I am trying to use either Perl or MATLAB to parse a few numbers out of a single line of text. My text line is:
t10_t20_t30_t40_
now in matlab, i used the following script
str = 't10_t20_t30_t40_';
a = regexp(str,'t(\d+)_t(\d+)','match')
and it returns
a =
't10_t20' 't30_t40'
What I want is for it to also return 't20_t30', since this obviously is a match. Why doesn't regexp scan it?
I thus turned to Perl, and wrote the following in Perl:
#!/usr/bin/perl -w
$str = "t10_t20_t30_t40_";
while($str =~ /(t\d+_t\d+)/g)
{
print "$1\n";
}
and the result is the same as matlab
t10_t20
t30_t40
but I really wanted "t20_t30" also be in the results.
Can anyone tell me how to accomplish that? Thanks!
[update with a solution]:
With help from colleagues, I identified a solution using the so-called "look-around assertion" afforded by Perl.
#!/usr/bin/perl -w
$str = "t10_t20_t30_t40_";
while($str =~ m/(?=(t\d+_t\d+))/g)
{print "$1\n";}
The key is to use "zero width look-ahead assertion" in Perl. When Perl (and other similar packages) uses regexp to scan a string, it does not re-scan what was already scanned in the last match. So in the above example, t20_t30 will never show up in the results. To capture that, we need to use a zero-width lookahead search to scan the string, producing matches that do not exclude any substrings from subsequent searches (see the working code above). The search will start from zero-th position and increment by one as many times as possible if "global" modifier is appended to the search (i.e. m//g), making it a "greedy" search.
This is explained in more detail in this blog post.
The expression (?=t\d+_t\d+) matches any 0-width string followed by t\d+_t\d+, and this creates the actual "sliding window". This effectively returns ALL t\d+_t\d+ patterns in $str without any exclusion since every position in $str is a 0-width string. The additional parenthesis captures the pattern while its doing sliding matching (?=(t\d+_t\d+)) and thus returns the desired sliding window outcome.
Using Perl:
#!/usr/bin/perl
use Data::Dumper;
use Modern::Perl;
my $re = qr/(?=(t\d+_t\d+))/;
my #l = 't10_t20_t30_t40' =~ /$re/g;
say Dumper(\#l);
Output:
$VAR1 = [
't10_t20',
't20_t30',
't30_t40'
];
Once the regexp algorithm has found a match, the matched characters are not considered for further matches (and usually, this is what one wants, e.g. .* is not supposed to match every conceivable contiguous substring of this post). A workaround would be to start the search again one character after the first match, and collect the results:
str = 't10_t20_t30_t40_';
sub_str = str;
reg_ex = 't(\d+)_t(\d+)';
start_idx = 0;
all_start_indeces = [];
all_end_indeces = [];
off_set = 0;
%// While there are matches later in the string and the first match of the
%// remaining string is not the last character
while ~isempty(start_idx) && (start_idx < numel(str))
%// Calculate offset to original string
off_set = off_set + start_idx;
%// extract string starting at first character after first match
sub_str = sub_str((start_idx + 1):end);
%// find further matches
[start_idx, end_idx] = regexp(sub_str, reg_ex, 'once');
%// save match if any
if ~isempty(start_idx)
all_start_indeces = [all_start_indeces, start_idx + off_set];
all_end_indeces = [all_end_indeces, end_idx + off_set];
end
end
display(all_start_indeces)
display(all_end_indeces)
matched_strings = arrayfun(#(st, en) str(st:en), all_start_indeces, all_end_indeces, 'uniformoutput', 0)
Related
If I have the lines:
'aslkdfjcacttlaksdjcacttlaksjdfcacttlskjdf'
'asfdcacttaskdfjcacttklasdjf'
'cksjdfcacttlkasdjf'
I want to match them by the number of times a repeating subunit (cactt) occurs. In other words, if I ask for n repeats, I want matches that contain n and ONLY n instances of the pattern.
