Regex and the characters case - regex

Okay, I got a rather simple one (at least seems simple). I have a multi lined string and I am just playing around with replacing different words with something else. Let me show you...
#!/usr/bin/perl -w
use strict;
$_ = "That is my coat.\nCoats are very expensive.";
s/coat/Hat/igm;
print;
The output would be
That is my Hat
Hats are very expensive...
The "hat" on the first line shouldn't be capitalized. Are there any tricks that can make the casing compliant with how english is written? Thanks :)

see how-to-replace-string-and-preserve-its-uppercase-lowercase
For more detail go to How do I substitute case insensitively on the LHS while preserving case on the RHS?

You can use the e modifier to s/// to do the trick:
s/(coat)/ucfirst($1) eq $1 ? 'Hat' : 'hat'/igme;

For one, you should use \b (word boundary) to match only the whole word. For example s/hat/coat/ would change That to Tcoat without leading \b. Now for your question. With the flag /e you can use Perl code in the replacement part of the regex. So you can write a Perl function that checks the case of the match and then set the case of the replacement properly:
my $s = "That is my coat.\nCoats are very expensive.";
$s =~ s/(\bcoat)/&same_case($1, "hat")/igme;
print $s, "\n";
sub same_case {
my ($match, $replacement) = #_;
# if match starts with uppercase character, apply ucfirst to replacement
if($match =~ /^[A-Z]/) {
return ucfirst($replacement);
}
else {
return $replacement;
}
}
Prints:
That is my hat.
Hats are very expensive.

This may solve your problem:
#!/usr/bin/perl -w
use strict;
sub smartSubstitute {
my $target = shift;
my $pattern = shift;
my $replacement = shift;
$pattern = ucfirst $pattern;
$replacement = ucfirst $replacement;
$target =~ s/$pattern/$replacement/gm;
$pattern = lcfirst $pattern;
$replacement = lcfirst $replacement;
$target =~ s/$pattern/$replacement/gm;
return $target;
}
my $x = "That is my coat.\nCoats are very expansive.";
my $y = smartSubstitute($x, "coat", "Hat");
print $y, "\n";

Related

Matching a variable in a string in Perl from the end

I want to match a variable character in a given string, but from the end.
Ideas on how to do this action?
for example:
sub removeCharFromEnd {
my $string = shift;
my $char = shift;
if($string =~ m/$char/){ // I want to match the char, searching from the end, $doesn't work
print "success";
}
}
Thank you for your assistance.
There is no regex modifier that would force Perl regex engine to parse the string from right to left. Thus, the most convenient way to achieve that is via a negative lookahead:
m/$char(?!.*$char)/
The (?!.*$char) negative lookahead will require the absence (=will fail the match if found) of a $char after any 0+ chars other than linebreak chars (use s modifier if you are running the regex against a multiline string input).
The regex engine works from left to right.
You can use the natural greediness of quantifiers to reach the end of the string and find the last char with the backtracking mechanism:
if($string =~ m/.*\K$char/s) { ...
\K marks the position of the match result beginning.
Other ways:
you can also reverse the string and use your previous pattern.
you can search all occurrences and take the last item in the list
I'm having trouble understanding what you want. Your subroutine is called removeCharFromEnd, so perhaps you want to remove $char from $string if it appears at the end of the string
You can do that like this
sub removeCharFromEnd {
my ( $string, $char ) = #_;
if ( $string =~ s/$char\z// ) {
print "success";
}
$string;
}
Or perhaps you want to remove the last occurrence of $char wherever it is. You can do that with
s/.*\K$char//
The subroutine I have written returns the modified string, so you would have to assign the result to a variable to save it. You can write
my $s = 'abc';
$s = removeCharFromEnd($s, 'c');
say $s;
output
ab
If you just want to modify the string in place then you should write
$ARGV[0] =~ s/$char\z//
using whichever substitution you choose. Then you can do this
my $s = 'abc';
removeCharFromEnd($s, 'c');
say $s;
This produces the same output
To get Perl to search from the end of a string, reverse the string.
sub removeCharFromEnd {
my $string = reverse shift #_;
my $char = quotemeta reverse shift #_;
$string =~ s/$char//;
$string = reverse $string;
return $string;
}
print removeCharFromEnd(qw( abcabc b )), "\n";
print removeCharFromEnd(qw( abcdefabcdef c )), "\n";
print removeCharFromEnd(qw( !"/$%?&*!"/$%?&* $ )), "\n";

