How can I extract and get just the numeric value after the hyphen in a string?
Here is the input string:
var x = "-2147467259"
After some processing.... return:
alert(2147467259)
How do I accomplish this?
You could replace away the hyphen:
alert(+x.replace("-", ""));
And yes, the + is important. It converts a string to a number; so you're removing the hypen by replacing it with nothing, and then essentially casting the result of that operation into a number. This operation will also work if no hyphen is present.
You could also use substr to achieve this:
alert(+x.substr(1));
You could also use parseInt to convert the string to a number (which will end up negative if a hyphen is persent), and then find its absolute value:
alert(Math.abs(parseInt(x, 10));
As Bergi notes, if you can be sure that the first character in the string is always a hyphen, you can simple return its negative, which will by default cast the value into a number and then perform the negative operation on it:
alert(-x);
You could also check to see if the number is negative or positive via a tertiary operator and then perform the respective operation on it to ensure that it is a positive Number:
x = x >= 0 ? +x : -x;
This may be cheaper in terms of performance than using Math.abs, but the difference will be minuscule either way.
As you can see, there really are a variety of ways to achieve this. I'd recommend reading up on JavaScript string functions and number manipulation in general, as well as examining JavaScript's Math object to get a feel for what tools are available to you when you go to solve a problem.
How about:
Math.abs(parseInt("-2147467259"))
Or
"-2147467259".replace('-','')
or
"-2147467259".replace(/\-/,'')
#1 option is converting the string to numbers. The #2 approach is removing all - from the string and the #3 option even though it will not be necessary on this example uses Regular Expression but I wanted to show the possibility of using RegEx in replace situations.
If you need a number as the final value #1 is your choice if you need strings #2 is your choice.
Related
I am facing issues with regex pattern for Float number -> that should not end or stars with decimal points..
I have tried following regex patter.. that is
regex = /^\d*\.?\d*$/
// on doing
regex.test(11.)
regex.test(.11)
// it is returning true in checking
// I need to make this as false, comment will be much helpful
thank you.
You should bear in mind that regex only works with strings. When you pass a non-string variable as input to a RegExp, it will first coerce it to a string type.
Have a look:
console.log(11. , 'and', .11); // => 11 and 0.11
So, the actual string values you pass to your ^\d*\.?\d*$ regex are 11 and 0.11 that can be matched with the given pattern. Actually, ^\d*\.?\d*$ is a regex that is usually used for a very loose live number input validation, e.g. see How to make proper Input validation with regex?.
What you want is to implement a final, on-submit validation pattern, so that it could not pass strings like 11. and .11. There have been lots of threads discussing this kind of regex:
Regular expression for floating point numbers
regular expression for finding decimal/float numbers?
Basically, for validation, you will need something like
/^\d+(?:\.\d+)?$/.test(input_string)
/^[0-9]+(?:\.[0-9]+)?$/.test(input_string)
/^[0-9]+(?:\.[0-9]{1,2})?$/.test(input_string) // Some need to only allow 1 or 2 fractional digits
/^[0-9]{1,3}(?:\.[0-9]{2})?$/.test(input_string) // 1-3 digits in the integer part and two required in the fractional part
This is an extension of this SO question
I made a function to see if i can correctly format any number. The answers below work on tools like https://regex101.com and https://regexr.com/, but not within my function(tried in node and browser):
const
const format = (num, regex) => String(num).replace(regex, '$1')
Basically given any whole number, it should not exceed 15 significant digits. Given any decimal, it should not exceed 2 decimal points.
so...
Now
format(0.12345678901234567890, /^\d{1,13}(\.\d{1,2}|\d{0,2})$/)
returns 0.123456789012345678 instead of 0.123456789012345
but
format(0.123456789012345,/^-?(\d*\.?\d{0,2}).*/)
returns number formatted to 2 deimal points as expected.
Let me try to explain what's going on.
For the given input 0.12345678901234567890 and the regex /^\d{1,13}(\.\d{1,2}|\d{0,2})$/, let's go step by step and see what's happening.
^\d{1,13} Does indeed match the start of the string 0
(\. Now you've opened a new group, and it does match .
