Here's is the scenario, i have these lines mentioned below i wanted to extract only the middle character in between two dots.
"scvmm.new.resources" --> This after an regular expression match should return only "new"
"sc.new1.rerces" --> This after an regular expression match should return only "new1"
What my basic requirement was to exract anything between two dots anything can come in prefix and suffix
(.*).<required code>.(.*)
Could anyone please help me out??
You can do that without using regex. Split the string on '.' and grab the middle element:
PS> "scvmm.new.resources".Split('.')[1]
new
Or this
'scvmm.new.resources' -replace '.*\.(.*)\..*', '$1'
Like this:
([regex]::Match("scvmm.new1.resources", '(?<=\.)([^\.]*)(?=\.)' )).value
You don't actually need regular expressions for such a trivial substring extraction. Like Shay's Split('.') one can use IndexOf() for similar effect like so,
$s = "scvmm.new.resources"
$l = $s.IndexOf(".")+1
$r = $s.IndexOf(".", $l)
$s.Substring($l, $r-$l) # Prints new
$s = "sc.new1.rerces"
$l = $s.IndexOf(".")+1
$r = $s.IndexOf(".", $l)
$s.Substring($l, $r-$l) # Prints new1
This looks the first occurence of a dot. Then it looks for first occurense of a dot after the first hit. Then it extracts the characters between the two locations. This is useful in, say, scenarios in which the separation characters are not the same (though the Split() way would work in many cases too).
Related
I have this code in perl where I want to extract the value of 'EUR_AF', in this case '0.39'.
Sometimes 'EUR_AF' ends with ';', sometimes it doesn't.
Alternatively, 'EUR_AF' may end with '=0' instead of '=0.39;' or '=0.39'.
How do I make the code handle that? Can't seem to find it online...I could of course wrap everything in an almost endless if-elsif-else statement, but that seems overkill.
Example text:
AVGPOST=0.9092;AN=2184;RSQ=0.5988;ERATE=0.0081;AC=144;VT=SNP;THETA=0.0045;AA=A;SNPSOURCE=LOWCOV;LDAF=0.0959;AF=0.07;ASN_AF=0.05;AMR_AF=0.10;AFR_AF=0.11;EUR_AF=0.039
Code: $INFO =~ m/\;EUR\_AF\=(.*?)(;)/
I did find that: $INFO =~ m/\;EUR\_AF\=(.*?0)/ handles the cases of EUR_AF=0, but how to handle alternative scenarios efficiently?
Extract one value:
my ($eur_af) = $s =~ /(?:^|;)EUR_AF=([^;]*)/;
my ($eur_af) = ";$s" =~ /;EUR_AF=([^;]*)/;
Extract all values:
my %rec = split(/[=;]/, $s);
my $eur_af = $rec{EUR_AF};
This regex should work for you: (?<=EUR_AF=)\d+(\.\d+)?
It means
(?<=EUR_AF=) - look for a string preceeded by EUR_AF=
\d+(\.\d+)? - consist of a digit, optionally a decimal digit
EDIT: I originally wanted the whole regex to return the correct result, not only the capture group. If you want the correct capture group edit it to (?<=EUR_AF=)(\d+(?:\.\d+)?)
I have found the answer. The code:
$INFO =~ m/(?:^|;)EUR_AF=([^;]*)/
seems to handle the cases where EUR_AF=0 and EUR_AF=0.39, ending with or without ;. The resulting $INFO will be 0 or 0.39.
Working with Windows PowerShell, I am trying to replace all occurrences of a certain word pattern in a big string, by appending a suffix.
$string = "The string that contains several occurrences of ab_abcde, ab_abcde_FOO, ab_vwxyz, ab_vwxyz_BAR, etc"
I want to append the suffix "_XXX" to each one of these matched words in $string
Since I expect all my matches to start with either ab_abcde or ab_vwxyz, I have tried:
$string -replace "(ab_abcde|ab_vwxyz)(.?)\s+", '$1$2_XXX'
and a million other variations, to no avail.
I have spent all day hunting at SO, but haven't found a solution that works well. I would appreciate a little assistance.
