Using regular expression how do I replace 1,186.55 with 1186.55?
My search string is
\b[1-9],[0-9][0-9][0-9].[0-9][0-9]
which works fine. I just can't seem to get the replacement part to work.
You are very sparse with information in your question. I try to answer as general as possible:
You can shorten the regex a bit by using quantifiers, I would make this in a first step
\b[1-9],[0-9]{3}.[0-9]{2}
Most probably you can also replace [0-9] by \d, is also more readable IMO.
\b\d,\d{3}.\d{2}
Now we can go to the replacement part. Here you need to store the parts you want to keep. You can do that by putting that part into capturing groups, by placing brackets around, this would be your search pattern:
\b(\d),(\d{3}.\d{2})
So, now you can access the matched content of those capturing groups in the replacement string. The first opening bracket is the first group the second opening bracket is the second group, ...
Here there are now two possibilities, either you can get that content by \1 or by $1
Your replacement string would then be
\1\2
OR
$1$2
Python:
def repl(initstr, unwanted=','):
res = set(unwanted)
return ''.join(r for r in initstr if r not in res)
Using regular expressions:
from re import compile
regex = compile(r'([\d\.])')
print ''.join(regex.findall('1,186.55'))
Using str.split() method:
num = '1,186.55'
print ''.join(num.split(','))
Using str.replace() method:
num = '1,186.55'
print num.replace(',', '')
if you just wanna remove the comma you can do(in java or C#):
str.Replace(",", "");
(in java it's replace)
Or in Perl:
s/(\d+),(\d+)/$1$2/
Related
For example we have a string:
asd/asd/asd/asd/1#s_
I need to match this part: /asd/1#s_ or asd/1#s_
How is it possible to do with plain regex?
I've tried negative lookahead like this
But it didn't work
\/(?:.(?!\/))?(asd)(\/(([\W\d\w]){1,})|)$
it matches this '/asd/asd/asd/asd/asd/asd/1#s_'
from this 'prefix/asd/asd/asd/asd/asd/asd/1#s_'
and I need to match '/asd/1#s_' without all preceding /asd/'s
Match should work with plain regex
Without any helper functions of any programming language
https://regexr.com/
I use this site to check if regex matches or not
here's the possible strings:
prefix/asd/asd/asd/1#s
prefix/asd/asd/asd/1s#
prefix/asd/asd/asd/s1#
prefix/asd/asd/asd/s#1
prefix/asd/asd/asd/#1s
prefix/asd/asd/asd/#s1
and asd part could be replaced with any word like
prefix/a1sd/a1sd/a1sd/1#s
prefix/a1sd/a1sd/a1sd/1s#
...
So I need to match last repeating part with everything to the right
And everything to the right could be character, not character, digit, in any order
A more complicated string example:
prefix/a1sd/a1sd/a1sd/1s#/ds/dsse/a1sd/22$$#!/123/321/asd
this should match that part:
/a1sd/22$$#!/123/321/asd
Try this one. This works in python.
import re
reg = re.compile(r"\/[a-z]{1,}\/\d+[#a-z_]{1,}")
s = "asd/asd/asd/asd/1#s_"
print(reg.findall(s))
# ['/asd/1#s_']
Update:
Since the question lacks clarity, this only works with the given order and hence, I suppose any other combination simply fails.
Edits:
New Regex
reg = r"\/\w+(\/\w*\d+\W*)*(\/\d+\w*\W*)*(\/\d+\W*\w*)*(\/\w*\W*\d+)*(\/\W*\d+\w*)*(\/\W*\w*\d+)*$"
Simple regex function that matches the start of a string "Bananas: " and returns the second part. I've done the regex, but it's not the way I expected it to work:
import re
def return_name(s):
m = re.match(r"^Bananas:\s?(.*)", s)
if m:
# print m.group(0)
# print m.group(1)
return m.group(1)
somestring = "Bananas: Gwen Stefani" # Bananas: + name
print return_name(somestring) # Gwen Stefani - correct!
However, I'm convinced that you don't have identify the group with (.*) in order to get the same results. ie match first part of string - return the remaining part. But I'm not sure how to do that.
Also I read somewhere that you should be being cautious using .* in a regex.
