I have a large body of text and I print only lines that contain one of several strings. Each line can contain more than one string.
Example of the rule:
(house|mall|building)
I want to mark the found string for making the result easier to read.
Example of the result I want:
New record: Two New York houses under contract for nearly $5 million each.
New record: Two New York #house#s under contract for nearly $5 million each.
I know I can find the location, trim, add marker, add string etc.
I am asking if there is a way to mark the found string in one command.
Thanks.
http://pubs.opengroup.org/onlinepubs/009695399/utilities/awk.html
gsub(ere, repl[, in])
Behave like sub (see below), except that it shall replace all occurrences of the regular expression ...
sub(ere, repl[, in ])
Substitute the string repl in place of the first instance of the
extended regular expression ERE in string in and return the number of
substitutions. An ampersand ( '&' ) appearing in the string repl shall
be replaced by the string from in that matches the ERE ...
BEGIN {
r = "house|mall|building"
s = "Two New York houses under contract for nearly $5 million each."
gsub(r, "#&#", s)
print s
}
Related
I am searching for a way to use a formatter to put a space between two characters. i thought it would be easy with a string formatter.
here is what i am trying to accomplish:
given: "AB" it will produce "A B"
Here is what i have tried so far:
"AB".format("%#s")
but this keep returning "AB" i want "A B". i thought the number sign could be used for space.
i also tried this:
"26".format("%#d") but its still prints "26"
is there anyway to do this with string.formatter.
It is kind of possible with the string formatter although not directly with a pattern.
jshell> String.format("%1$c %2$c", "AB".chars().boxed().toArray())
$10 ==> "A B"
We need to turn the string into an object array so it can be passed in as varargs and the formatter pattern can extract characters based on index (1$ and 2$) and format them as characters (c).
A much simpler regex solution is the following which scales to any number of characters:
jshell> "ABC^&*123".replaceAll(".", "$0 ").trim()
$3 ==> "A B C ^ & * 1 2 3"
All single characters are replaced with them-self ($0) followed by a space. Then the last extra space is removed with the trim() call.
I could not find way to do this using String#format. But here is a way to accomplish this using regex replacement:
String input = "AB";
String output = input.replaceAll("(?<=[A-Z])(?=[A-Z])", " ");
System.out.println(output);
The regex pattern (?<=[A-Z])(?=[A-Z]) will match every position in between two capital letters, and interpolate a space at that point. The above script prints:
A B
I'm having issues when trying to remove the first space of a string if that string has 2 spaces in it. For example it should be turning "Fully Functional Method" into "FullyFunctional Method", but "Functional Method" should not be changed because it only has 1 space. I can't really think of a way to remove first space if the string contains 2 spaces.
I don't know exactly what you want to do, but you may search into RegExp and String.replace() to replace some stuff in a String.
Here is another link to understand the Characters, metacharacters, and metasequences.
var myPattern1:RegExp = / /g;
var str1:String = "This is a string that contains double spaces.";
trace(str1.replace(myPattern1, " "));
//this replaces all " " by " "...
//outputs : This is a string that contains double spaces.
Or in your case (I suppose) something like this
var myPattern2:RegExp = / /;
var str2:String = "Fully Functional Method";
trace(str2.replace(myPattern2, ""));
//If you omit the g, only the first space will be replaced by ""
//outputs : FullyFunctional Method
There is so much things you can do by using RegExp, that I will not explain this here...
Just check on the Adobe website...
This is a quick and efficient way to work on Strings.
I hope this will help.
Since you check at those links, you will understand that my example is pure rough and should be modified to have a FullyFunctional Method. :D
Do a linear scan through the string. Count the number of spaces and record the index of the first space, if any. If there are two spaces, return a string that is the concatenation of the characters up to but not including the first space, and the characters after the first space.
Keep it simple. It is possible to solve your problem with regex, but keep in mind that the worst case time complexity of finding a particular character in an unsorted set is always going to be O(N), so it won't be faster.
I have a problem with splitting string into two parts on special character.
For example:
12345#data
or
1234567#data
I have 5-7 characters in first part separated with "#" from second part, where are another data (characters,numbers, doesn't matter what)
I need to store two parts on each side of # in two variables:
x = 12345
y = data
without "#" character.
I was looking for some Lua string function like splitOn("#") or substring until character, but I haven't found that.
Use string.match and captures.
Try this:
s = "12345#data"
a,b = s:match("(.+)#(.+)")
print(a,b)
See this documentation:
First of all, although Lua does not have a split function is its standard library, it does have string.gmatch, which can be used instead of a split function in many cases. Unlike a split function, string.gmatch takes a pattern to match the non-delimiter text, instead of the delimiters themselves
It is easily achievable with the help of a negated character class with string.gmatch:
local example = "12345#data"
for i in string.gmatch(example, "[^#]+") do
print(i)
end
See IDEONE demo
The [^#]+ pattern matches one or more characters other than # (so, it "splits" a string with 1 character).
I am trying to solve this problem on hackerrank:
So the problem is:
Jack and Daniel are friends. Both of them like letters, especially upper-case ones.
They are cutting upper-case letters from newspapers, and each one of them has their collection of letters stored in separate stacks.
One beautiful day, Morgan visited Jack and Daniel. He saw their collections. Morgan wondered what is the lexicographically minimal string, made of that two collections. He can take a letter from a collection when it is on the top of the stack.
Also, Morgan wants to use all the letters in the boys' collections.
