I have thousands of article descriptions containing numbers.
they look like:
ca.2760h3x1000.5DIN345x1500e34
the resulting numbers should be:
2760
1000.5
1500
h3 or 3 shall not be a result of the parsing, since h3 is a tolerance only
same for e34
DIN345 is a norm an needs to be excluded (every number with a trailing DIN or BN)
My current REGEX is:
[^hHeE]([-+]?([0-9]+\.[0-9]+|[0-9]+))
This solves everything BUT the norm. How can I get this "DIN" and "BN" treated the same way as a single character ?
Thanx, TomE
Try using this regular expression:
(?<=x)[+-]?0*[0-9]+(?:\.[0-9]+)?|[+-]?0*[0-9]+(?:\.[0-9]+)?(?=h|e)
It looks like every number in your testcase you want to match exept the first number is starting with x.This is what the first part of the regex matches. (?<=x)[+-]?0*[0-9]+(?:\.[0-9]+)?The second part of the regex matches the number until h or e. [+-]?0*[0-9]+(?:\.[0-9]+)?(?=h|e)
The two parts [+-]?0*[0-9]+(?:\.[0-9]+)? in the regex is to match the number.
If we can assume that the numbers are always going to be four digits long, you can use the regex:
(\d{4}\.\d+|\d{4})
DEMO
Depending on the language you might need to replace \d with [0-9].
Related
I came across the regular expression not containing 101 as follows:
0∗1∗0∗+(1+00+000)∗+(0+1+0+)∗
I was unable to understand how the author come up with this regex. So I just thought of string which did not contain 101:
01000100
I seems that above string will not be matched by above regex. But I was unsure. So tried translating to equivalent pcre regex on regex101.com, but failed there too (as it can be seen my regex does not even matches string containing single 1.
Whats wrong with my translation? Is above regex indeed correct? If not what will be the correct regex?
Here is a bit shorter expression ^0*(1|00+)*0*$
https://www.regex101.com/r/gG3wP5/1
Explanation:
(1|00+)* we can mix zeroes and ones as long as zeroes occur in groups
^0*...0*$ we can have as many zeroes as we want in prefix/suffix
Direct translation of the original regexp would be like
^(0*1*0*|(1|00|000)*|(0+1+0+)*)$
Update
This seems like artificially complicated version of the above regexp:
(1|00|000)* is the same as (1|00+)*
it is almost the solution, but it does not match strings 0, 01.., and ..10
0*1*0* doesn't match strings with 101 inside, but matches 0 and some of 01.., and ..10
we still need to match those of 01.., and ..10 which have 0 & 1 mixed inside, e.g. 01001.. or ..10010
(0+1+0+)* matches some of the remaining cases but there are still some valid strings unmatched
e.g. 10010 is the shortest string that is not matched by all of the cases.
So, this solution is overly complicated and not complete.
read the explanation in the right side tab in regex101 it tells you what your regex does( I think you misunderstood what list operator does) , inside a list operator ( [ ) , the other characters such as ( won't be metacharacters anymore so the expression [(0*1*0*)[1(00)(000)] will be equivalent to [01()*[] which means it matches 0 or 1 or ( or ) or [
The correct translation of the regular expression 0∗1∗0∗+(1+00+000)∗+(0+1+0+)∗
will be as follows:
^((?:0*1*0*)|(?:1|00|000)*|(?:0+1+0+)*)$
regex101
Debuggex Demo
What your regex [(0*1*0*)[1(00)(000)]*(0+1+0+)*] does:
[(0*1*0*)[1(00)(000)]* -> matches any of characters 0,(,),*,[ zero or more times followed by
(0+1+0+)* --> matches the pattern 0+1+0+ 0 or more times followed by
] --> matches the character ]
so you expression is equivalent to
[([)01](0+1+0+)*] which is not a regular expression to match strings that do not contain 101
0* 1* ( (00+000)* 1*)* (ε+0)
i think this expression covers all cases because --
any number apart from 1 can be broken into constituent 2's and 3's i.e. any number n=2*i+3*j. So there can be any number of 0's between 2 consecutive 1's apart from one 0.Hence, 101 cannot be obtained.
ε+0 for expressions ending in one 0.
The RE for language not containing 101 as sub-string can also be written as (0*1*00)*.0*.1*.0*
This may me a smaller one then what you are using. Try to make use of this.
Regular Expression I got (0+10)1. (looks simple :P)
I just considered all cases to make this.
you consider two 1's we have to end up with continuous 1's
case 1: 11111111111111...
case 2: 0000000011111111111111...(once we take two 1's we cant accept 0's so one and only chance is to continue with 1's)
if you consider only one 1 which was followed by 0 So, no issue and after one 1 we can have any number of 0's.
case 3: 00000000 10100100010000100000100000 1111111111
=>(0*+10*)1
final answer (0+10)1.