You could use a lookbehind ((?<=)):
(?<=^Bananas:\s).*
Remember to use re.search instead of re.match as the latter will try to match at the start of the string (aka implicit ^).
As for the .* concerns - it can cause a lot of backtracking if you don't have a clear understanding of how regexes work, but in this case it is guaranteed to be a linear search.
Using the alternate regular expression module "regex" you could use perl's \K meta-character, which makes it able to discard previously matched content and only Keep the following.
I'm not really recommending this, I think your solution is good enough, and the lookbehind answer is also probably better than using another module just for that.
Within Notepad++, I want to replace many instances of the type ``string'' by \command{string} where string can be any string of characters. I am fairly close to what I want to achieve with:
Find: (?<=``)(.*?)(?='')
Replace: \\command{\1}
There is still a problem. With the regex code above, instead of \command{string} I get ``\command{string}'' and I am not sure why the `` and '' are not removed?
It is because you are using lookaround assertions. Lookaround (zero-width) assertions only assert that a position can be matched and do not "consume" any characters on the string. You can use the below regular expression.
Find: ``([^']+)''
Replace: \\command{\1}
You need to wrap everything into a capture group and use that. NP++ seems to not support lookahead/behind, but you dont need that for this specific case anyway:
``([^']+)'' -> \\command{\1}
This will make sure it does not match two commands (longest match) in something like:
run ``ls -l'' or ``ls -a''
I have this type of text:
string1_dog_bit_johny_bit_string2
string1_cat_bit_johny_bit_string2
string1_crocodile_bit_johny_bit_string2
string3_crocodile_bit_johny_bit_string4
string4_crocodile_bit_johny_bit_string5
I want to find all occurrences of “bit” that occur only between string1 and string2. How do I do this with regex?
I found the question Regex Match all characters between two strings, but the regex there matches the entire string between string1 and string2, whereas I want to match just parts of that string.
I am doing a global replacement in Notepad++. I just need regex, code will not work.
Thank you in advance.
Roman
If I understand correctly here a code to do what you want
var intput = new List<string>
{
"string1_dog_bit_johny_bit_string2",
"string1_cat_bit_johny_bit_string2",
"string1_crocodile_bit_johny_bit_string2",
"string3_crocodile_bit_johny_bit_string4",
"string4_crocodile_bit_johny_bit_string5"
};
Regex regex = new Regex(#"(?<bitGroup>bit)");
var allMatches = new List<string>();
foreach (var str in intput)
{
if (str.StartsWith("string1") && str.EndsWith("string2"))
{
var matchCollection = regex.Matches(str);
allMatches.AddRange(matchCollection.Cast<Match>().Select(match => match.Groups["bitGroup"].Value));
}
}
Console.WriteLine("All matches {0}", allMatches.Count);
This regex will do the job:
^string1_(?:.*(bit))+.*_string2$
^ means the start of the text (or line if you use the m option like so: /<regex>/m )
$ means the end of the text
. means any character
* means the previous character/expression is repeated 0 or more times
(?:<stuff>) means a non-capturing group (<stuff> won't be captured as a result of the matching)
You could use ^string1_(.*(bit).*)*_string2$ if you don't care about performance or don't have large/many strings to check. The outer parenthesis allow multiple occurences of "bit".
If you provide us with the language you want to use, we could give more specific solutions.
edit: As you added that you're trying a replacement in Notepad++ I propose the following:
Use (?<=string1_)(.*)bit(.*)(?=_string2) as regex and $1xyz$2 as replacement pattern (replace xyz with your string). Then perform an "replace all" operation until N++ doesn't find any more matches. The problem here is that this regex will only match 1 bit per line per iteration - and therefore needs to be applied repeatedly.
Btw. even if a regexp matches the whole line, you can still only replace parts of it using capturing groups.
You can use the regex:
(?:string1|\G)(?:(?!string2).)*?\Kbit
regex101 demo. Tried it on notepad++ as well and it's working.
There're description in the demo site, but if you want more explanations, let me know and I'll elaborate!
I need to extract from a string a set of characters which are included between two delimiters, without returning the delimiters themselves.
A simple example should be helpful:
Target: extract the substring between square brackets, without returning the brackets themselves.