\d{1,2} It does find the digits 1 and 2
|\d{0,2} So this part is skipped
) So this is the end of your capture group.
$ This indicates the end of the string, but it won't match, because you've still got 345678901234567890 remaining.
Javascript returns the whole string because the match failed in the end.
Let's try removing $ at the end, to become /^\d{1,13}(\.\d{1,2}|\d{0,2})/
You'd get back ".12345678901234567890". This generates a couple of questions.
Why did the preceding 0 get removed?
Because it was not part of your matching group, enclosed with ().
Why did we not get only two decimal places, i.e. .12?
Remember that you're doing a replace. Which means that by default, the original string will be kept in place, only the parts that match will get replaced. Since 345678901234567890 was not part of the match, it was left intact. The only part that matched was 0.12.
Answer to title question: your function doesn't replace, because there's nothing to replace - the regex doesn't match anything in the string. csb's answer explains that in all details.
But that's perhaps not the answer you really need.
Now, it seems like you have an XY problem. You ask why your call to .replace() doesn't work, but .replace() is definitely not a function you should use. Role of .replace() is replacing parts of string, while you actually want to create a different string. Moreover, in the comments you suggest that your formatting is not only for presenting data to user, but you also intend to use it in some further computation. You also mention cryptocurriencies.
Let's cope with these problems one-by-one.
What to do instead of replace?
Well, just produce the string you need instead of replacing something in the string you don't like. There are some edge cases. Instead of writing all-in-one regex, just handle them one-by-one.
The following code is definitely not best possible, but it's main aim is to be simple and show exactly what is going on.
function format(n) {
const max_significant_digits = 15;
const max_precision = 2;
let digits_before_decimal_point;
if (n < 0) {
// Don't count minus sign.
digits_before_decimal_point = n.toFixed(0).length - 1;
} else {
digits_before_decimal_point = n.toFixed(0).length;
}
if (digits_before_decimal_point > max_significant_digits) {
throw new Error('No good representation for this number');
}
const available_significant_digits_for_precision =
Math.max(0, max_significant_digits - digits_before_decimal_point);
const effective_max_precision =
Math.min(max_precision, available_significant_digits_for_precision);
const with_trailing_zeroes = n.toFixed(effective_max_precision);
// I want to keep the string and change just matching part,
// so here .replace() is a proper method to use.
const withouth_trailing_zeroes = with_trailing_zeroes.replace(/\.?0*$/, '');
return withouth_trailing_zeroes;
}
So, you got the number formatted the way you want. What now?
What can you use this string for?
Well, you can display it to the user. And that's mostly it. The value was rounded to (1) represent it in a different base and (2) fit in limited precision, so it's pretty much useless for any computation. And, BTW, why would you convert it to String in the first place, if what you want is a number?
Was the value you are trying to print ever useful in the first place?
Well, that's the most serious question here. Because, you know, floating point numbers are tricky. And they are absolutely abysmal for representing money. So, most likely the number you are trying to format is already a wrong number.
What to use instead?
Fixed-point arithmetic is the most obvious answer. Works most of the time. However, it's pretty tricky in JS, where number may slip into floating-point representation almost any time. So, it's better to use decimal arithmetic library. Optionally, switch to a language that has built-in bignums and decimals, like Python.
I am just dipping my toe into the confusing world of javascript, more out of necessity than desire and I have come across a problem of adding two integers.
1,700.00 + 500.00
returns 1,700.00500.00
So after some research I see that 1,700.00 is being treated as a string and that I need to convert it.
The most relevant pages I read to resolve this were this question and this page. However when I use
parseInt(string, radix)
it returns 1. Am I using the wrong function or the an incorrect radix (being honest I can't get my head around how I decide which radix to use).
var a="1,700.00";
var b=500.00;
parseInt(a, 10);
Basic Answer
The reason parseInt is not working is because of the comma. You could remove the comma using a regex such as:
var num = '1,700.00';
num = num.replace(/\,/g,'');
This will return a string with a number in it. Now you can parseInt. If you do not choose a radix it will default to 10 which was the correct value to use here.
num = parseInt(num);
Do this for each of your string numbers before adding them and everything should work.