This is my attempt in Perl:
#!/usr/bin/perl
use strict;
use warnings;
chomp(my $n=<>);
while($n>0){
chomp(my $string1=<>);
chomp(my $string2=<>);
lexi($string1,$string2);
$n--;
}
sub lexi{
my($str1,$str2)=#_;
my #str1=split(//,$str1);
my #str2=split(//,$str2);
my $final_string="";
while(#str2 && #str1){
my $st2=$str2[0];
my $st1=$str1[0];
if($st1 le $st2){
$final_string.=$st1;
shift #str1;
}
else{
$final_string.=$st2;
shift #str2;
}
}
if(#str1){
$final_string=$final_string.join('',#str1);
}
else{
$final_string=$final_string.join('',#str2);
}
print $final_string,"\n";
}
Sample Input:
2
JACK
DANIEL
ABACABA
ABACABA
The first line contains the number of test cases, T.
Every next two lines have such format: the first line contains string A, and the second line contains string B.
Sample Output:
DAJACKNIEL
AABABACABACABA
But for Sample test-case it is giving right results while it is giving wrong results for other test-cases. One case for which it gives an incorrect result is
1
AABAC
AACAB
It outputs AAAABACCAB instead of AAAABACABC.
I don't know what is wrong with the algorithm and why it is failing with other test cases?
Update:
As per #squeamishossifrage comments If I add
($str1,$str2)=sort{$a cmp $b}($str1,$str2);
The results become same irrespective of user-inputs but still the test-case fails.
The problem is in your handling of the equal characters. Take the following example:
ACBA
BCAB
When faced with two identical characters (C in my example), you naïvely chose the one from the first string, but that's not always correct. You need to look ahead to break ties. You may even need to look many characters ahead. In this case, next character after C of the second string is lower than the next character of the first string, so you should take the C from the second string first.
By leaving the strings as strings, a simple string comparison will compare as many characters as needed to determine which character to consume.
sub lexi {
my ($str1, $str2) = #_;
utf8::downgrade($str1); # Makes sure length() will be fast
utf8::downgrade($str2); # since we only have ASCII letters.
my $final_string = "";
while (length($str2) && length($str1)) {
$final_string .= substr($str1 le $str2 ? $str1 : $str2, 0, 1, '');
}
$final_string .= $str1;
$final_string .= $str2;
print $final_string, "\n";
}
Too little rep to comment thus the answer:
What you need to do is to look ahead if the two characters match. You currently do a simple le match and in the case of
ZABB
ZAAA
You'll get ZABBZAA since the first match Z will be le Z. So what you need to do (a naive solution which most likely won't be very effective) is to keep looking as long as the strings/chars match so:
Z eq Z
ZA eq ZA
ZAB gt ZAA
and at that point will you know that the second string is the one you want to pop from for the first character.
Edit
You updated with sorting the strings, but like I wrote you still need to look ahead. The sorting will solve the two above strings but will fail with these two:
ZABAZA
ZAAAZB
ZAAAZBZABAZA
Because here the correct answer is ZAAAZABAZAZB and you can't find that will simply comparing character per character
Basically, I have a string that consists of multiple, space-separated words. The thing is, however, that there can be multiple spaces instead of just one separating the words. This is why [split] does not do what I want:
split "a b"
gives me this:
{a {} {} {} b}
instead of this:
{a b}
Searching Google, I found a page on the Tcler's wiki, where a user asked more or less the same question.
One proposed solution would look like this:
split [regsub -all {\s+} "a b" " "]
which seems to work for simple string. But a test string such as [string repeat " " 4] (used string repeat because StackOverflow strips multiple spaces) will result in regsub returning " ", which split would again split up into {{} {}} instead of an empty list.
Another proposed solution was this one, to force a reinterpretation of the given string as a list:
lreplace "a list with many spaces" 0 -1
But if there's one thing I've learned about TCL, it is that you should never use list functions (starting with l) on strings. And indeed, this one will choke on strings containing special characters (namely { and }):
lreplace "test \{a b\}"
returns test {a b} instead of test \{a b\} (which would be what I want, every space-separated word split up into a single element of the resulting list).
Yet another solution was to use a 'filter':
proc filter {cond list} {
set res {}
foreach element $list {if [$cond $element] {lappend res $element}}
set res
}
You'd then use it like this:
filter llength [split "a list with many spaces"]
Again, same problem. This would call llength on a string, which might contain special characters (again, { and }) - passing it "\{a b\}" would result in TCL complaining about an "unmatched open brace in list".
I managed to get it to work by modifying the given filter function, adding a {*} in front of $cond in the if, so I could use it with string length instead of llength, which seemed to work for every possible input I've tried to use it on so far.
Is this solution safe to use as it is now? Would it choke on some special input I didn't test so far? Or, is it possible to do this right in a simpler way?
The easiest way is to use regexp -all -inline to select and return all words. For example:
# The RE matches any non-empty sequence of non-whitespace characters
set theWords [regexp -all -inline {\S+} $theString]
If instead you define words to be sequences of alphanumerics, you instead use this for the regular expression term: {\w+}
You can use regexp instead:
From tcl wiki split:
Splitting by whitespace: the pitfalls
split { abc def ghi}
{} abc def {} ghi
Usually, if you are splitting by whitespace and do not want those blank fields, you are better off doing:
regexp -all -inline {\S+} { abc def ghi}
abc def ghi