PS: Since my $string is actually the output of mysqldump, some of my search words will be wrapped in between grave-accent(`) like so: `ab_abcde`, and I would like to match them. I am not sure how much complexity this might bring, considering that this is also the PowerShell escape character.
Thank you.
Something like
$string -replace "((?:ab_abcde|ab_vwxyz)\w*)", '$1_XXX'
Regex Demo
Like this?
$string = "The string that contains several occurrences of ab_abcde, ab_abcde_FOO, ab_vwxyz, ab_vwxyz_BAR, etc"
$string -replace '(ab_abcde.*?\b|ab_vwxyz.*?\b)', '$1_XXX'
The string that contains several occurrences of ab_abcde_XXX, ab_abcde_FOO_XXX, ab_vwxyz_XXX, ab_vwxyz_BAR_XXX, etc
Edit: updated after additional comments posted.
I am trying to get the a sub string from a string using regular expression but it getting error as my regular expression is not working. Can any one help me out in writing correct one :
Here is the Pattern on which i am trying to write the regular expression :
MSM8_BD_V4.3_1-1_idle-Kr_Run3.xlsx
MSM8_BD_V4.3_2-6_mp3-Kr_Run2.xlsx
MSM8_BD_V4.3_Camera_snap-7.xlsx
MSM8_BD_V4.3_Camera_snap-8.xlsx
MSM8_BD_V4.3_Radio_202.16-0.xlsx
I am trying to get the bold part of the substring .
below is the Regular expression i tried:
my $line = "MSM8939_BD_V4.3_1-1_idle-Kratos_Run3.xlsx";
my ($captured) = $line =~ /MSM8939_BD_V4\.\3\_[d]*(.+?)\w/gx;
print "$captured\n";
[d] matches nothing but the literal letter d. You want \d, without the brackets, to match a digit. However, it looks like you also want to include underscores. That would be [\d_].
Try this:
/^MSM8_BD_V4\.3_[\d_]*-?([^-]+)/
If I run this on your input (with e.g. perl -nE 'say $1 if /^MSM8_BD_V4\.3_[\d_]*-?([^-]+)/'), I get this output:
1_idle
6_mp3
Camera_snap
Camera_snap
Radio_202.16
my $line = "MSM8939_BD_V4.3_1-1_idle-Kratos_Run3.xlsx";
for (qw(
MSM8939_BD_V4.3_1-1_idle-Kratos_Run3.xlsx
MSM8939_BD_V4.3_2-6_mp3-Kratos_Run2.xlsx
MSM8939_BD_V4.3_Camera_snap-7.xlsx
MSM8939_BD_V4.3_Camera_snap-8.xlsx
MSM8939_BD_V4.3_Radio_202.16-0.xlsx
)) {
my ($captured) = ($_ =~ /.*[-_]([^\W_]+_[\w.]+)-/gx);
print "$captured\n";
}
Use a greedy pattern to go as far as possible, then grab the last two strings that look like what you want which are still followed by a hyphen.
As does the other answer which was just edited while I was typing, this produces:
1_idle
6_mp3
Camera_snap
Camera_snap
Radio_202.16
This one may be more general in that the beginning of the substring is not hard-coded, i.e., you could use it in other cases which did not necessarily start with MSM8_BD_V4.3.
How would I go about removing all characters before a "_" in perl? So if I had a string that was "124312412_hithere" it would replace the string as just "hithere". I imagine there is a very simple way to do this using regex, but I am still new dealing with that so I need help here.
Remove all characters up to and including "_":
s/^[^_]*_//;
Remove all characters before "_":
s/^[^_]*(?=_)//;
Remove all characters before "_" (assuming the presence of a "_"):
s/^[^_]*//;
This is a bit more verbose than it needs to be, but would be probably more valuable for you to see what's going on:
my $astring = "124312412_hithere";
my $find = "^[^_]*_";
my $replace = "_";
$astring =~ s/$find/$replace/;
print $astring;
Also, there's a bit of conflicting requirements in your question. If you just want hithere (without the leading _), then change it to:
$astring =~ s/$find//;
I know it's slightly different than what was asked, but in cases like this (where you KNOW the character you are looking for exists in the string) I prefer to use split:
$str = '124312412_hithere';
$str = (split (/_/, $str, 2))[1];
Here I am splitting the string into parts, using the '_' as a delimiter, but to a maximum of 2 parts. Then, I am assigning the second part back to $str.
