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)
Related
I have a string that sometimes contains a certain substring at the end and sometimes does not. When the string is present I want to update its value. When it is absent I want to add it at the end of the existing string.
For example:
int _newCount = 7;
_myString = 'The count is: COUNT=1;'
_myString2 = 'The count is: '
_rRuleString.replaceAllMapped(RegExp('COUNT=(.*?)\;'), (match) {
//if there is a match (like in _myString) update the count to value of _newCount
//if there is no match (like in _myString2) add COUNT=1; to the string
}
I have tried using a return of:
return "${match.group(1).isEmpty ? _myString + ;COUNT=1;' : 'COUNT=$_newCount;'}";
But it is not working.
Note that replaceAllMatched will only perform a replacement if there is a match, else, there will be no replacement (insertion is still a replacement of an empty string with some string).
Your expected matches are always at the end of the string, and you may leverage this in your current code. You need a regex that optionally matches COUNT= and then some text up to the first ; including the char and then checks if the current position is the end of string.
Then, just follow the logic: if Group 1 is matched, set the new count value, else, add the COUNT=1; string:
The regex is
(COUNT=[^;]*;)?$
See the regex demo.
Details
(COUNT=[^;]*;)? - an optional group 1: COUNT=, any 0 or more chars other than ; and then a ;
$ - end of string.
Dart code:
_myString.replaceFirstMapped(RegExp(r'(COUNT=[^;]*;)?$'), (match) {
return match.group(0).isEmpty ? "COUNT=1;" : "COUNT=$_newCount;" ; }
)
Note the use of replaceFirstMatched, you need to replace only the first match.
I need to manully hyphante words that are too long. Using hyphen.js, I get soft hyphens between every syllable, like below.
I want to find the hyphen closes to the middle. All words will be more than 14 characters long. Regex that works in https://regex101.com/ or node/js example.
Basically, find the middle character excluding hyphens, check if there is a hyphen there, then step backwards one step and then forwards one step, then backwards to steps etc.
re-spon-si-bil-i-ties => [re-spon-si,-bil-i-ties]
com-pe-ten-cies. => [com-pe,-ten-cies.]
ini-tia-tives. => [ini-tia,-tives]
vul-ner-a-bil-i-ties => [vul-ner-a,-bil-i-ties]
Here's a simple js approach based on string splitting. There could be a binary search style algorithm as you mentioned which would avoid the array allocation but that seems overkill for these small data sets.
function halve(str) {
var right = str.split('-');
var left = right.splice(0, Math.ceil(right.length / 2));
return right.length > 0 ? [left.join('-'), '-' + right.join('-')] : left;
}
console.log(halve('re-spon-si-bil-i-ties'));
console.log(halve('com-pe-ten-cies.'));
console.log(halve('ini-tia-tives.'));
console.log(halve('vul-ner-a-bil-i-ties'));
console.log(halve('none')); // no hyphens returns ["none"]
You can work this out with this method:
Get middle point of string
From the middle point, and checking each character in both directions (left from middle, right from middle) check if that position is the - character. Set the index to the first such match.
If it matches that character, stop the loop and split the string on that index, otherwise return the original word.
words = [
're-spon-si-bil-i-ties',
'com-pe-ten-cies.',
'ini-tia-tives.',
'vul-ner-a-bil-i-ties',
'test',
'-aa',
'aa-'
];
split = '-'
for(word of words) {
m=Math.floor(word.length/2),offset=0,i=null
do{
if(word[m-offset] == split) i = m-offset
else if(word[m+offset] == split) i = m+offset
else offset++
}while(offset<=m && i == null)
if(i!=null && i>0) console.log([word.substring(0,i),word.substring(i)])
else console.log(word)
}
You can achieve this with:
var words = [
're-spon-si-bil-i-ties',
'com-pe-ten-cies.',
'ini-tia-tives.',
'vul-ner-a-bil-i-ties',
're-ports—typ-i-cal-ly',
'none'
];
for(var i = 0; i < words.length; ++i){
var matches = words[i]
.match(
new RegExp(
'^((?:[^-]+?-?){' // Start the regex
+parseInt(
words[i].replace( /-/g, '' ).length/2 // Round down the halfway point of this word's length without the hyphens
)
+'})(-.+)?$' // End the regex
)
)
.slice( 1 ); // Remove position 0 because it is the entire word
console.log( matches );
}
Regex explanation for re-spon-si-bil-i-ties:
^((?:[^-]+?-?){8})(-.+)$
^( - start the capture group leading up to the half way point
(?:[^-]+?-?) - find everything not a hyphen with an optional hyphen after it. Make the hyphen optional so that the second capture group can greedily claim it
{8} - 8 times; this will get us half way
) - close the half way capture group
(-.+)?$ - greedily get the hyphen and everything after it till the end of the string
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++
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)
I was wondering if there was a way to do pattern matching in Octave / matlab? I know Maple 10 has commands to do this but not sure what I need to do in Octave / Matlab. So if a number was 12341234123412341234 the pattern match would be 1234. I'm trying to find the shortest pattern that upon repetiton generates the whole string.