Perl Grepping from an Array

I need to grep a value from an array.
For example i have a values
#a=('branches/Soft/a.txt', 'branches/Soft/h.cpp', branches/Main/utils.pl');
#Array = ('branches/Soft/a.txt', 'branches/Soft/h.cpp', branches/Main/utils.pl','branches/Soft/B2/c.tct', 'branches/Docs/A1/b.txt');
Now, i need to loop #a and find each value matches to #Array. For Example
It works for me with grep. You'd do it the exact same way as in the More::ListUtils example below, except for having grep instead of any. You can also shorten it to
my $got_it = grep { /$str/ } #paths;
my #matches = grep { /$str/ } #paths;
This by default tests with /m against $_, each element of the list in turn. The $str and #paths are the same as below.
You can use the module More::ListUtils as well. Its function any returns true/false depending on whether the condition in the block is satisfied for any element in the list, ie. whether there was a match in this case.
use warnings;
use strict;
use Most::ListUtils;
my $str = 'branches/Soft/a.txt';
my #paths = ('branches/Soft/a.txt', 'branches/Soft/b.txt',
'branches/Docs/A1/b.txt', 'branches/Soft/B2/c.tct');
my $got_match = any { $_ =~ m/$str/ } #paths;
With the list above, containing the $str, the $got_match is 1.
Or you can roll it by hand and catch the match as well
foreach my $p (#paths) {
print "Found it: $1\n" if $p =~ m/($str)/;
}
This does print out the match.
Note that the strings you show in your example do not contain the one to match. I added it to my list for a test. Without it in the list no match is found in either of the examples.
To test for more than one string, with the added sample
my #strings = ('branches/Soft/a.txt', 'branches/Soft/h.cpp', 'branches/Main/utils.pl');
my #paths = ('branches/Soft/a.txt', 'branches/Soft/h.cpp', 'branches/Main/utils.pl',
'branches/Soft/B2/c.tct', 'branches/Docs/A1/b.txt');
foreach my $str (#strings) {
foreach my $p (#paths) {
print "Found it: $1\n" if $p =~ m/($str)/;
}
# Or, instead of the foreach loop above use
# my $match = grep { /$str/ } #paths;
# print "Matched for $str\n" if $match;
}
This prints
Found it: branches/Soft/a.txt
Found it: branches/Soft/h.cpp
Found it: branches/Main/utils.pl
When the lines with grep are uncommented and foreach ones commented out I get the corresponding prints for the same strings.
The slashes dot in $a will pose a problem so you either have to escape them it when doing regex match or use a simple eq to find the matches:
Regex match with $a escaped:
my #matches = grep { /\Q$a\E/ } #array;
Simple comparison with "equals":
my #matches = grep { $_ eq $a } #array;
With your sample data both will give an empty array #matches because there is no match.
This Solved My Question. Thanks to all especially #zdim for the valuable time and support
my #SVNFILES = ('branches/Soft/a.txt', 'branches/Soft/b.txt');
my #paths = ('branches/Soft/a.txt', 'branches/Soft/b.txt',
'branches/Docs/A1/b.txt', 'branches/Soft/B2/c.tct');
foreach my $svn (#SVNFILES)
{
chomp ($svn);
my $m = grep { /$svn/ } (#paths);
if ( $m eq '0' ) {
print "Files Mismatch\n";
exit 1;
}
}
You should escape characters like '/' and '.' in any regex when you need it as a character.
Likewise :
$a="branches\/Soft\/a\.txt"
Retry whatever you did with either grep or perl with that. If it still doesn't work, tell us precisely what you tried.

how to extract string with any operator between?