My initial attempt was implemented in perl and looks like this:
sub MATCHER {
print "matches with $_ CACTT's\n";
my $pattern = "^(.*?CACTT.+?){$_}(?!.*?CACTT).*\$";
my #grep_matches = grep(/$pattern/, #matching);
print "$_\n" for #grep_matches;
my #copy = #grep_matches;
my $squashed = #copy;
print "number of rows total: $squashed\n";
}
for (2...6) {
MATCHER($_);
}
Notes:
#matching contains the strings from 1, 2, and 3 in an array.
the for loop is set from integers 2-6 because I have a separate regex that works to forbid duplicate occurrences of the pattern.
This loop ALMOST works except that for n=2, matches containing 3 occurrences of the "cactt" pattern are returned. In fact, for any string containing n+1 matches (where n>=2), lines with n+1 occurrences are also returned by the match. I though the negative lookahead could prevent this behavior in perl. If anyone could give me thoughts, I would be appreciative.
Also, I have thought of getting a count per line and separating them by count; I dislike the approach because it requires two steps when one should accomplish what I want.
I would be okay with a:
foreach (#matches) { $_ =~ /$pattern/; push(#selected_by_n, $1);}
The regex seems like it should be similar, but for whatever reason in practice the results differ dramatically.
Thanks in advance!
Your code is sort of strange. This regex
my $pattern = "^(.*?CACTT.+?){$_}(?!.*?CACTT).*\$";
..tries to match first beginning of string ^, then a minimal match of any character .*?, followed by your sequence CACTT, followed by a minimal match (but slightly different from .*?) .+?. And you want to match these $_ times. You assume $_ will be correct when calling the sub (this is bad). Then you have a look-ahead assumption that wants to make sure that there is no minimal match of any char .*? followed by your sequence, followed by any char of any length followed by end of line $.
First off, this is always redundant: ^.*. Beginning of line anchor followed by any character any number of times. This actually makes the anchor useless. Same goes for .*$. Why? Because any match that will occur, will occur anyway at the first possible time. And .*$ matches exactly the same thing that the empty string does: Anything.
For example: the regex /^.*?foo.*?$/ matches exactly the same thing as /foo/. (Excluding cases of multiline matching with strings that contain newlines).
In your case, if you want to count the occurrences of a string inside a string, you can just match them like this:
my $count = () = $str =~ /CACTT/gi;
This code:
my #copy = #grep_matches;
my $squashed = #copy;
Is completely redundant. You can just do my $squashed = #grep_matches. It makes little to no sense to first copy the array.
This code:
MATCHER($_);
Does the same as this: MATCHER("foo") or MATCHER(3.1415926536). You are not using the subroutine argument, you are ignoring it, and relying on the fact that $_ is global and visible inside the sub. What you want to do is
sub MATCHER {
my $number = shift; # shift argument from #_
Now you have encapsulated the code and all is well.
What you want to do in your case, I assume, is to count the occurrences of the substring inside your strings, then report them. I would do something like this
use strict;
use warnings;
use Data::Dumper;
my %data;
while (<DATA>) {
chomp;
my $count = () = /cactt/gi; # count number of matches
push #{ $data{$count} }, $_; # store count and original
}
print Dumper \%data;
__DATA__
aslkdfjcacttlaksdjcacttlaksjdfcacttlskjdf
asfdcacttaskdfjcacttklasdjf
cksjdfcacttlkasdjf
This will print
$VAR1 = {
'2' => [
'asfdcacttaskdfjcacttklasdjf'
],
'3' => [
'aslkdfjcacttlaksdjcacttlaksjdfcacttlskjdf'
],
'1' => [
'cksjdfcacttlkasdjf'
]
};
This is just to demonstrate how to create the data structure. You can now access the strings in the order of matches. For example:
for (#$data{3}) { # print strings with 3 matches
print;
}
Would you just do something like this:
use warnings;
use strict;
my $n=2;
my $match_line_cnt=0;
my $line_cnt=0;
while (<DATA>) {
my $m_cnt = () = /cactt/g;
if ($m_cnt>=$n){
print;
$match_line_cnt++;
}
$line_cnt++;
}
print "total lines: $line_cnt\n";
print "matched lines: $match_line_cnt\n";
print "squashed: ",$line_cnt-$match_line_cnt;
__DATA__
aslkdfjcacttlaksdjcacttlaksjdfcacttlskjdf
asfdcacttaskdfjcacttklasdjf
cksjdfcacttlkasdjf
prints:
aslkdfjcacttlaksdjcacttlaksjdfcacttlskjdf
asfdcacttaskdfjcacttklasdjf
total lines: 3
matched lines: 2
squashed: 1
I think you're unintentionally asking two seperate questions.
