I store some parameters client-side in HTML and then need to compare them as integers. Unfortunately I have come across a serious bug that I cannot explain. The bug seems to be that my JS reads parameters as strings rather than integers, causing my integer comparisons to fail.
I have generated a small example of the error, which I also can't explain. The following returns 'true' when run:
console.log("2" > "10")
Parse the string into an integer using parseInt:
javascript:alert(parseInt("2", 10)>parseInt("10", 10))
Checking that strings are integers is separate to comparing if one is greater or lesser than another. You should always compare number with number and string with string as the algorithm for dealing with mixed types not easy to remember.
'00100' < '1' // true
as they are both strings so only the first zero of '00100' is compared to '1' and because it's charCode is lower, it evaluates as lower.
However:
'00100' < 1 // false
as the RHS is a number, the LHS is converted to number before the comparision.
A simple integer check is:
function isInt(n) {
return /^[+-]?\d+$/.test(n);
}
It doesn't matter if n is a number or integer, it will be converted to a string before the test.
If you really care about performance, then:
var isInt = (function() {
var re = /^[+-]?\d+$/;
return function(n) {
return re.test(n);
}
}());
Noting that numbers like 1.0 will return false. If you want to count such numbers as integers too, then:
var isInt = (function() {
var re = /^[+-]?\d+$/;
var re2 = /\.0+$/;
return function(n) {
return re.test((''+ n).replace(re2,''));
}
}());
Once that test is passed, converting to number for comparison can use a number of methods. I don't like parseInt() because it will truncate floats to make them look like ints, so all the following will be "equal":
parseInt(2.9) == parseInt('002',10) == parseInt('2wewe')
and so on.
Once numbers are tested as integers, you can use the unary + operator to convert them to numbers in the comparision:
if (isInt(a) && isInt(b)) {
if (+a < +b) {
// a and b are integers and a is less than b
}
}
Other methods are:
Number(a); // liked by some because it's clear what is happening
a * 1 // Not really obvious but it works, I don't like it
Comparing Numbers to String Equivalents Without Using parseInt
console.log(Number('2') > Number('10'));
console.log( ('2'/1) > ('10'/1) );
var item = { id: 998 }, id = '998';
var isEqual = (item.id.toString() === id.toString());
isEqual;
use parseInt and compare like below:
javascript:alert(parseInt("2")>parseInt("10"))
Always remember when we compare two strings.
the comparison happens on chacracter basis.
so '2' > '12' is true because the comparison will happen as
'2' > '1' and in alphabetical way '2' is always greater than '1' as unicode.
SO it will comeout true.
I hope this helps.
You can use Number() function also since it converts the object argument to a number that represents the object's value.
Eg: javascript:alert( Number("2") > Number("10"))
+ operator will coerce the string to a number.
console.log( +"2" > +"10" )
The answer is simple. Just divide string by 1.
Examples:
"2" > "10" - true
but
"2"/1 > "10"/1 - false
Also you can check if string value really is number:
!isNaN("1"/1) - true (number)
!isNaN("1a"/1) - false (string)
!isNaN("01"/1) - true (number)
!isNaN(" 1"/1) - true (number)
!isNaN(" 1abc"/1) - false (string)
But
!isNaN(""/1) - true (but string)
Solution
number !== "" && !isNaN(number/1)
The alert() wants to display a string, so it will interpret "2">"10" as a string.
Use the following:
var greater = parseInt("2") > parseInt("10");
alert("Is greater than? " + greater);
var less = parseInt("2") < parseInt("10");
alert("Is less than? " + less);
I need to sort a list by the second batch number with JavaScript.
This is how it looks like now:
1101.19
1201.17
1301.09
What I need is:
1301.09
1201.17
1101.19
As I am still learning to program I can't figure out the issue. But need it at work.
Can someone help me understand the process of how to do it?
Sort the array depending on the decimal part. Here is the solution
Sort the array by selecting the decimal part of the number inside the sort function.