I have an array contain #arr = { "a=b", "a>b", "a<b", "a!=b", "a-b" }. What is the best way to get a and b with any operator between. I can extract by
for($i=0; $i<=$#arr; $i++){
$str = $arr[$i];
if($str =~ m/^(.*?)(\s*=\s*)(.*)(;)/g){
my $d = $1;
my $e = $3;
}
Follow by all if statement with the possible operator like "!=", "<" etc. But this will make my code look messy. Any better solution for this?
You could try something like this one liner
perl -e '#a = ("a=b","a>b","a<b","a!=b","a-b"); for $l (#a) { $l =~ s/(.).*(.)/$1/; print "$1$2\n"};'
The key thing is the greedy match ie "(.*)" between the two single character matches ie "(.)". To really make sure that you start at the start and end of the strings you could use this
perl -e '#a = ("a=b","a>b","a<b","a!=b","a-b"); for $l (#a) { $l =~ s/^(.).*(.)$/$1/; print "$1$2\n"};'
A complete working example that demonstrates the whole thing would be
#!/usr/bin/perl
use strict;
use warnings;
my #expressions = ("a=b","a>b","a<b","a!=b","a-b");
for my $exp (#expressions) {
$exp =~ s/^(.).*(.)$/$1$2/;
print "$1$2 is the same as $exp\n";
};
A very simple regex might be
/^(\w+)\s*(\W+)\s*(\w+)$/
Or you enumerate possible operators
/^(\w+)\s*(=|!=|<|>|<=|>=|\+|-|\*|\/|==)\s*(\w+)$/
It depends whether the input can be trusted or not. If not, you might have to be more meticulous w.r.t. the identifiers, too. Here's a simpler loop, and no need to use m//g(lobal). Not sure about the semicolon - omitted it.
my #arr = ( "a=b", "a>b", "a<b", "a!=b", "a-b" );
for my $str (#arr){
if($str =~ /^(\w+)\s*(=|!=|<|>|<=|>=|\+|-|\*|\/|==)\s*(\w+)$/ ){
my $d = $1;
my $e = $3;
print "d=$d e=$e\n";
}
}
Later If you enumerate the operators, you can also add word symbols:
if($str =~ /^(\w+)\s*(=|!=|<|>|<=|>=|\+|-|\*|\/|==|x?or|and)\s*(\w+)$/ ){
...
if there always 'a' and 'b' at the beginning and the end you could try:
my $str = 'a<b';
my( $op ) = $str =~ /^a(.*)b$/;
Not a well thought out answer. Will reconsider the problem.