If you want to directly capture the number of times a pattern matches in a string, this one liner is all you need.
$string = 'aslkdfjcacttlaksdjcacttlaksjdfcacttlskjdf';
$pattern = qr/cactt/;
print $count = () = $string =~ m/$pattern/g;
-> 3
That last line is as if you had written $count = #junk = $string =~ m/$pattern/g; but without needing an intermediate array variable. () = is the null list assignment and it throws away whatever is assigned to it just like scalar undef = throws away its right hand side. But, the null list assignment still returns the number of things thrown away when its left hand side is in scalar context. It returns an empty list in list context.
If you want to match strings that only contain some number of pattern matches, then you want to stop matching once too many are found. If the string is large (like a document) then you would waste a lot of time counting past n.
Try this.
sub matcher {
my ($string, $pattern, $n) = #_;
my $c = 0;
while ($string =~ m/$pattern/g) {
$c++;
return if $c > $n;
}
return $c == $n ? 1 : ();
}
Now there is one more option but if you call it over and over again it gets inefficient. You can build a custom regex that matches only n times on the fly. If you only build this once however, it's just fine and speedy. I think this is what you originally had in mind.
$regex = qr/^(?:(?:(?!$pattern).)*$pattern){$n}(?:(?!$pattern).)*$/;
I'll leave the rest of that one to you. Check for n > 1 etc. The key is understanding how to use lookahead. You have to match all the NOT THINGS before you try to match THING.
https://perldoc.perl.org/perlre
Need to count the number of "$0.00" in a string. I'm using:
my $zeroDollarCount = ("\Q$menu\E" =~ tr/\$0\.00//);
but it doesn't work. The issue is the $ sign is throwing the regex off. It works if I just want to count the number of $, but fails to find $0.00.
How is this a duplicate? Your solution does not address dollar sign which is an issue for me.
You are using the transliteration operator tr///. That doesn't have anything to do with a pattern. You need the match operator m// instead. And because you want it to find all occurances of the pattern, use the /g modifier.
my $count = () = $menu =~ m/\$0\.00/g;
If we run this program, the output is 2.
use strict;
use warnings;
my $menu = '$0.00 and $0.00';
my $count = () = $menu =~ m/\$0\.00/g;
print $count;
Now lets take a look at what is going on. First, the pattern of the match.
/\$0\.00/
This is fairly straight-forward. There is a literal $, which we need to escape with a backslash \. The zero is followed by a literal dot ., which again we need to escape, because like the $ it has special meanings in regular expressions.
my $count = () = $menu =~ m/\$0\.00/g;
This whole line looks weird. We can break it up into a few lines to make it more readable.
my #matches = ( $menu =~ m/\$0\.00/g );
my $count = scalar #matches;
We need the /g switch on the regular expression match to make it match all occurrences. In list context, the match operation returns all matches (which will be the string "$0.00" a number of times). Because we want the count, we then force that into scalar context, which gives us the number of elements. That can be shortened to one line by the idiom shown above.
Using just one Perl substitute regular expression statement (s///), how can we write below:
Every success match contains just a string of Alphabetic characters A..Z. We need to substitute the match string with a substitution that will be the sum of character index (in alphabetical order) of every character in the match string.
Note: For A, character index would be 1, for B, 2 ... and for Z would be 26.
Please see example below:
success match: ABCDMNA
substitution result: 38
Note:
1 + 2 + 3 + 4 + 13 + 14 + 1 = 38;
since
A = 1, B = 2, C = 3, D = 4, M = 13, N = 14 and A = 1.
I will post this as an answer, I guess, though the credit for coming up with the idea should go to abiessu for the idea presented in his answer.
perl -ple'1 while s/(\d*)([A-Z])/$1+ord($2)-64/e'
Since this is clearly homework and/or of academic interest, I will post the explanation in spoiler tags.
- We match an optional number (\d*), followed by a letter ([A-Z]). The number is the running sum, and the letter is what we need to add to the sum.