You can get the decimal part of any number by taking modulus operation with 0.1. Link.
const arr = [1101.19, 1201.17, 1301.09, 1201.20];
arr.sort((a, b) => {return (a % 1 - b % 1)});
console.log(arr);
You need to split each element before sort and compare second parts
let array = ["1101.69", "1701.57", "1301.09"];
array.sort((a,b)=>{
let pair1 = a.split('.');
let pair2 = b.split('.');
return ( parseInt(pair1[1]) < parseInt(pair2[1])) ? -1 : 1;
});
console.log(array);
I have a list of objects I wish to sort based on a field attr of type string. I tried using -
list.sort(function (a, b) {
return a.attr - b.attr
})
but found that - doesn't appear to work with strings in JavaScript. How can I sort a list of objects based on an attribute with type string?
Use String.prototype.localeCompare as per your example:
list.sort(function (a, b) {
return ('' + a.attr).localeCompare(b.attr);
})
We force a.attr to be a string to avoid exceptions. localeCompare has been supported since Internet Explorer 6 and Firefox 1. You may also see the following code used that doesn't respect a locale:
if (item1.attr < item2.attr)
return -1;
if ( item1.attr > item2.attr)
return 1;
return 0;
An updated answer (October 2014)
I was really annoyed about this string natural sorting order so I took quite some time to investigate this issue.
Long story short
localeCompare() character support is badass, just use it.
As pointed out by Shog9, the answer to your question is:
return item1.attr.localeCompare(item2.attr);
Bugs found in all the custom JavaScript "natural string sort order" implementations
There are quite a bunch of custom implementations out there, trying to do string comparison more precisely called "natural string sort order"
When "playing" with these implementations, I always noticed some strange "natural sorting order" choice, or rather mistakes (or omissions in the best cases).
Typically, special characters (space, dash, ampersand, brackets, and so on) are not processed correctly.
You will then find them appearing mixed up in different places, typically that could be:
some will be between the uppercase 'Z' and the lowercase 'a'
some will be between the '9' and the uppercase 'A'
some will be after lowercase 'z'
When one would have expected special characters to all be "grouped" together in one place, except for the space special character maybe (which would always be the first character). That is, either all before numbers, or all between numbers and letters (lowercase & uppercase being "together" one after another), or all after letters.
My conclusion is that they all fail to provide a consistent order when I start adding barely unusual characters (i.e., characters with diacritics or characters such as dash, exclamation mark and so on).
Research on the custom implementations:
Natural Compare Lite https://github.com/litejs/natural-compare-lite : Fails at sorting consistently https://github.com/litejs/natural-compare-lite/issues/1 and http://jsbin.com/bevututodavi/1/edit?js,console, basic Latin characters sorting http://jsbin.com/bevututodavi/5/edit?js,console
Natural Sort https://github.com/javve/natural-sort : Fails at sorting consistently, see issue https://github.com/javve/natural-sort/issues/7 and see basic Latin characters sorting http://jsbin.com/cipimosedoqe/3/edit?js,console
JavaScript Natural Sort https://github.com/overset/javascript-natural-sort: seems rather neglected since February 2012, Fails at sorting consistently, see issue https://github.com/overset/javascript-natural-sort/issues/16
Alphanum http://www.davekoelle.com/files/alphanum.js , Fails at sorting consistently, see http://jsbin.com/tuminoxifuyo/1/edit?js,console
Browsers' native "natural string sort order" implementations via localeCompare()
localeCompare() oldest implementation (without the locales and options arguments) is supported by Internet Explorer 6 and later, see http://msdn.microsoft.com/en-us/library/ie/s4esdbwz(v=vs.94).aspx (scroll down to localeCompare() method).
The built-in localeCompare() method does a much better job at sorting, even international & special characters.
The only problem using the localeCompare() method is that "the locale and sort order used are entirely implementation dependent". In other words, when using localeCompare such as stringOne.localeCompare(stringTwo): Firefox, Safari, Chrome, and Internet Explorer have a different sort order for Strings.
