Regex isn't my strongest point. Let's say I need a custom parser for strings which strips the string of any letters and multiple decimal points and alphabets.
For example, input string is "--1-2.3-gf5.47", the parser would return
"-12.3547".
I could only come up with variations of this :
string.replaceAll("[^(\\-?)(\\.?)(\\d+)]", "")
which removes the alphabets but retains everything else. Any pointers?
More examples:
Input: -34.le.78-90
Output: -34.7890
Input: df56hfp.78
Output: 56.78
Some rules:
Consider only the first negative sign before the first number, everything else can be ignored.
I'm trying to do this using Java.
Assume the -ve sign, if there is one, will always occur before the
decimal point.
Just tested this on ideone and it seemed to work. The comments should explain the code well enough. You can copy/paste this into Ideone.com and test it if you'd like.
It might be possible to write a single regex pattern for it, but you're probably better off implementing something simpler/more readable like below.
The three examples you gave prints out:
--1-2.3-gf5.47 -> -12.3547
-34.le.78-90 -> -34.7890
df56hfp.78 -> 56.78
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Ideone
{
public static void main (String[] args) throws java.lang.Exception
{
System.out.println(strip_and_parse("--1-2.3-gf5.47"));
System.out.println(strip_and_parse("-34.le.78-90"));
System.out.println(strip_and_parse("df56hfp.78"));
}
public static String strip_and_parse(String input)
{
//remove anything not a period or digit (including hyphens) for output string
String output = input.replaceAll("[^\\.\\d]", "");
//add a hyphen to the beginning of 'out' if the original string started with one
if (input.startsWith("-"))
{
output = "-" + output;
}
//if the string contains a decimal point, remove all but the first one by splitting
//the output string into two strings and removing all the decimal points from the
//second half
if (output.indexOf(".") != -1)
{
output = output.substring(0, output.indexOf(".") + 1)
+ output.substring(output.indexOf(".") + 1, output.length()).replaceAll("[^\\d]", "");
}
return output;
}
}
In terms of regex, the secondary, tertiary, etc., decimals seem tough to remove. However, this one should remove the additional dashes and alphas: (?<=.)-|[a-zA-Z]. (Hopefully the syntax is the same in Java; this is a Python regex but my understanding is that the language is relatively uniform).
That being said, it seems like you could just run a pretty short "finite state machine"-type piece of code to scan the string and rebuild the reduced string yourself like this:
a = "--1-2.3-gf5.47"
new_a = ""
dash = False
dot = False
nums = '0123456789'
for char in a:
if char in nums:
new_a = new_a + char # record a match to nums
dash = True # since we saw a number first, turn on the dash flag, we won't use any dashes from now on
elif char == '-' and not dash:
new_a = new_a + char # if we see a dash and haven't seen anything else yet, we append it
dash = True # activate the flag
elif char == '.' and not dot:
new_a = new_a + char # take the first dot
dot = True # put up the dot flag
(Again, sorry for the syntax, I think you need some curly backets around the statements vs. Python's indentation only style)
Related
I faced with the following problem. I need to remove all underscores between the start of the string and last digit in string (like was: 123_456__ - became: 123456__). I used the usual loop for it, which goes through string.length - 1 down to 0 and when the symbol is digit I start the new loop from the 0 to the i, where i is position of the found digit and forming new string skipping underscores. But it seems that there are some ways to replace it with regex or more "Kotlin-style" code, but I do not know how to do it. Is it possible to do it in more convenient way?
One way to to this is to use string functions like takeLastWhile / drop etc.
val s = "123_456__"
val end = s.takeLastWhile { !it.isDigit() }
val start = s.dropLast(end.length).filter { it != '_' } // or replace("_", "")
val result = start + end
println(result)
I have a C++ function that accepts strings in below format:
<WORD>: [VALUE]; <ANOTHER WORD>: [VALUE]; ...
This is the function:
std::wstring ExtractSubStringFromString(const std::wstring String, const std::wstring SubString) {
std::wstring S = std::wstring(String), SS = std::wstring(SubString), NS;
size_t ColonCount = NULL, SeparatorCount = NULL; WCHAR Separator = L';';
ColonCount = std::count(S.begin(), S.end(), L':');
SeparatorCount = std::count(S.begin(), S.end(), Separator);
if ((SS.find(Separator) != std::wstring::npos) || (SeparatorCount > ColonCount))
{
// SEPARATOR NEED TO BE ESCAPED, BUT DON'T KNOW TO DO THIS.