- By using the /e modifier, we can do the math, which is add the captured number to the ord() value of the captured letter, minus 64. The sum is returned and inserted instead of the number and the letter.
- We use a while loop to rinse and repeat until all letters have been replaced, and all that is left is a number. We use a while loop instead of the /g modifier to reset the match to the start of the string.
Just split, translate, and sum:
use strict;
use warnings;
use List::Util qw(sum);
my $string = 'ABCDMNA';
my $sum = sum map {ord($_) - ord('A') + 1} split //, $string;
print $sum, "\n";
Outputs:
38
Can you use the /e modifier in the substitution?
$s = "ABCDMNA";
$s =~ s/(.)/$S += ord($1) - ord "#"; 1 + pos $s == length $s ? $S : ""/ge;
print "$s\n"
Consider the following matching scenario:
my $text = "ABCDMNA";
my $val = $text ~= s!(\d)*([A-Z])!($1+ord($2)-ord('A')+1)!gr;
(Without having tested it...) This should repeatedly go through the string, replacing one character at a time with its ordinal value added to the current sum which has been placed at the beginning. Once there are no more characters the copy (/r) is placed in $val which should contain the translated value.
Or an short alternative:
echo ABCDMNA | perl -nlE 'm/(.)(?{$s+=-64+ord$1})(?!)/;say$s'
or readable
$s = "ABCDMNA";
$s =~ m/(.)(?{ $sum += ord($1) - ord('A')+1 })(?!)/;
print "$sum\n";
prints
38
Explanation:
trying to match any character what must not followed by "empty regex". /.(?!)/
Because, an empty regex matches everything, the "not follow by anything", isn't true ever.
therefore the regex engine move to the next character, and tries the match again
this is repeated until is exhausted the whole string.
because we want capture the character, using capture group /(.)(?!)/
the (?{...}) runs the perl code, what sums the value of the captured character stored in $1
when the regex is exhausted (and fails), the last say $s prints the value of sum
from the perlre
(?{ code })
This zero-width assertion executes any embedded Perl code. It always
succeeds, and its return value is set as $^R .
WARNING: Using this feature safely requires that you understand its
limitations. Code executed that has side effects may not perform
identically from version to version due to the effect of future
optimisations in the regex engine. For more information on this, see
Embedded Code Execution Frequency.
I have a question I am hoping someone could help with...
I have a variable that contains the content from a webpage (scraped using WWW::Mechanize).
The variable contains data such as these:
$var = "ewrfs sdfdsf cat_dog,horse,rabbit,chicken-pig"
$var = "fdsf iiukui aawwe dffg elephant,MOUSE_RAT,spider,lion-tiger hdsfds jdlkf sdf"
$var = "dsadp poids pewqwe ANTELOPE-GIRAFFE,frOG,fish,crab,kangaROO-KOALA sdfdsf hkew"
The only bits I am interested in from the above examples are:
#array = ("cat_dog","horse","rabbit","chicken-pig")
#array = ("elephant","MOUSE_RAT","spider","lion-tiger")
#array = ("ANTELOPE-GIRAFFE","frOG","fish","crab","kangaROO-KOALA")
The problem I am having:
I am trying to extract only the comma-separated strings from the variables and then store these in an array for use later on.
But what is the best way to make sure that I get the strings at the start (ie cat_dog) and end (ie chicken-pig) of the comma-separated list of animals as they are not prefixed/suffixed with a comma.
Also, as the variables will contain webpage content, it is inevitable that there may also be instances where a commas is immediately succeeded by a space and then another word, as that is the correct method of using commas in paragraphs and sentences...
For example:
Saturn was long thought to be the only ringed planet, however, this is now known not to be the case.
^ ^
| |
note the spaces here and here
I am not interested in any cases where the comma is followed by a space (as shown above).
I am only interested in cases where the comma DOES NOT have a space after it (ie cat_dog,horse,rabbit,chicken-pig)
I have a tried a number of ways of doing this but cannot work out the best way to go about constructing the regular expression.
How about
[^,\s]+(,[^,\s]+)+
which will match one or more characters that are not a space or comma [^,\s]+ followed by a comma and one or more characters that are not a space or comma, one or more times.