Please note: the numbers (only numbers will be used) won't be this simple. Also, I won't know the pattern ahead of time (that's what I'm trying to find). Please see the Maple 10 example below which shows that the pattern isn't known ahead of time but the command finds the pattern.
Example of Maple 10 pattern matching:
ns:=convert(12341234123412341234,string);
ns := "12341234123412341234"
StringTools:-PrimitiveRoot(ns);
"1234"
How can I do this in Octave / Matlab?
Ps: I'm using Octave 3.8.1
To find the shortest pattern that upon repetition generates the whole string, you can use regular expressions as follows:
result = regexp(str, '^(.+?)(?=\1*$)', 'match');
Some examples:
>> str = '12341234123412341234';
>> result = regexp(str, '^(.+?)(?=\1*$)', 'match')
result =
'1234'
>> str = '1234123412341234123';
>> result = regexp(str, '^(.+?)(?=\1*$)', 'match')
result =
'1234123412341234123'
>> str = 'lullabylullaby';
>> result = regexp(str, '^(.+?)(?=\1*$)', 'match')
result =
'lullaby'
>> str = 'lullaby1lullaby2lullaby1lullaby2';
>> result = regexp(str, '^(.+?)(?=\1*$)', 'match')
result =
'lullaby1lullaby2'
I'm not sure if this can be accomplished with regular expressions. Here is a script that will do what you need in the case of a repeated word called pattern.
It loops through the characters of a string called str, trying to match against another string called pattern. If matching fails, the pattern string is extended as needed.
EDIT: I made the code more compact.
str = 'lullabylullabylullaby';
pattern = str(1);
matchingState = false;
sPtr = 1;
pPtr = 1;
while sPtr <= length(str)
if str(sPtr) == pattern(pPtr) %// if match succeeds, keep looping through pattern string
matchingState = true;
pPtr = pPtr + 1;
pPtr = mod(pPtr-1,length(pattern)) + 1;
else %// if match fails, extend pattern string and start again
if matchingState
sPtr = sPtr - 1; %// don't change str index when transitioning out of matching state
end
matchingState = false;
pattern = str(1:sPtr);
pPtr = 1;
end
sPtr = sPtr + 1;
end
display(pattern);
The output is:
pattern =
lullaby
Note:
This doesn't allow arbitrary delimiters between occurrences of the pattern string. For example, if str = 'lullaby1lullaby2lullaby1lullaby2';, then
pattern =
lullaby1lullaby2
This also allows the pattern to end mid-way through a cycle without changing the result. For example, str = 'lullaby1lullaby2lullaby1'; would still result in
pattern =
lullaby1lullaby2
To fix this you could add the lines
if pPtr ~= length(pattern)
pattern = str;
end
Another approach is as follows:
determine length of string, and find all possible factors of the string length value
for each possible factor length, reshape the string and check
for a repeated substring
To find all possible factors, see this solution on SO. The next step can be performed in many ways, but I implement it in a simple loop, starting with the smallest factor length.
function repeat = repeats_in_string(str);
ns = numel(str);
nf = find(rem(ns, 1:ns) == 0);
for ii=1:numel(nf)
repeat = str(1:nf(ii));
if all(ismember(reshape(str,nf(ii),[])',repeat));
break;
end
end
This problem is a great Rorschach test for your approach to problem solving. I'll add a signal engineering solution, which should be simple since the signal is expected to be perfectly repetitive, assuming this holds: find the shortest pattern that upon repetition generates the whole string.
In the following str fed to the function is actually a column vector of floats, not a string, the original string having been converted with str2num(str2mat(str)'):
function res=findshortestrepel(str);
[~,ii] = max(fft(str-mean(str)));
res = str(1:round(numel(str)/(ii-1)));
I performed a small test, comparing this to the regexp solution and found it to be faster overall (blue squares), although somewhat inconsistently, and only if you don't consider the time required to convert the string into a vector of floats (green squares). However I did not pursue this further (not breaking records with this):
Times in sec.