supports a variable number of arguments).
function mergeTokens() {
var data = {};
for (var i = 0; i < arguments.length; ++i) {
var d = splitTokens(arguments[i]);
for (var key in d) {
data[key] = d[key];
}
}
return joinTokens(data);
}
The utility functions are also useful if you want to extract some keys (say,"test") and/or check for existence:
var data = splitTokens(str);
if (data["test"] === undefined) {
// Does not exist
} else {
alert("Value of 'test': " + data["test"]);
}
The following is what I ended thiking about. What do you guys recon?
Thanks
Anthony
function Tokenizer(input, tokenSpacer, tokenValueSpacer) {
this.Tokenizer = {};
this.TokenSpacer = tokenSpacer;
this.TokenValueSpacer = tokenValueSpacer;
if (input) {
var TokenizerParts = input.split(this.TokenSpacer);
var i, nv;
for (i = 0; i < TokenizerParts.length; i++) {
nv = TokenizerParts[i].split(this.TokenValueSpacer);
this.Tokenizer[nv[0]] = nv[1];
}
}
}
Tokenizer.prototype.add = function(name, value) {
if (arguments.length == 1 && arguments[0].constructor == Object) {
this.addMany(arguments[0]);
return;
}
this.Tokenizer[name] = value;
}
Tokenizer.prototype.addMany = function(newValues) {
for (nv in newValues) {
this.Tokenizer[nv] = newValues[nv];
}
}
Tokenizer.prototype.remove = function(name) {
if (arguments.length == 1 && arguments[0].constructor == Array) {
this.removeMany(arguments[0]);
return;
}
delete this.Tokenizer[name];
}
Tokenizer.prototype.removeMany = function(deleteNames) {
var i;
for (i = 0; i < deleteNames.length; i++) {
delete this.Tokenizer[deleteNames[i]];
}
}
Tokenizer.prototype.MergeTokenizers = function(newTokenizer) {
this.addMany(newTokenizer.Tokenizer);
}
Tokenizer.prototype.getTokenString = function() {
var nv, q = [];
for (nv in this.Tokenizer) {
q[q.length] = nv + this.TokenValueSpacer + this.Tokenizer[nv];
}
return q.join(this.TokenSpacer);
}
Tokenizer.prototype.toString = Tokenizer.prototype.getTokenString;
i am a few years late, but i think this is what you are looking for:
function MergeTokens(input, replace){
var replaceTokens = replace.split(";");
for(i=0; i<replaceTokens.length; i++){
var pair = replaceTokens[i].split(":");
var result = input;
regString = "\\b" + pair[0] + ":[\\w]*\\b";
var reg = new RegExp(regString);
if(reg.test(result)){
result = result.replace(reg, replaceTokens[i]);
}
else{
result = result + replaceTokens[i];
}
}
return result;
}