Further to comments
To match more than one sequence add the g modifier for global matching.
The following splits each match $& on a , and pushes the results to #matches.
my $str = "sdfds cat_dog,horse,rabbit,chicken-pig then some more pig,duck,goose";
my #matches;
while ($str =~ /[^,\s]+(,[^,\s]+)+/g) {
push(#matches, split(/,/, $&));
}
print join("\n",#matches),"\n";
Though you can probably construct a single regex, a combination of regexs, splits, grep and map looks decently
my #array = map { split /,/ } grep { !/^,/ && !/,$/ && /,/ } split
Going from right to left:
Split the line on spaces (split)
Leave only elements having no comma at the either end but having one inside (grep)
Split each such element into parts (map and split)
That way you can easily change the parts e.g. to eliminate two consecutive commas add && !/,,/ inside grep.
I hope this is clear and suits your needs:
#!/usr/bin/perl
use warnings;
use strict;
my #strs = ("ewrfs sdfdsf cat_dog,horse,rabbit,chicken-pig",
"fdsf iiukui aawwe dffg elephant,MOUSE_RAT,spider,lion-tiger hdsfds jdlkf sdf",
"dsadp poids pewqwe ANTELOPE-GIRAFFE,frOG,fish,crab,kangaROO-KOALA sdfdsf hkew",
"Saturn was long thought to be the only ringed planet, however, this is now known not to be the case.",
"Another sentence, although having commas, should not confuse the regex with this: a,b,c,d");
my $regex = qr/
\s #From your examples, it seems as if every
#comma separated list is preceded by a space.
(
(?:
[^,\s]+ #Now, not a comma or a space for the
#terms of the list
, #followed by a comma
)+
[^,\s]+ #followed by one last term of the list
)
/x;
my #matches = map {
$_ =~ /$regex/;
if ($1) {
my $comma_sep_list = $1;
[split ',', $comma_sep_list];
}
else {
[]
}
} #strs;
$var =~ tr/ //s;
while ($var =~ /(?<!, )\b[^, ]+(?=,\S)|(?<=,)[^, ]+(?=,)|(?<=\S,)[^, ]+\b(?! ,)/g) {
push (#arr, $&);
}
the regular expression matches three cases :
(?<!, )\b[^, ]+(?=,\S) : matches cat_dog
(?<=,)[^, ]+(?=,) : matches horse & rabbit
(?<=\S,)[^, ]+\b(?! ,) : matches chicken-pig
I'm writing regular expression for checking if there is a substring, that contains at least 2 repeats of some pattern next to each other. I'm matching the result of regex with former string - if equal, there is such pattern. Better said by example: 1010 contains pattern 10 and it is there 2 times in continuous series. On other hand 10210 wouldn't have such pattern, because those 10 are not adjacent.
What's more, I need to find the longest pattern possible, and it's length is at least 1. I have written the expression to check for it ^.*?(.+)(\1).*?$. To find longest pattern, I've used non-greedy version to match something before patter, then pattern is matched to group 1 and once again same thing that has been matched for group1 is matched. Then the rest of string is matched, producing equal string. But there's a problem that regex is eager to return after finding first pattern, and don't really take into account that I intend to make those substrings before and after shortest possible (leaving the rest longest possible). So from string 01011010 I get correctly that there's match, but the pattern stored in group 1 is just 01 though I'd except 101.
As I believe I can't make pattern "more greedy" or trash before and after even "more non-greedy" I can only come whit an idea to make regex less eager, but I'm not sure if this is possible.
Further examples:
56712453289 - no pattern - no match with former string
22010110100 - pattern 101 - match with former string (regex resulted in 22010110100 with 101 in group 1)
5555555 - pattern 555 - match
1919191919 - pattern 1919 - match
191919191919 - pattern 191919 - match
2323191919191919 - pattern 191919 - match
What I would get using current expression (same strings used):
no pattern - no match
pattern 2 - match
pattern 555 - match
pattern 1919 - match
pattern 191919 - match
pattern 23 - match
In Perl you can do it with one expression with help of (??{ code }):
$_ = '01011010';
say /(?=(.+)\1)(?!(??{ '.+?(..{' . length($^N) . ',})\1' }))/;
Output:
101
What happens here is that after a matching consecutive pair of substrings, we make sure with a negative lookahead that there is no longer pair following it.