More information
How the replace works:
More information on replace at mdn:
`/` - start
`\,` - escaped comma
`/` - end
`g` - search globally
The global search will look for all matches (it would stop after the first match without this)
'' replace the matched sections with an empty string, essentially deleting them.
Regular Expressions
A great tool to test regular expressions: Rubular and more info about them at mdn
If you are looking for a good tutorial here is one.
ParseInt and Rounding, parseFloat
parseInt always rounds to the nearest integer. If you need decimal places there are a couple of tricks you can use. Here is my favorite:
2 places: `num = parseInt(num * 100) / 100;`
3 places: `num = parseInt(num * 1000) / 1000;`
For more information on parseInt look at mdn.
parseFloat could also be used if you do not want rounding. I assumed you did as the title was convert to an integer. A good example of this was written by #fr0zenFry below. He pointed out that parseFloat also does not take a radix so it is always in base10. For more info see mdn.
Try using replace() to replace a , with nothing and then parseFloat() to get the number as float. From the variables in OP, it appears that there may be fractional numbers too, so, parseInt() may not work well in such cases(digits after decimal will be stripped off).
Use regex inside replace() to get rid of each appearance of ,.
var a = parseFloat('1,700.00'.replace(/,/g, ''));
var b = parseFloat('500.00'.replace(/,/g, ''));
var sum = a+b;
This should give you correct result even if your number is fractional like 1,700.55.
If I go by the title of your question, you need an integer. For this you can use parseInt(string, radix). It works without a radix but it is always a good idea to specify this because you never know how browsers may behave(for example, see comment #Royi Namir). This function will round off the string to nearest integer value.
var a = parseInt('1,700.00'.replace(/,/g, ''), 10); //radix 10 will return base10 value
var b = parseInt('500.00'.replace(/,/g, ''), 10);
var sum = a+b;
Note that a radix is not required in parseFloat(), it will always return a decimal/base10 value. Also, it will it will strip off any extra zeroes at the end after decimal point(ex: 17500.50 becomes 17500.5 and 17500.00 becomes 17500). If you need to get 2 decimal places always, append another function toFixed(decimal places).
var a = parseFloat('1,700.00'.replace(/,/g, ''));
var b = parseFloat('500.00'.replace(/,/g, ''));
var sum = (a+b).toFixed(2); //change argument in toFixed() as you need
// 2200.00
Another alternative to this was given by #EpiphanyMachine which will need you to multiply and then later divide every value by 100. This may become a problem if you want to change decimal places in future, you will have to change multiplication/division factor for every variable. With toFixed(), you just change the argument. But remember that toFixed() changes the number back to string unlike #EpiphanyMachine solution. So you will be your own judge.
try this :
parseFloat(a.replace(/,/g, ''));
it will work also on : 1,800,300.33
Example :
parseFloat('1,700,800.010'.replace(/,/g, '')) //1700800.01
Javascript doesn't understand that comma. Remove it like this:
a.replace(',', '')
Once you've gotten rid of the comma, the string should be parsed with no problem.
Whats a simple way to delete the last two characters of a string?
To convert 245px in 245 just run:
parseInt('245px', 10);
It retains only leading numbers and discards all the rest.
use
var size = parseInt('245px', 10);
where 10 is the radix defining parseInt is parsing to a decimal value
use parseInt but don't use parseInt without a radix
The parseInt() function parses a string and returns an integer.
The signature is parseInt(string, radix)
The second argument forces parseInt to use a base ten numbering system.
The default input type for ParseInt() is decimal (base 10).
If the number begins in "0", it is assumed to be octal (base 8).
If it begins in "0x", it is assumed to be hexadecimal
why? if $(this).attr('num') would be "08" parsInt without a radix would become 0
To convert a pixel value without the "px" at the end. use parseFloat.
parseFloat('245px'); // returns 245
Note: If you use parseInt, the value will be correct if the value is an integer. If the value is a decimal one like 245.50px, then the value will be rounded to 245.
This does exactly what you ask: remove last two chars of a string:
s.substr(0, s.length-2);
Surprisingly, the substring method s.substr(0, s.length-2); is actually quite a bit faster for removing the px (yes it isn't as clean looking, and if there is a space it will remain -- "250px" -> "250" vs "250 px" -> "250 ").