Base string: This is a test string [more or less]
If I use the following reg. ex.
\[.*?\]
The match is [more or less]. I need to get only more or less (without the brackets).
Is it possible to do it?
Easy done:
(?<=\[)(.*?)(?=\])
Technically that's using lookaheads and lookbehinds. See Lookahead and Lookbehind Zero-Width Assertions. The pattern consists of:
is preceded by a [ that is not captured (lookbehind);
a non-greedy captured group. It's non-greedy to stop at the first ]; and
is followed by a ] that is not captured (lookahead).
Alternatively you can just capture what's between the square brackets:
\[(.*?)\]
and return the first captured group instead of the entire match.
If you are using JavaScript, the solution provided by cletus, (?<=\[)(.*?)(?=\]) won't work because JavaScript doesn't support the lookbehind operator.
Edit: actually, now (ES2018) it's possible to use the lookbehind operator. Just add / to define the regex string, like this:
var regex = /(?<=\[)(.*?)(?=\])/;
Old answer:
Solution:
var regex = /\[(.*?)\]/;
var strToMatch = "This is a test string [more or less]";
var matched = regex.exec(strToMatch);
It will return:
["[more or less]", "more or less"]
So, what you need is the second value. Use:
var matched = regex.exec(strToMatch)[1];
To return:
"more or less"
Here's a general example with obvious delimiters (X and Y):
(?<=X)(.*?)(?=Y)
Here it's used to find the string between X and Y. Rubular example here, or see image:
You just need to 'capture' the bit between the brackets.
\[(.*?)\]
To capture you put it inside parentheses. You do not say which language this is using. In Perl for example, you would access this using the $1 variable.
my $string ='This is the match [more or less]';
$string =~ /\[(.*?)\]/;
print "match:$1\n";
Other languages will have different mechanisms. C#, for example, uses the Match collection class, I believe.
[^\[] Match any character that is not [.
+ Match 1 or more of the anything that is not [. Creates groups of these matches.
(?=\]) Positive lookahead ]. Matches a group ending with ] without including it in the result.
Done.
[^\[]+(?=\])
Proof.
http://regexr.com/3gobr
Similar to the solution proposed by null. But the additional \] is not required. As an additional note, it appears \ is not required to escape the [ after the ^. For readability, I would leave it in.
Does not work in the situation in which the delimiters are identical. "more or less" for example.
Most updated solution
If you are using Javascript, the best solution that I came up with is using match instead of exec method.
Then, iterate matches and remove the delimiters with the result of the first group using $1
const text = "This is a test string [more or less], [more] and [less]";
const regex = /\[(.*?)\]/gi;
const resultMatchGroup = text.match(regex); // [ '[more or less]', '[more]', '[less]' ]
const desiredRes = resultMatchGroup.map(match => match.replace(regex, "$1"))
console.log("desiredRes", desiredRes); // [ 'more or less', 'more', 'less' ]
As you can see, this is useful for multiple delimiters in the text as well
PHP:
$string ='This is the match [more or less]';
preg_match('#\[(.*)\]#', $string, $match);
var_dump($match[1]);
This one specifically works for javascript's regular expression parser /[^[\]]+(?=])/g
just run this in the console
var regex = /[^[\]]+(?=])/g;
var str = "This is a test string [more or less]";
var match = regex.exec(str);
match;
To remove also the [] use:
\[.+\]
I had the same problem using regex with bash scripting.
I used a 2-step solution using pipes with grep -o applying
'\[(.*?)\]'
first, then
'\b.*\b'
Obviously not as efficient at the other answers, but an alternative.
I wanted to find a string between / and #, but # is sometimes optional. Here is the regex I use:
(?<=\/)([^#]+)(?=#*)
Here is how I got without '[' and ']' in C#:
var text = "This is a test string [more or less]";
// Getting only string between '[' and ']'
Regex regex = new Regex(#"\[(.+?)\]");
var matchGroups = regex.Matches(text);
for (int i = 0; i < matchGroups.Count; i++)
{
Console.WriteLine(matchGroups[i].Groups[1]);
}
The output is:
more or less
If you need extract the text without the brackets, you can use bash awk
echo " [hola mundo] " | awk -F'[][]' '{print $2}'
result:
hola mundo