I'm trying to find a regular expression for a float with a fixed maximum (for example 4) number of significant figures.
this should match with:
- 123.4
- 12.34
- 1.2
- 223
- 0.1234
- 0.000001234
the problem is that the number of non-zeros before and after the dot has to be at most 4 in total.
I tried to split the problem and found solutions for the cases:
- 0.xxxx
- 0.000xxx
- xxxx
But I didn't find a solution for the case that significant digits are found before and after the dot. (examples: 1.23 2.345 )
update:
I think I found a solution:
^(?!(?:.*[1-9](\.?[0-9]){4,}))([-+]?\d+\.?\d*?)$
^(?!(?:.*?[1-9]){5,})([-+]?\s*\d+\.?\d*?)$
Try this.This will match only 4 or less significant digits.Do not forget to put flags g and m.See demo.
http://regex101.com/r/hQ1rP0/28
I think you want something like this,
^0*(?:[1-9]\d{0,3})?(?:\.0*(?:[1-9]\d{0,3})?)?$
DEMO
Related
Hi I'm looking for a regular expression that would match only numbers that represent percentages with up to 4 decimals. meaning for example:
should match
should not match
100
100.0001
99.9999
101
0.0001
100.00001
0
Iv'e tried this but it does not work (does not match 0.0004 for example):
"^\\d{1,3}\\.\\d{1,4}$"
Thanks!
I think the best way to aproach this is by considering 100% to be a special case.
^(100(\.0{1,4})?|\d{1,2}(\.\d{1,4})?)$
That way you can match 0->99.9999 or 100 with up to 4 decimal 0s after it.
demo
Use alterations to capture the different possibilities.
This works:
^((?:100)|(?:100\.0{1,4})|(?:\d{1,2})|(?:\d{1,2}\.\d{0,4}))$
Demo
Need regular expression which have:
Maximum 8 digits before decimal(.) point
Maximum 4 digits after decimal point
Decimal point is optional
Maximum valid decimal is 8 digits before decimal and 4 digits after decimal
So 99999999.9999
The regular rexpression I have tried ^\d{0,8}[.]?\d{1,4}$ is failing for 123456789
and more than this. means it is taking more than 8 digits if decimal point is not available.
Tested here : http://regexpal.com/
Many many thanks in advance!
^\d{0,8}(\.\d{1,4})?$
You can make the entire decimal optional
You can try this:
^\d{1,8}(?:\.\d{1,4})?$
or
^[1-9]\d{0,7}(?:\.\d{1,4})?$
If you don't want to have a zero as first digit.
You can allow this if you want: (.1234)
^[1-9]\d{0,7}(?:\.\d{1,4})?|\.\d{1,4}$
Any of the above did not work for me.
Only this works for me
^([0-9]{0,2}((.)[0-9]{0,2}))$
This regex is working for most cases even negative prices,
(\-?\d+\.?\d{0,2})
Tested with the following,
9
9.97
37.97
132.97
-125.55
12.2
1000.00
10000.00
100000.00
1000000.00
401395011
If there is a price of $9.97, £9.97 or €9.97 it will validate 9.97 removing the symbol.
1-(\$+.[1-9])
2-(\£+.[1-9])
You can use this expression for complete price digits.
I'm using this:
^[1-9]\d{0,7}(\.\d{1-4})$
^ = the start of the string
[1-9] = at least the string has to begin with 1 number between 1 and 9
\d{0,7} = optional or max 7 times d (digit: a number between 0 and 9)
() = create a group like a substring
. = need a .
\d{1-4} = digit repited max 4 time
$ end of the string
For price validation we can not allow inputs with leading repeating zeros like 0012 etc.
My solution check for any cases. Also it allows maximum 2 decimal point after the dot.
^(?:0\.[0-9]{1,2}|[1-9]{1}[0-9]*(\.[0-9]{1,2})?|0)$
I am having a bit of difficulty with the following:
I need to allow any positive numeric value up to four decimal places. Here are some examples.
Allowed:
123
12345.4
1212.56
8778787.567
123.5678
Not allowed:
-1
12.12345
-12.1234
I have tried the following:
^[0-9]{0,2}(\.[0-9]{1,4})?$|^(100)(\.[0]{1,4})?$
However this doesn't seem to work, e.g. 1000 is not allowed when it should be.
Any ideas would be greatly appreciated.
Thanks
To explain why your attempt is not working for a value of 1000, I'll break down the expression a little:
^[0-9]{0,2} # Match 0, 1, or 2 digits (can start with a zero)...