Research on the browser-native implementations:
http://jsbin.com/beboroyifomu/1/edit?js,console - basic Latin characters comparison with localeCompare()
http://jsbin.com/viyucavudela/2/ - basic Latin characters comparison with localeCompare() for testing on Internet Explorer 8
http://jsbin.com/beboroyifomu/2/edit?js,console - basic Latin characters in string comparison : consistency check in string vs when a character is alone
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/localeCompare - Internet Explorer 11 and later supports the new locales & options arguments
Difficulty of "string natural sorting order"
Implementing a solid algorithm (meaning: consistent but also covering a wide range of characters) is a very tough task. UTF-8 contains more than 2000 characters and covers more than 120 scripts (languages).
Finally, there are some specification for this tasks, it is called the "Unicode Collation Algorithm", which can be found at http://www.unicode.org/reports/tr10/. You can find more information about this on this question I posted https://softwareengineering.stackexchange.com/questions/257286/is-there-any-language-agnostic-specification-for-string-natural-sorting-order
Final conclusion
So considering the current level of support provided by the JavaScript custom implementations I came across, we will probably never see anything getting any close to supporting all this characters and scripts (languages). Hence I would rather use the browsers' native localeCompare() method. Yes, it does have the downside of being non-consistent across browsers but basic testing shows it covers a much wider range of characters, allowing solid & meaningful sort orders.
So as pointed out by Shog9, the answer to your question is:
return item1.attr.localeCompare(item2.attr);
Further reading:
https://softwareengineering.stackexchange.com/questions/257286/is-there-any-language-agnostic-specification-for-string-natural-sorting-order
How to sort strings in JavaScript
Natural sort of alphanumerical strings in JavaScript
Sort Array of numeric & alphabetical elements (Natural Sort)
Sort mixed alpha/numeric array
https://web.archive.org/web/20130929122019/http://my.opera.com/GreyWyvern/blog/show.dml/1671288
https://web.archive.org/web/20131005224909/http://www.davekoelle.com/alphanum.html
http://snipplr.com/view/36012/javascript-natural-sort/
http://blog.codinghorror.com/sorting-for-humans-natural-sort-order/
Thanks to Shog9's nice answer, which put me in the "right" direction I believe.
Answer (in Modern ECMAScript)
list.sort((a, b) => (a.attr > b.attr) - (a.attr < b.attr))
Or
list.sort((a, b) => +(a.attr > b.attr) || -(a.attr < b.attr))
Description
Casting a boolean value to a number yields the following:
true -> 1
false -> 0
Consider three possible patterns:
x is larger than y: (x > y) - (y < x) -> 1 - 0 -> 1
x is equal to y: (x > y) - (y < x) -> 0 - 0 -> 0
x is smaller than y: (x > y) - (y < x) -> 0 - 1 -> -1
(Alternative)
x is larger than y: +(x > y) || -(x < y) -> 1 || 0 -> 1
x is equal to y: +(x > y) || -(x < y) -> 0 || 0 -> 0
x is smaller than y: +(x > y) || -(x < y) -> 0 || -1 -> -1
So these logics are equivalent to typical sort comparator functions.
if (x == y) {
return 0;
}
return x > y ? 1 : -1;
Since strings can be compared directly in JavaScript, this will do the job:
list.sort(function (a, b) {
return a.attr < b.attr ? -1: 1;
})
This is a little bit more efficient than using
return a.attr > b.attr ? 1: -1;
because in case of elements with same attr (a.attr == b.attr), the sort function will swap the two for no reason.
For example
var so1 = function (a, b) { return a.atr > b.atr ? 1: -1; };
var so2 = function (a, b) { return a.atr < b.atr ? -1: 1; }; // Better
var m1 = [ { atr: 40, s: "FIRST" }, { atr: 100, s: "LAST" }, { atr: 40, s: "SECOND" } ].sort (so1);
var m2 = [ { atr: 40, s: "FIRST" }, { atr: 100, s: "LAST" }, { atr: 40, s: "SECOND" } ].sort (so2);
// m1 sorted but ...: 40 SECOND 40 FIRST 100 LAST
// m2 more efficient: 40 FIRST 40 SECOND 100 LAST
You should use > or < and == here. So the solution would be:
list.sort(function(item1, item2) {
var val1 = item1.attr,
val2 = item2.attr;
if (val1 == val2) return 0;
if (val1 > val2) return 1;
if (val1 < val2) return -1;
});
Nested ternary arrow function
(a,b) => (a < b ? -1 : a > b ? 1 : 0)
I had been bothered about this for long, so I finally researched this and give you this long winded reason for why things are the way they are.