If you want to account for spaces (which you probably should) then using the .trim() function will actually slow down the substr test enough that the parseInt method actually becomes superior.
An added benefit of using parseInt(s, 10) is that you also get a type conversion and can immediately start to apply it to mathematical functions.
So in the end, it really depends on what you plan on doing with the result.
If it is display only, then using the substr method without a trim would probably be your best bet.
If you're just trying to see if the value without px is the same as another value s.substr(0, s.length-2) == 0, then using the substr method would be best, as "250 " == 250 (even with the space) will result as true
If you want to account for the possibility of a space, add it to another value, or to compute something with it, then you may want to consider going with the parseInt route.
http://jsperf.com/remove-px-from-coord
The tests on jspref account for a space. I also tried a s.split('px')[0] and s.replace(/ *px/g, '') function, both found to be slower.
Feel free to add additional test cases.
Although parseInt() is a good option but still it is good to have many other solutions
var pixels = '245px';
Number(pixels.replace('px', ''));
substr() is now a legacy feature; use substring() instead: (syntax is the same in this case)
str.substring(0, str.length-2);
Or, use slice():
str.slice(0, -2);
slice() looks much cleaner, IMO. Negative values count back from the end.
Check http://www.w3schools.com/jsref/jsref_substr.asp
In your case would be something like
string.substr(0, string.length - 2)
I prefer:
"245px".replace(/px/,'')*1
since it's not surrounding the input.
Also, the *1 is for casting it to int.
A thought struck me as I was writing a piece of JavaScript code that processed some floating point values. What is the decimal point symbol in JavaScript? Is it always .? Or is it culture-specific? And what about .toFixed() and .parseFloat()? If I'm processing a user input, it's likely to include the local culture-specific decimal separator symbol.
Ultimately I'd like to write code that supports both decimal points in user input - culture-specific and ., but I can't write such a code if I don't know what JavaScript expects.
Added: OK, Rubens Farias suggests to look at similar question which has a neat accepted answer:
function whatDecimalSeparator() {
var n = 1.1;
n = n.toLocaleString().substring(1, 2);
return n;
}
That's nice, it lets me get the locale decimal point. A step towards the solution, no doubt.
Now, the remaining part would be to determine what the behavior of .parseFloat() is. Several answers point out that for floating point literals only . is valid. Does .parseFloat() act the same way? Or might it require the local decimal separator in some browser? Are there any different methods for parsing floating point numbers as well? Should I roll out my own just-to-be-sure?
According to the specification, a DecimalLiteral is defined as:
DecimalLiteral ::
DecimalIntegerLiteral . DecimalDigitsopt ExponentPartopt
. DecimalDigits ExponentPartopt
DecimalIntegerLiteral ExponentPartopt
and for satisfying the parseFloat argument:
Let inputString be ToString(string).
Let trimmedString be a substring of inputString consisting of the leftmost character that is not a StrWhiteSpaceChar and all characters to the right of that character.(In other words, remove leading white space.)
If neither trimmedString nor any prefix of trimmedString satisfies the syntax of a StrDecimalLiteral (see 9.3.1), return NaN.
Let numberString be the longest prefix of trimmedString, which might be trimmedString itself, that satisfies the syntax of a StrDecimalLiteral.
Return the Number value for the MV
So numberString becomes the longest prefix of trimmedString that satisfies the syntax of a StrDecimalLiteral, meaning the first parseable literal string number it finds in the input. Only the . can be used to specify a floating-point number. If you're accepting inputs from different locales, use a string replace:
function parseLocalNum(num) {
return +(num.replace(",", "."));
}
The function uses the unary operator instead of parseFloat because it seems to me that you want to be strict about the input. parseFloat("1ABC") would be 1, whereas using the unary operator +"1ABC" returns NaN. This makes it MUCH easier to validate the input. Using parseFloat is just guessing that the input is in the correct format.
use:
theNumber.toLocaleString();
to get a properly formatted string with the right decimal and thousands separators
As far as I'm aware, javascript itself only knows about the . separator for decimals. At least one person whose judgement I trust on JS things concurs:
http://www.merlyn.demon.co.uk/js-maths.htm#DTS