A non-greedy Perl regular expression

I need to write a script which does the following:
$ cat testdata.txt
this is my file containing data
for checking pattern matching with a patt on the back!
only one line contains the p word.
$ ./mygrep5 pat th testdata.txt
this is my file containing data
for checking PATTERN MATCHING WITH a PATT ON THe back!
only one line contains the p word.
I have been able to print the line which is amended with the "a" capitalized as well. I have no idea how to only take what is needed.
I have been messing around (below is my script so far) and all I manage to return is the "PATT ON TH" part.
#!/usr/bin/perl
use strict;
use warnings;
use feature 'say';
use Data::Dump 'pp';
my ($f, $s, $t) = #ARGV;
my #output_lines;
open(my $fh, '<', $t);
while (my $line = <$fh>) {
if ($line =~ /$f/ && $line =~ /$s/) {
$line =~ s/($f.+?$s)/$1/g;
my $sub_phrase = uc $1;
$line =~ s/$1/$sub_phrase/g;
print $line;
}
#else {
# print $line;
#}
}
close($fh);
which returns: "for checking pattern matching with a PATT ON THe back!"
How can I fix this problem?
It sounds like you want to capitalize from pat to th except for instances of a surrounded by spaces. The easiest way is to uppercase the whole thing, and then fix any instances of A surrounded by spaces.
sub capitalize {
my $s = shift;
my $uc = uc($s);
$uc =~ s/ \s \K A (?=\s) /a/xg;
return $uc;
}
s{ ( \Q$f\E .* \Q$s\E ) }{ capitalize($1) }xseg;
The downside is that will replacing any existing A surrounded by spaces with a. The following is more complicated, but it doesn't suffer from that problem:
sub capitalize {
my $s = shift;
my #parts = $s =~ m{ \G ( \s+ | \S+ ) }xg;
for (#parts) {
$_ = uc($_) if $_ ne "a";
}
return join('', #parts);
}
s{ ( \Q$f\E .* \Q$s\E ) }{ capitalize($1) }xseg;
The rest of the code can be simplified:
#!/usr/bin/perl
use strict;
use warnings;
sub capitalize { ... }
my $f = shift;
my $s = shift;
while (<>) {
s{ ( \Q$f\E .* \Q$s\E ) }{ capitalize($1) }xseg;
print;
}
So, if you want to match each sequence that starts with pat and ends with th, non-greedily, and uppercase that sequence, you can simply use an expression on the right side of your substitution:
$line =~ s/($f.+?$s)/uc($1)/eg;
And that's it.

pattern matching in regular expression (Perl)

Make a pattern that will match three consecutive copies of whatever is currently contained in $what. That is, if $what is fred, your pattern should match fredfredfred. If $what is fred|barney, your pattern should match fredfredbarney, barneyfredfred, barneybarneybarney, or many other variations. (Hint: You should set $what at the top of the pattern test program with a statement like my $what = 'fred|barney';)
But my solution to this is just too easy so I'm assuming its wrong. My solution is:
#! usr/bin/perl
use warnings;
use strict;
while (<>){
chomp;
if (/fred|barney/ig) {
print "pattern found! \n";
}
}
It display what I want. And I didn't even have to save the pattern in a variable. Can someone help me through this? Or enlighten me if I'm doing/understanding the problem wrong?
This example should clear up what was wrong with your solution:
my #tests = qw(xxxfooxx oofoobar bar bax rrrbarrrrr);
my $str = 'foo|bar';
for my $test (#tests) {
my $match = $test =~ /$str/ig ? 'match' : 'not match';
print "$test did $match\n";
}
OUTPUT
xxxfooxx did match
oofoobar did match
bar did match
bax did not match
rrrbarrrrr did match
SOLUTION
#!/usr/bin/perl
use warnings;
use strict;
# notice the example has the `|`. Meaning
# match "fred" or "barney" 3 times.
my $str = 'fred|barney';
my #tests = qw(fred fredfredfred barney barneybarneybarny barneyfredbarney);
for my $test (#tests) {
if( $test =~ /^($str){3}$/ ) {
print "$test matched!\n";
} else {
print "$test did not match!\n";
}
}
OUTPUT
$ ./test.pl
fred did not match!
fredfredfred matched!
barney did not match!
barneybarneybarny did not match!
barneyfredbarney matched!
use strict;
use warnings;
my $s="barney/fred";
my #ra=split("/", $s);
my $test="barneybarneyfred"; #etc, this will work on all permutations
if ($test =~ /^(?:$ra[0]|$ra[1]){3}$/)
{
print "Valid\n";
}
else
{
print "Invalid\n";
}
Split delimited your string based off of "/". (?:$ra[0]|$ra[1]) says group, but do not extract, "barney" or "fred", {3} says exactly three copies. Add an i after the closing "/" if the case doesn't matter. The ^ says "begins with," and the $ says "ends with."
EDIT:
If you need the format to be barney\fred, use this:
my $s="barney\\fred";
my #ra=split(/\\/, $s);
If you know that the matching will always be on fred and barney, then you can just replace $ra[0], $ra[1] with fred and barney.