From the spec:
Section 11.9.4 The Strict Equals Operator ( === )
The production EqualityExpression : EqualityExpression === RelationalExpression
is evaluated as follows:
- Let lref be the result of evaluating EqualityExpression.
- Let lval be GetValue(lref).
- Let rref be the result of evaluating RelationalExpression.
- Let rval be GetValue(rref).
- Return the result of performing the strict equality comparison
rval === lval. (See 11.9.6)
So now we go to 11.9.6
11.9.6 The Strict Equality Comparison Algorithm
The comparison x === y, where x and y are values, produces true or false.
Such a comparison is performed as follows:
- If Type(x) is different from Type(y), return false.
- If Type(x) is Undefined, return true.
- If Type(x) is Null, return true.
- If Type(x) is Number, then
...
- If Type(x) is String, then return true if x and y are exactly the
same sequence of characters (same length and same characters in
corresponding positions); otherwise, return false.
That's it. The triple equals operator applied to strings returns true iff the arguments are exactly the same strings (same length and same characters in corresponding positions).
So === will work in the cases when we're trying to compare strings which might have arrived from different sources, but which we know will eventually have the same values - a common enough scenario for inline strings in our code. For example, if we have a variable named connection_state, and we wish to know which one of the following states ['connecting', 'connected', 'disconnecting', 'disconnected'] is it in right now, we can directly use the ===.
But there's more. Just above 11.9.4, there is a short note:
NOTE 4
Comparison of Strings uses a simple equality test on sequences of code
unit values. There is no attempt to use the more complex, semantically oriented
definitions of character or string equality and collating order defined in the
Unicode specification. Therefore Strings values that are canonically equal
according to the Unicode standard could test as unequal. In effect this
algorithm assumes that both Strings are already in normalized form.
Hmm. What now? Externally obtained strings can, and most likely will, be weird unicodey, and our gentle === won't do them justice. In comes localeCompare to the rescue:
15.5.4.9 String.prototype.localeCompare (that)
...
The actual return values are implementation-defined to permit implementers
to encode additional information in the value, but the function is required
to define a total ordering on all Strings and to return 0 when comparing
Strings that are considered canonically equivalent by the Unicode standard.
We can go home now.
tl;dr;
To compare strings in javascript, use localeCompare; if you know that the strings have no non-ASCII components because they are, for example, internal program constants, then === also works.
An explanation of why the approach in the question doesn't work:
let products = [
{ name: "laptop", price: 800 },
{ name: "phone", price:200},
{ name: "tv", price: 1200}
];
products.sort( (a, b) => {
{let value= a.name - b.name; console.log(value); return value}
});
> 2 NaN
Subtraction between strings returns NaN.
Echoing Alejadro's answer, the right approach is:
products.sort( (a,b) => a.name > b.name ? 1 : -1 )
A typescript sorting method modifier using a custom function to return a sorted string in either ascending or descending order
const data = ["jane", "mike", "salome", "ababus", "buisa", "dennis"];
const sortStringArray = (stringArray: string[], mode?: 'desc' | 'asc') => {
if (!mode || mode === 'asc') {
return stringArray.sort((a, b) => a.localeCompare(b))
}
return stringArray.sort((a, b) => b.localeCompare(a))
}
console.log(sortStringArray(data, 'desc'));// [ 'salome', 'mike', 'jane', 'dennis', 'buisa', 'ababus' ]
console.log(sortStringArray(data, 'asc')); // [ 'ababus', 'buisa', 'dennis', 'jane', 'mike', 'salome' ]
There should be ascending and descending orders functions
if (order === 'asc') {
return a.localeCompare(b);
}
return b.localeCompare(a);
If you want to control locales (or case or accent), then use Intl.collator:
const collator = new Intl.Collator();
list.sort((a, b) => collator.compare(a.attr, b.attr));
You can construct a collator like:
new Intl.Collator("en");
new Intl.Collator("en", {sensitivity: "case"});
...