There's still a regex in this solution (the /_/) but I think this is a much simpler solution to read and understand than regexes full of character classes, conditional matches, etc.
You can try out this: -
$_ = "124312412_hithere";
s/^[^_]*_//;
print $_; # hithere
Note that this will also remove the _(as I infer from your sample output). If you want to keep the _ (as it seems doubtful what you want as per your first statement), you would probably need to use look-ahead as in #ikegami's answer.
Also, just to make it little more clear, any substitution and matching in regex is applied by default on $_. So, you don't need to bind it to $_ explicitly. That is implied.
So, s/^[^_]*_//; is essentially same as - $_ =~ s/^[^_]*_//;, but later one is not really required.
How do I return the first 5 digits of a string of characters in Regular Expressions?
For example, if I have the following text as input:
15203 Main Street
Apartment 3 63110
How can I return just "15203".
I am using C#.
This isn't really the kind of problem that's ideally solved by a single-regex approach -- the regex language just isn't especially meant for it. Assuming you're writing code in a real language (and not some ill-conceived embedded use of regex), you could do perhaps (examples in perl)
# Capture all the digits into an array
my #digits = $str =~ /(\d)/g;
# Then take the first five and put them back into a string
my $first_five_digits = join "", #digits[0..4];
or
# Copy the string, removing all non-digits
(my $digits = $str) =~ tr/0-9//cd;
# And cut off all but the first five
$first_five_digits = substr $digits, 0, 5;
If for some reason you really are stuck doing a single match, and you have access to the capture buffers and a way to put them back together, then wdebeaum's suggestion works just fine, but I have a hard time imagining a situation where you can do all that, but don't have access to other language facilities :)
it would depend on your flavor of Regex and coding language (C#, PERL, etc.) but in C# you'd do something like
string rX = #"\D+";
Regex.replace(input, rX, "");
return input.SubString(0, 5);
Note: I'm not sure about that Regex match (others here may have a better one), but basically since Regex itself doesn't "replace" anything, only match patterns, you'd have to look for any non-digit characters; once you'd matched that, you'd need to replace it with your languages version of the empty string (string.Empty or "" in C#), and then grab the first 5 characters of the resulting string.
You could capture each digit separately and put them together afterwards, e.g. in Perl:
$str =~ /(\d)\D*(\d)\D*(\d)\D*(\d)\D*(\d)/;
$digits = $1 . $2 . $3 . $4 . $5;
I don't think a regular expression is the best tool for what you want.
Regular expressions are to match patterns... the pattern you are looking for is "a(ny) digit"
Your logic external to the pattern is "five matches".
Thus, you either want to loop over the first five digit matches, or capture five digits and merge them together.
But look at that Perl example -- that's not one pattern -- it's one pattern repeated five times.
Can you do this via a regular expression? Just like parsing XML -- you probably could, but it's not the right tool.
Not sure this is best solved by regular expressions since they are used for string matching and usually not for string manipulation (in my experience).
However, you could make a call to:
strInput = Regex.Replace(strInput, "\D+", "");
to remove all non number characters and then just return the first 5 characters.
If you are wanting just a straight regex expression which does all this for you I am not sure it exists without using the regex class in a similar way as above.
A different approach -
#copy over
$temp = $str;
#Remove non-numbers
$temp =~ s/\D//;
#Get the first 5 numbers, exactly.
$temp =~ /\d{5}/;
#Grab the match- ASSUMES that there will be a match.
$first_digits = $1
result =~ s/^(\d{5}).*/$1/
Replace any text starting with a digit 0-9 (\d) exactly 5 of them {5} with any number of anything after it '.*' with $1, which is the what is contained within the (), that is the first five digits.
if you want any first 5 characters.
result =~ s/^(.{5}).*/$1/
Use whatever programming language you are using to evaluate this.
ie.
regex.replace(text, "^(.{5}).*", "$1");