Splitting string into words with regex
You can use re.split: import re s = “ads1323z123123c123123890sdfakslk123klaad,313ks” results = list(filter(lambda x:len(x) <= 3, re.split(‘[^a-zA-Z]+’, s))) Output: [‘ads’, ‘z’, ‘c’, ‘ks’]
regex
Regex pattern can't figure out
This will work ^\d{4}-\d{3,4}-[23]$ Regex Demo Regex Breakdown ^ #Start of string \d{4} #Match 4 digits – #Match – literally \d{3,4} #Match 3 or 4 digits – #Match – literally [23] #Match 2 or 3 $ #End of string
How to determine if a number is binary and has an even amount of zeros with only one regex?
^[1]*(0[1]*0[1]*)*$ which can be interpreted in the following way: find any number of 1 followed by pairs of 00 with optional one or more 1s in between.
Removing contents within brackets in string [duplicate]
Try this: import re s = “I want to remove all words in brackets( like (this) and ((this)) and ((even) this)).” while True: s_new = re.sub(r’\([^\(]*?\)’, r”, s) if s_new == s: break s = s_new print(s_new) # I want to remove all words in brackets.
regular expression [closed]
If the rest of the url never changes, then you don’t need a regular expression. Just store the first part of the url in a string: $url_prefix = “http:// www.badoobook.com/clips/index.php?page=videos§ion=view&vid_id=”; then append your identifier to it: $id = 100162; $full_url = $url_prefix + $id;
Regex to allow numbers only between 1 and 20000
Although I’m doubtful this is the most-efficient solution, this should do the trick. /^((1[0-9]{0,4})|([1-9][0-9]{0,3})|20000)$/ Be warned that this is string-matching, not type-casting, so things like decimal places and zero-filling will not be matched. Regex Explained: Unit test I used: for (var i = 0; i < 100000; i++) { var should = i > 0 … Read more
Regex – To allow numeric, only characters R, N, X, B, @ [closed]
You want the token [\dRNXB@$]. Add to the ending + if you need it to match 1 or more times or a * if you need it to match 0 or more times. \d is special and matches any digit character. Make sure that if you’re in a context where backslashes are being used for … Read more
Regular expression [ 1 ( 0 1* 0)* 1 ]* DFA
The * at the end can be taken to mean the initial state is accepting and that the automaton returns to this state whenever it accepts anything. Call the initial state q1. To accept 1(01*0)1 we must first consume a 1 and go to a new state, say q2. From there, we can self-loop on … Read more
regex except string
Try this: https?:\/\/example[.]it\/(?!hello).* Working example: http://regexr.com/3dc97 I’m guessing the follow up question would be if you can make domain dynamic but I’m sure you can figure that out.
I need a Javascript REGEX for Integers between 18 and 140 inclusive [closed]
When Answering your question, I am not the person to ask why you need regex for this. And the regex you want is /^(1[89]|[2-9][0-9]|1([0-3][0-9]|40))$/ Sample var age=/^(1[89]|[2-9][0-9]|1([0-3][0-9]|40))$/; console.log(age.test(18)); console.log(age.test(140)); console.log(age.test(12)); console.log(age.test(142)); But, you can simply use the following code to test if(age>=18 && age<=140) That is function test(age){ return age>=18 && age<=140; } console.log(test(18)); console.log(test(140)); … Read more