(\.[0-9]{1,4})?$ # ... optionally followed by (a decimal, then 1-4 digits)
| # -OR-
^(100) # Capture 100...
(\.[0]{1,4})?$ # ... optionally followed by (a decimal, then 1-4 ZEROS)
There is no room for 4 digits of any sort, much less 1000 (theres only room for a 0-2 digit number or the number 100)
^\d* # Match any number of digits (can start with a zero)
(\.\d{1,4})?$ # ...optionally followed by (a decimal and 1-4 digits)
This expression will pass any of the allowed examples and reject all of the Not Allowed examples as well, because you (and I) use the beginning-of-string assertion ^.
It will also pass these numbers:
.2378
1234567890
12374610237856987612364017826350947816290385
000000000000000000000.0
0
... as well as a completely blank line - which might or might not be desired
to make it reject something that starts with a zero, use this:
^(?!0\d)\d* # Match any number of digits (cannot "START" with a zero)
(\.\d{1,4})?$ # ...optionally followed by (a decimal and 1-4 digits)
This expression (which uses a negative lookahead) has these evaluations:
REJECTED Allowed
--------- -------
0000.1234 0.1234
0000 0
010 0.0
You could also test for a completely blank line in other ways, but if you wanted to reject it with the regex, use this:
^(?!0\d|$)\d*(\.\d{1,4})?$
Try this:
^[0-9]*(?:\.[0-9]{0,4})?$
Explanation: match only if starting with a digit (excluding negative numbers), optionally followed by (non-capturing group) a dot and 0-4 digits.
Edit: With this pattern .2134 would also be matched. To only allow 0 < x < 1 of format 0.2134, replace the first * with a + above.
This regex would do the trick:
^\d+(?:\.\d{1,4})?$
From the beginning of the string search for one or more digits. If there's a . it must be followed with atleast one digit but a maximum of 4.
^(?<!-)\+?\d+(\.?\d{0,4})?$
The will match something with doesn't start with -, maybe has a + followed by an integer part with at least one number and an optional floating part of maximum 4 numbers.
Note: Regex does not support scientific notation. If you want that too let me know in a comment.
Well asked!!
You can try this:
^([0-9]+[\.]?[0-9]?[0-9]?[0-9]?[0-9]?|[0-9]+)$
If you have a double value but it goes to more decimal format and you want to shorter it to 4 then !
double value = 12.3457652133
value =Double.parseDouble(new DecimalFormat("##.####").format(value));
What is the RegEx for value Range from 1- 365
Try this:
^(?:[1-9]\d?|[12]\d{2}|3[0-5]\d|36[0-5])$
The start anchor ^ and end anchor
$ are to match the whole input and
not just part of it.
(? ) is for grouping.
| is for alternation
[1-9]\d? matches 1 to 99
[12]\d{2} matches 100 to 299
3[0-5]\d matches 300 to 359
36[0-5] matches 360 to 365
You would have to list the possible combinations 1-9, 10-99, 100-299, 300-359, 360-365:
^([1-9]\d?|[12]\d\d|3[0-5]\d|36[0-5])$
Not really a good fit for regex, but if you insist:
^(?:36[0-5]|3[0-5][0-9]|[12][0-9][0-9]|[1-9][0-9]|[1-9])$
This is not allowing leading zeroes. If you wish to allow those, let me know.
The expression above can be shortened a little to
^(?:36[0-5]|3[0-5]\d|[12]\d{2}|[1-9]\d?)$
but I find the first solution to be a bit more readable. YMMV.
A general solution for matching the numbers from 1 to XYZ
^(?!0)(?!\d{4}$)(?![X+1-9]\d{2}$)(?!X[Y+1-9]\d$)(?!XY[Z+1-9]$)\d+$
Notes:
If any of X, Y or Z are 9 that will make X+1 etc. be 10. If that happens the regex part that would require using the 10 should be left out.
This can be extended to numbers with more or less digits following the same principles.
It does not allow left-padding 0es.
Applied to your case:
^(?!0)(?!\d{4}$)(?![4-9]\d{2}$)(?!3[7-9]\d$)(?!36[6-9]$)\d+$
Lets explain:
(?!0\d*) - does not start with 0
(?!\d{4}$) - does not have 4 digits, i.e. between 1000 and infinity
(?![4-9]\d{2}$) - it's not between 400 and 999
(?!3[7-9]\d$) - it's not between 370 and 399
(?!36[6-9]$) - it's not between 366 and 369
Test it.