}
if (S.find(SS) != std::wstring::npos)
{
NS = S.substr(S.find(SS) + SS.length() + 1);
if (NS.find(Separator) != std::wstring::npos) { NS = NS.substr(NULL, NS.find(Separator)); }
if (NS[NS.length() - 1] == L']') { NS.pop_back(); }
return NS;
}
return L"";
}
Above function correctly outputs MANGO if I use it like:
ExtractSubStringFromString(L"[VALUE: MANGO; DATA: NOTHING]", L"VALUE")
However, if I have two escape separators in following string, I tried doubling like ;;, but I am still getting MANGO instead ;MANGO;:
ExtractSubStringFromString(L"[VALUE: ;;MANGO;;; DATA: NOTHING]", L"VALUE")
Here, value assigner is colon and separator is semicolon. I want to allow users to pass colons and semicolons to my function by doubling extra ones. Just like we escape double quotes, single quotes and many others in many scripting languages and programming languages, also in parameters in many commands of programs.
I thought hard but couldn't even think a way to do it. Can anyone please help me on this situation?
Thanks in advance.
You should search in the string for ;; and replace it with either a temporary filler char or string which can later be referenced and replaced with the value.
So basically:
1) Search through the string and replace all instances of ;; with \tempFill- It would be best to pick a combination of characters that would be highly unlikely to be in the original string.
2) Parse the string
3) Replace all instances of \tempFill with ;
Note: It would be wise to run an assert on your string to ensure that your \tempFill (or whatever you choose as the filler) is not in the original string to prevent an bug/fault/error. You could use a character such as a \n and make sure there are non in the original string.
Disclaimer:
I can almost guarantee there are cleaner and more efficient ways to do this but this is the simplest way to do it.
First as the substring does not need to be splitted I assume that it does not need to b pre-processed to filter escaped separators.
Then on the main string, the simplest way IMHO is to filter the escaped separators when you search them in the string. Pseudo code (assuming the enclosing [] have been removed):
last_index = begin_of_string
index_of_current_substring = begin_of_string
loop: search a separator starting at last index - if not found exit loop
ok: found one at ix
if char at ix+1 is a separator (meaning with have an escaped separator
remove character at ix from string by copying all characters after it one step to the left
last_index = ix+1
continue loop
else this is a true separator
search a column in [ index_of_current_substring, ix [
if not found: error incorrect string
say found at c
compare key_string with string[index_of_current_substring, c [
if equal - ok we found the key
value is string[ c+2 (skip a space after the colum), ix [
return value - search is finished
else - it is not our key, just continue searching
index_of_current_substring = ix+1
last_index = index_of_current_substring
continue loop
It should now be easy to convert that to C++
I would like to convert the following C++ method to a regular expression match/replace string pair. Is it possible to do this in a single pass, i.e. with a single call to a regex replace method? (such as this one)
std::string f(std::string value)
{
if (value.length() < 3)
{
value = std::string("0") + value;
}
value = value.substr(0, value.length() - 2) + std::string(".") + value.substr(value.length() - 2, 2);
return value;
}
The input is a string of one or more digits.
Some examples:
f("1234") = "12.34"
f("123") = "1.23"
f("12") = "0.12"
f("1") = ".01"
The only way I've been able to achieve this so far is by using 2 steps:
1. Apply a prefix of "00" to the input string.
2. Use the following regex match/replace pair:
Match: (0*)(\d+)(\d{2})
Replace: $2.$3
My question is, can this be done in a single "pass" by only calling the Regex replace method once and without prepending anything to the string beforehand.
I believe this isn't possible with a single expression/replacement, but I'd just like someone to confirm that (or otherwise provide a solution :) ).
I hope this will help. (Change a bit again) x3.
string a_="123456";
a_="14";
a_="9";
string a = regex_replace(a_,regex("(.*)(.{2})|()"),string("$1.$2."));
//a = regex_replace(regex_replace(a,regex("^"),string("00$1$2")),regex("(.+)(.{2})"),string("$1.$2"));
//a = regex_replace("00"+a,regex("(.+)(.{2})"),string("$1.$2"));
float i=atof(a.c_str());
if(!(i))//just go here for 0-9
{
i=atof((string("0.0")+a_).c_str());
}
cout<<i<<endl;
return 0;
I wanted to match the words in string with reverse order.
We wanted to put validation to prompt user, if name exists in reverse order.
For example:
If name column has the value, 'Viral,Tennis'
Now if user enters a new name with the value, 'Tennis,Viral'
Then how can we match reverse order of word using regex or some other way?
I am using C#.net for development.
You could take a look at the Regex.Split(String input, String regex) and do something like so:
String[] userEntry = Regex.Split(userString, "\\s+");
StringBuilder sb = new StringBuilder()
for (int i = userEntry.Length -1; i >= 0; i--)
{
sb.append(userEntry[i]).append(" ");
}
String result = sb.ToString();
//Do Validation
That would do the trick, however, you need to keep in mind that things will get a little bit messy if you do not want to change the order of special symbols such as the comma. You could easily remove those and do any validation without special symbols.
EDIT: It depends on what you mean by special symbols. The regex [^a-zA-z0-9]+ will match any character which is not a letter (upper or lower case) and which is also not a number. So you could easily do something like so:
string input = ...
string pattern = "[^a-zA-z0-9]+";
string replacement = "";
Regex rgx = new Regex(pattern);
string result = rgx.Replace(input, replacement);
The above should yield a string which is only made from letters and digits. White spaces will also be removed.