See the above link for documentation.
Note: unlike some other solutions, it handles null, undefined the JavaScript way, i.e., moves them to the end.
Use sort() straightforward without any - or <
const areas = ['hill', 'beach', 'desert', 'mountain']
console.log(areas.sort())
// To print in descending way
console.log(areas.sort().reverse())
In your operation in your initial question, you are performing the following operation:
item1.attr - item2.attr
So, assuming those are numbers (i.e. item1.attr = "1", item2.attr = "2") You still may use the "===" operator (or other strict evaluators) provided that you ensure type. The following should work:
return parseInt(item1.attr) - parseInt(item2.attr);
If they are alphaNumeric, then do use localCompare().
list.sort(function(item1, item2){
return +(item1.attr > item2.attr) || +(item1.attr === item2.attr) - 1;
})
How they work samples:
+('aaa'>'bbb')||+('aaa'==='bbb')-1
+(false)||+(false)-1
0||0-1
-1
+('bbb'>'aaa')||+('bbb'==='aaa')-1
+(true)||+(false)-1
1||0-1
1
+('aaa'>'aaa')||+('aaa'==='aaa')-1
+(false)||+(true)-1
0||1-1
0
var str = ['v','a','da','c','k','l']
var b = str.join('').split('').sort().reverse().join('')
console.log(b)
<!doctype html>
<html>
<body>
<p id = "myString">zyxtspqnmdba</p>
<p id = "orderedString"></p>
<script>
var myString = document.getElementById("myString").innerHTML;
orderString(myString);
function orderString(str) {
var i = 0;
var myArray = str.split("");
while (i < str.length){
var j = i + 1;
while (j < str.length) {
if (myArray[j] < myArray[i]){
var temp = myArray[i];
myArray[i] = myArray[j];
myArray[j] = temp;
}
j++;
}
i++;
}
var newString = myArray.join("");
document.getElementById("orderedString").innerHTML = newString;
}
</script>
</body>
</html>
This is my data ["2","1","10"]. When I am trying to sort these numbers my output is ["1","10","2"], but my expected output is ["1","2","10] . I am using orderBy to sort these numbers. It sorts depending upon 0 to 9 only.
How can I sort these numbers using js?
Here is my code:
ng-options="option.id as option.roomno for option in frData
| unique:'roomno' | orderBy:'roomno' "
function sortNumStr(m,n) {
return m - n;
}
var numArray = ["2", "1", "10"];
numArray.sort(sortNumStr);
alert(numArray)
You can use the string's native prototype localeCompare function like so:
["2","1","10"].sort((a,b) => a.localeCompare(b, undefined, { numeric: true }))
It works well with numbers among other chars too.
This question already has an answer here:
Why won't Safari 5 sort an array of objects?
(1 answer)
Closed 8 years ago.
var myArray = [{date:"2013.03.01"},{date:"2013.03.08"},{date:"2013.03.19"}];
I tried:
function(a,b){
return b.date > a.date;
}
and
function(a,b){
return b.date - a.date;
}
The console.log in Chrome and Firefox give me the desired output:
"2013.03.19", "2013.03.08", "2013.03.01"
but Safari give the original sorting:
"2013.03.01", "2013.03.08", "2013.03.19"
Why?
A sort function in JavaScript is supposed to return a real number -- not true or false or a string or date. Whether that number is positive, negative, or zero affects the sort result.
Try this sort function (which will also correctly sort any strings in reverse-alphabetical order):
myArray.sort(function(a,b){
return (b.date > a.date) ? 1 : (b.date < a.date) ? -1 : 0;
});
"2013.03.01" is not a date. It's a string.
In order to correctly sort by dates, you need to convert these to dates (timestamps).
var myArray = [{date:"2013.03.01"},{date:"2013.03.08"},{date:"2013.03.19"}];
myArray.sort(function(a,b){
return Date.parse(b.date) - Date.parse(a.date);
});
You might also be able to sort them using direct string comparasions too:
myArray.sort(function(a,b){
return b.date.localeCompare(a.date);
});