^36[0-5]|(3[0-5]|[12]?[0-9])[0-9]$
^3(6[0-5]|[0-5]\d)|[12]\d\d|[1-9]\d|[1-9]$
Or if numbers like 05 can not be in input:
^3(6[0-5]|[0-5]\d)|[12]?\d?\d$
P.S.: Anyway no need of regex here. Use ToInt(), <=, >=
It really depends on your regex engine since they may not all be PCRE-style. I usually work to the lowest common denominator unless I know it will be targeting a minimum engine.
To that end, I'd just use something like:
^[1-9]|[1-9][0-9]|[1-2][0-9]{2}|3[0-5][0-9]|36[0-5]$
This will take care of (in order):
1-9.
10-99.
100-299.
300-359.
360-365.
However, unless you're absolutely required to use just a regex, I wouldn't. It's like trying to kill a fly with a thermo-nuclear warhead.
Just use the much simpler ^[0-9]{1,3}$ then use whatever language features you have to convert it to an integer and check it's between 1 and 365 inclusive:
def isValidDayOtherThanLeapYear (s):
if not s.matches ("^[0-9]{1,3}$"):
return false
n = s.toInteger()
if n < 1 or n > 365:
return false
return true
Your code will be more readable that way and I tend to rethink the use of regular expressions the second they start looking like they may be hard to read six months down the track.
This worked for me...
^[1-3][0-6]?[0-5]?$
I have tried 2 questions, could you tell me whether I am right or not?
Regular expression of nonnegative integer constants in C, where numbers beginning with 0 are octal constants and other numbers are decimal constants.
I tried 0([1-7][0-7]*)?|[1-9][0-9]*, is it right? And what string could I match? Do you think 034567 will match and 000083 match?
What is a regular expression for binary numbers x such that hx + ix = jx?
I tried (0|1){32}|1|(10)).. do you think a string like 10 will match and 11 won’t match?
Please tell me whether I am right or not.
You can always use http://www.spaweditor.com/scripts/regex/ for a quick test on whether a particular regex works as you intend it to. This along with google can help you nail the regex you want.
0([1-7][0-7])?|[1-9][0-9] is wrong because there's no repetition - it will only match 1 or 2-character strings. What you need is something like 0[0-7]*|[1-9][0-9]*, though that doesn't take hexadecimal into account (as per spec).
This one is not clear. Could you rephrase that or give some more examples?
Your regex for integer constants will not match base-10 numbers longer than two digits and octal numbers longer than three digits (2 if you don't count the leading zero). Since this is a homework, I leave it up to you to figure out what's wrong with it.
Hint: Google for "regular expression repetition quantifiers".
Question 1:
Octal numbers:
A string that start with a [0] , then can be followed by any digit 1, 2, .. 7 [1-7](assuming no leading zeroes) but can also contain zeroes after the first actual digit, so [0-7]* (* is for repetition, zero or more times).
So we get the following RegEx for this part: 0 [1-7][0-7]*
Decimal numbers:
Decimal numbers must not have a leading zero, hence start with all digits from 1 to 9 [1-9], but zeroes are allowed in all other positions as well hence we need to concatenate [0-9]*
So we get the following RegEx for this part: [1-9][0-9]*
Since we have two options (octal and decimal numbers) and either one is possible we can use the Alternation property '|' :
L = 0[1-7][0-7]* | [1-9][0-9]*
Question 2:
Quickly looking at Fermat's Last Theorem:
In number theory, Fermat's Last Theorem (sometimes called Fermat's conjecture, especially in older texts) states that no three positive integers a, b, and c can satisfy the equation an + bn = cn for any integer value of n greater than two.
(http://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem)
Hence the following sets where n<=2 satisfy the equation: {0,1,2}base10 = {0,1,10}base2
If any of those elements satisfy the equation, we use the Alternation | (or)
So the regular expression can be: L = 0 | 1 | 10 but can also be L = 00 | 01 | 10 or even be L = 0 | 1 | 10 | 00 | 01
Or can be generalized into:
{0} we can have infinite number of zeroes: 0*
{1} we can have infinite number of zeroes followed by a 1: 0*1
{10} we can have infinite number of zeroes followed by 10: 0*10
So L = 0* | 0*1 | 0*10
max answered the first question.
the second appears to be the unsolvable diophantine equation of fermat's last theorem. if h,i,j are non-zero integers, x can only be 1 or 2, so you're looking for
^0*10?$
does that help?
There are several tool available to test regular expressions, such as The Regulator.
If you search for "regular expression test" you will find numerous links to online testers.