I have been looking for a regular expression with Google for an hour or so now and can't seem to work this one out :(
If I have a number, say:
2345
and I want to find any other number with the same digits but in a different order, like this:
2345
For example, I match
3245 or 5432 (same digits but different order)
How would I write a regular expression for this?
There is an "elegant" way to do it with a single regex:
^(?:2()|3()|4()|5()){4}\1\2\3\4$
will match the digits 2, 3, 4 and 5 in any order. All four are required.
Explanation:
(?:2()|3()|4()|5()) matches one of the numbers 2, 3, 4, or 5. The trick is now that the capturing parentheses match an empty string after matching a number (which always succeeds).
{4} requires that this happens four times.
\1\2\3\4 then requires that all four backreferences have participated in the match - which they do if and only if each number has occurred once. Since \1\2\3\4 matches an empty string, it will always match as long as the previous condition is true.
For five digits, you'd need
^(?:2()|3()|4()|5()|6()){5}\1\2\3\4\5$
etc...
This will work in nearly any regex flavor except JavaScript.
I don't think a regex is appropriate. So here is an idea that is faster than a regex for this situation:
check string lengths, if they are different, return false
make a hash from the character (digits in your case) to integers for counting
loop through the characters of your first string:
increment the counter for that character: hash[character]++
loop through the characters of the second string:
decrement the counter for that character: hash[character]--
break if any count is negative (or nonexistent)
loop through the entries, making sure each is 0:
if all are 0, return true
else return false
EDIT: Java Code (I'm using Character for this example, not exactly Unicode friendly, but it's the idea that matters now):
import java.util.*;
public class Test
{
public boolean isSimilar(String first, String second)
{
if(first.length() != second.length())
return false;
HashMap<Character, Integer> hash = new HashMap<Character, Integer>();
for(char c : first.toCharArray())
{
if(hash.get(c) != null)
{
int count = hash.get(c);
count++;
hash.put(c, count);
}
else
{
hash.put(c, 1);
}
}
for(char c : second.toCharArray())
{
if(hash.get(c) != null)
{
int count = hash.get(c);
count--;
if(count < 0)
return false;
hash.put(c, count);
}
else
{
return false;
}
}
for(Integer i : hash.values())
{
if(i.intValue()!=0)
return false;
}
return true;
}
public static void main(String ... args)
{
//tested to print false
System.out.println(new Test().isSimilar("23445", "5432"));
//tested to print true
System.out.println(new Test().isSimilar("2345", "5432"));
}
}
This will also work for comparing letters or other character sequences, like "god" and "dog".
Put the digits of each number in two arrays, sort the arrays, find out if they hold the same digits at the same indices.
RegExes are not the right tool for this task.
You could do something like this to ensure the right characters and length
[2345]{4}
Ensuring they only exist once is trickier and why this is not suited to regexes
(?=.*2.*)(?=.*3.*)(?=.*4.*)(?=.*5.*)[2345]{4}
The simplest regular expression is just all 24 permutations added up via the or operator:
/2345|3245|5432|.../;
That said, you don't want to solve this with a regex if you can get away with it. A single pass through the two numbers as strings is probably better:
1. Check the string length of both strings - if they're different you're done.
2. Build a hash of all the digits from the number you're matching against.
3. Run through the digits in the number you're checking. If you hit a match in the hash, mark it as used. Keep going until you don't get an unused match in the hash or run out of items.
I think it's very simple to achieve if you're OK with matching a number that doesn't use all of the digits. E.g. if you have a number 1234 and you accept a match with the number of 1111 to return TRUE;
Let me use PHP for an example as you haven't specified what language you use.
$my_num = 1245;
$my_pattern = '/[' . $my_num . ']{4}/'; // this resolves to pattern: /[1245]{4}/
$my_pattern2 = '/[' . $my_num . ']+/'; // as above but numbers can by of any length
$number1 = 4521;
$match = preg_match($my_pattern, $number1); // will return TRUE
$number2 = 2222444111;
$match2 = preg_match($my_pattern2, $number2); // will return TRUE
$number3 = 888;
$match3 = preg_match($my_pattern, $number3); // will return FALSE
$match4 = preg_match($my_pattern2, $number3); // will return FALSE
Something similar will work in Perl as well.
Regular expressions are not appropriate for this purpose. Here is a Perl script:
#/usr/bin/perl
use strict;
use warnings;
my $src = '2345';
my #test = qw( 3245 5432 5542 1234 12345 );
my $canonical = canonicalize( $src );
for my $candidate ( #test ) {
next unless $canonical eq canonicalize( $candidate );
print "$src and $candidate consist of the same digits\n";
}
sub canonicalize { join '', sort split //, $_[0] }
Output:
C:\Temp> ks
2345 and 3245 consist of the same digits
2345 and 5432 consist of the same digits