How to print hyphens to an output like this for ex, 344-34-4333. If this ID is read from a file that has the number without hyphens, how can I get it to print xxx-xx-xxxx 3 to 2 to 4 ?
The std::string class has a lot of constructors to help you with these problems. The substr() member function is also useful.
A quick and dirty example:
std::string x("344344333");
std::string res = x.substr(0, 3) + '-' + s.substr(3, 2) + '-' + x.substr(5,4);
For more complex strings, you'll probably prefer to use the std::ostringstream class.
Related
Inside my data flow pipeline I would like to add a derived column and its datatype is array. I would like to split the existing column with 1000 characters without breaking words. I think we can use regexSplit,
regexSplit(<string to split> : string, <regex expression> : string) => array
But I do not know which regular expression I can use for split the existing column without breaking words.
Please help me to figure it out.
I created a workaround for this and it works fine for me.
filter(split(regexReplace(regexReplace(text, `[\t\n\r]`, ``), `(.{1,1000})(?:\s|$)`, `$1~~`), '~~'), #item !="")
I think, we have a better solution than this.
I wouldn't use a regex for this, but a truncating function like this one, courtesy of TimS:
public static string TruncateAtWord(this string input, int length)
{
if (input == null || input.Length < length)
return input;
int iNextSpace = input.LastIndexOf(" ", length, StringComparison.Ordinal);
return string.Format("{0}…", input.Substring(0, (iNextSpace > 0) ? iNextSpace : length).Trim());
}
Translated into expression functions it would look something* like this.
substring(Input, 1, iif(locate(Input, ' ', 1000) > 0, locate(Input, ' ', 1000) , length(Input)) )
Since you don't have a lastIndexOf available as an expression function, you would have to default to locate, which means that this expression truncates the string at the first space after the 1000th character.
*I don't have an environment where I can test this.
This question already has answers here:
How to find sum of integers in a string using JavaScript
(3 answers)
Closed 3 years ago.
I am getting a string back "1+2" and would like to remove the "+" and then add the numbers together.
Is this possible using Regex? So far I have:
let matches = pattern.exec(this.expression);
matches.input.replace(/[^a-zA-Z ]/g, "")
I am now left with two numbers. How would I add together?
"this.a + this.b"
Assuming the string returned only has '+' operation how about:
const sum = str.split('+').reduce((sumSoFar, strNum) => sumSoFar + parseInt(strNum), 0);
You cannot add two numbers using regex.
If what you have is a string of the form "1+2", why not simply split the string on the + symbol, and parseInt the numbers before adding them?
var str = "1+2";
var parts = str.split("+"); //gives us ["1", "2"]
console.log(parseInt(parts[0]) + parseInt(parts[1]));
If you don't always know what the delimiter between the two numbers is going to be you could use regex to get your array of numbers, and then reduce or whatever from there.
var myString = '1+2 and 441 with 9978';
var result = myString.match(/\d+/g).reduce((a,n)=> a+parseInt(n),0);
console.log(result); // 1 + 2 + 441 + 9978 = 10422
*Edit: If you actually want to parse the math operation contained in the string, there are a couple of options. First, if the string is from a trusted source, you could use a Function constructor. But this can be almost as dangerous as using eval, so it should be used with great caution. You should NEVER use this if you are dealing with a string entered by a user through the web page.
var myFormula = '1+2 * 441 - 9978';
var fn = new Function('return ' + myFormula);
var output = fn();
console.log(myFormula, ' = ', output); //1+2 * 441 - 9978 = -9095
A safer (but more difficult) course would be to write your own math parser which would detect math symbols and numbers, but would prevent someone from injecting other random commands that could affect global scope variables and such.
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 have the following strings in a long string:
a=b=c=d;
a=b;
a=b=c=d=e=f;
I want to first search for above mentioned pattern (X=Y=...=Z) and then output like the following for each of the above mentioned strings:
a=d;
b=d;
c=d;
a=b;
a=f;
b=f;
c=f;
d=f;
e=f;
In general, I want all the variables to have an equal sign with the last variable on the extreme right of the string. Is there a way I can do it using regexprep in MATLAB. I am able to do it for a fixed length string, but for variable length, I have no idea how to achieve this. Any help is appreciated.
My attempt for the case of two equal signs is as follows:
funstr = regexprep(funstr, '([^;])+\s*=\s*+(\w+)+\s*=\s*([^;])+;', '$1 = $3; \n $2 = $3;\n');
Not a regexp but if you stick to Matlab you can make use of the cellfun function to avoid loop:
str = 'a=b=c=d=e=f;' ; %// input string
list = strsplit(str,'=') ;
strout = cellfun( #(a) [a,'=',list{end}] , list(1:end-1), 'uni', 0).' %'// Horchler simplification of the previous solution below
%// this does the same than above but more convoluted
%// strout = cellfun( #(a,b) cat(2,a,'=',b) , list(1:end-1) , repmat(list(end),1,length(list)-1) , 'uni',0 ).'
Will give you:
strout =
'a=f;'
'b=f;'
'c=f;'
'd=f;'
'e=f;'
Note: As Horchler rightly pointed out in comment, although the cellfun instruction allows to compact your code, it is just a disguised loop. Moreover, since it runs on cell, it is notoriously slow. You won't see the difference on such simple inputs, but keep this use when super performances are not a major concern.
Now if you like regex you must like black magic code. If all your strings are in a cell array from the start, there is a way to (over)abuse of the cellfun capabilities to obscure your code do it all in one line.
Consider:
strlist = {
'a=b=c=d;'
'a=b;'
'a=b=c=d=e=f;'
};
Then you can have all your substring with:
strout = cellfun( #(s)cellfun(#(a,b)cat(2,a,'=',b),s(1:end-1),repmat(s(end),1,length(s)-1),'uni',0).' , cellfun(#(s) strsplit(s,'=') , strlist , 'uni',0 ) ,'uni',0)
>> strout{:}
ans =
'a=d;'
'b=d;'
'c=d;'
ans =
'a=b;'
ans =
'a=f;'
'b=f;'
'c=f;'
'd=f;'
'e=f;'
This gives you a 3x1 cell array. One cell for each group of substring. If you want to concatenate them all then simply: strall = cat(2,strout{:});
I haven't had much experience w/ Matlab; but your problem can be solved by a simple string split function.
[parts, m] = strsplit( funstr, {' ', '='}, 'CollapseDelimiters', true )
Now, store the last part of parts; and iterate over parts until that:
len = length( parts )
for i = 1:len-1
print( strcat(parts(i), ' = ', parts(len)) )
end
I do not know what exactly is the print function in matlab. You can update that accordingly.
There isn't a single Regex that you can write that will cover all the cases. As posted on this answer:
https://stackoverflow.com/a/5019658/3393095
However, you have a few alternatives to achieve your final result:
You can get all the values in the line with regexp, pick the last value, then use a for loop iterating throughout the other values to generate the output. The regex to get the values would be this:
matchStr = regexp(str,'([^=;\s]*)','match')
If you want to use regexprep at any means, you should write a pattern generator and a replace expression generator, based on number of '=' in the input string, and pass these as parameters of your regexprep func.
You can forget about Regex and Split the input to generate the output looping throughout the values (similarly to alternative #1) .
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;