Thanks for your patience.
There's a long natural number that can be grouped to smaller numbers by the 0 (zero) delimiter.
Example: 4201100370880
This would divide to Group1: 42, Group2: 110, Group3: 370880
There are 3 groups, groups never start with 0 and are at least 1 char long. Also the last groups is "as is", meaning it's not terminated by a tailing 0.
This is what I came up with, but it only works for certain inputs (like 420110037880):
(\d+)0([1-9][0-9]{1,2})0([1-9]\d+)
This shows I'm attempting to declare the 2nd group's length to min2 max3, but I'm thinking the correct solution should not care about it. If the delimiter was non-numeric I could probably tackle it, but I'm stumped.
All right, factoring in comment information, try splitting on a regex (this may vary based on what language you're using - .split(/.../) in JavaScript, preg_split in PHP, etc.)
The regex you want to split on is: 0(?!0). This translates to "a zero that is not followed by a zero". I believe this will solve your splitting problem.
If your language allows a limit parameter (PHP does), set it to 3. If not, you will need to do something like this (JavaScript):
result = input.split(/0(?!0)/);
result = result.slice(0,2).concat(result.slice(2).join("0"));
The following one should suit your needs:
^(.*?)0(?!0)(.*?)0(?!0)(.*)$
Visualization by Debuggex
The following regex works:
(\d+?)0(?!0) with the g modifier
Demo: http://regex101.com/r/rS4dE5
For only three matches, you can do:
(\d+?)0(?!0)(\d+?)0(?!0)(.*)
I have the requirement to validate a String containing two numbers separated by a dash(-) or a comma(,). Valid values are :
23.98-34.76 or 23.98,34.76
23-34 or 23,34
5-6 or 5,6
I have the following regex which is a slight modification of the answer that I received here in SO. It is covering the 1st and 2nd case above but not the third case involving single digits only.
The modified regex String that I am using is :
(\d+\.?\d+?)([-,])(\d+\.?\d+?)
Where did my regex go wrong?
Correct regex should be like this:
(\d+(\.\d+)?)[-,](\d+(\.\d+)?)
i.e. if there is a period then it is always followed by 1 or more digits.
Otherwise in your regex it will also match strings like 123.,789.
I need a regX which can match like 123,123 only. My regX is
var regX = /^\d*[0-9](|.\d*[0-9]|,\d*[0-9])*$/;
but its currently matching 123,123 and 123, as well
Valid case: 123,123 or 123,000 or 000,000
Invalid case: 123.123 or 123?123 or '123'.'123'
you should use this regex = \d+(,\d+)+
You might want to use the {x,y} quantifier. I matches at least X of the item, and at most Y. If you leave one out, it has no limit in that direction. If you just have one number, with no comma it matches exactly that amount.
Exactly three digits:
(\d{3}),(\d{3})
Three or more
(\d{3,}),(\d{3,})
Between 2 and 7 digits:
(\d{2,7}),(\d{2,7})
And so on...
It looks like you're actually trying to match a number with thousand separators.
Try this: /\d{1,3}(?:,\d{3})*/
If your numbers are positive integers you can use: \d+,\d+
If you want floating point numbers as well: (\d|.)+,(\d|.)+
although this will also match malformed numbers with multiple or misplaced decimal points including .,. etc.
I'm working with some pretty funky HTML markup that I inherited, and I need to remove the following attributes from about 72 td elements.
sdval="285"
I know I can do this with find/replace in my code editor, except since the value of each attribute is different by 5 degree increments, I can't match them all without a Regular Expression. (FYI I'm using Esspress and it does support RegExes in it's Find/Replace tool)
Only trouble is, I really can't figure out how to write a RegEx for this value. I understand the concept of RegExes, but really don't know how to use them.
So how would I write the following with a Regular Expression in place of the digits so that it would match any three digit value?
sdval="285"
/sdval="\d{3}"/
EDIT:
To answer your comment, \d in regular expressions means match any digit, and the {n} construct means repeat the previous item n times.
Easiest, most portable: [0-9][0-9][0-9]
More "modern": \d{3}
This should do (ignores leading zeros):
[1-9][0-9]{0,2}
import re
data = "719"
data1 = "79"
# This expression will match any single, double or triple digit Number
expression = '[\d]{1,3}'
print(re.search(expression, data).string)
# This expression will match only triple digit Number
expression1 = '[\d]{3}'
print(re.search(expression1, data1).string)
Output :
expression : 719
expression1 : 79
It sounds like you're trying to do a find / replace in Visual Studio of a 3 digit number (references to Express and Find/Replace tool). If that's the case the regex to find a 3 digit number in Visual Studio is the following
<:d:d:d>
Breakdown
The < and > establish a word boundary to make sure we don't match a number subset.
Each :d entry matches a single digit.