To make the expression for the longer pair a postponed subexpression construct is used (??{ code }), which evaluates the code inside (every time) and uses the returned string as an expression.
The subexpression it constructs has the form .+?(..{N,})\1, where N is the current length of the first capturing group (length($^N), $^N contains the current value of the previous capturing group).
Thus the full expression would have the form:
(?=(.+)\1)(?!.+?(..{N,})\2}))
With the magical N (and second capturing group not being a "real"/proper capturing group of the original expression).
Usage example:
use v5.10;
sub longest_rep{
$_[0] =~ /(?=(.+)\1)(?!(??{ '.+?(..{' . length($^N) . ',})\1' }))/;
}
say longest_rep '01011010';
say longest_rep '010110101000110001';
say longest_rep '2323191919191919';
say longest_rep '22010110100';
Output:
101
10001
191919
101
You can do it in a single regex, you just have to pick the longest match from the list of results manually.
def longestrepeating(strg):
regex = re.compile(r"(?=(.+)\1)")
matches = regex.findall(strg)
if matches:
return max(matches, key=len)
This gives you (since re.findall() returns a list of the matching capturing groups, even though the matches themselves are zero-length):
>>> longestrepeating("yabyababyab")
'abyab'
>>> longestrepeating("10100101")
'010'
>>> strings = ["56712453289", "22010110100", "5555555", "1919191919",
"191919191919", "2323191919191919"]
>>> [longestrepeating(s) for s in strings]
[None, '101', '555', '1919', '191919', '191919']
Here's a long-ish script that does what you ask. It basically goes through your input string, shortens it by one, then goes through it again. Once all possible matches are found, it returns one of the longest. It is possible to tweak it so that all the longest matches are returned, instead of just one, but I'll leave that to you.
It's pretty rudimentary code, but hopefully you'll get the gist of it.
use v5.10;
use strict;
use warnings;
while (<DATA>) {
chomp;
print "$_ : ";
my $longest = foo($_);
if ($longest) {
say $longest;
} else {
say "No matches found";
}
}
sub foo {
my $num = shift;
my #hits;
for my $i (0 .. length($num)) {
my $part = substr $num, $i;
push #hits, $part =~ /(.+)(?=\1)/g;
}
my $long = shift #hits;
for (#hits) {
if (length($long) < length) {
$long = $_;
}
}
return $long;
}
__DATA__
56712453289
22010110100
5555555
1919191919
191919191919
2323191919191919
Not sure if anyone's thought of this...
my $originalstring="pdxabababqababqh1234112341";
my $max=int(length($originalstring)/2);
my #result;
foreach my $n (reverse(1..$max)) {
#result=$originalstring=~m/(.{$n})\1/g;
last if #result;
}
print join(",",#result),"\n";
The longest doubled match cannot exceed half the length of the original string, so we count down from there.
If the matches are suspected to be small relative to the length of the original string, then this idea could be reversed... instead of counting down until we find the match, we count up until there are no more matches. Then we need to back up 1 and give that result. We would also need to put a comma after the $n in the regex.
my $n;
foreach (1..$max) {
unless (#result=$originalstring=~m/(.{$_,})\1/g) {
$n=--$_;
last;
}
}
#result=$originalstring=~m/(.{$n})\1/g;
print join(",",#result),"\n";
Regular expressions can be helpful in solving this, but I don't think you can do it as a single expression, since you want to find the longest successful match, whereas regexes just look for the first match they can find. Greediness can be used to tweak which match is found first (earlier vs. later in the string), but I can't think of a way to prefer an earlier, longer substring over a later, shorter substring while also preferring a later, longer substring over an earlier, shorter substring.
One approach using regular expressions would be to iterate over the possible lengths, in decreasing order, and quit as soon as you find a match of the specified length:
my $s = '01011010';
my $one = undef;
for(my $i = int (length($s) / 2); $i > 0; --$i)
{
if($s =~ m/(.{$i})\1/)
{
$one = $1;
last;
}
}
# now $one is '101'