I try to locate a special part in a string.
The example of string as follow:
22.21594087,1.688530832,0
I want to locate 1.688530832 out.
I tried
temp.substr(temp.find(",")+1,temp.rfind(","));
and got 1.688530832,0.
I replaced rfind() with find_last_of() but still got the same result.
temp.substr(temp.find(",")+1,temp.find_last_of(","));
I know this is a simple problem and there are other solutions.But I just want to know why the rfind did not work.
Thank you very much!
The second argument for substr is not the ending index, but rather the length of the substring you want. Simply throw in the length of 1.688530832 and you'll be fine.
If the length of the search string is not available, then you can find the position of the last comma and subtract that from the position of the first character of the special word:
auto beginning_index = temp.find(",") + 1;
auto last_comma_index = temp.rfind(",");
temp.substr(beginning_index, last_comma_index - beginning_index);
I see what you are doing. You are trying to have kind of iterators to the beginning and the end of a substring. Unfortunately, substr does not work that way, and instead expects an index and an offset from that index to select the substring.
What you were trying to achieve can be done with std::find, which does work with iterators:
auto a = std::next(std::find(begin(temp), end(temp), ','));
auto b = std::next(std::find(rbegin(temp), rend(temp), ',')).base();
std::cout << std::string(a, b);
Live demo
Related
I am struggling to find a working regex pattern to match all instances of a string except the first.
I am using std::regex_replace to add a new line before each instance of a substring with a new line. however none of the googling I have found so far has produced a working regex pattern for this.
outputString = std::regex_replace(outputString, std::regex("// ~"), "\n// ~");
So all but the first instance of // ~ should be changed to \n// ~
In this case, I'd advise being the anti-Nike: "Just don't do it!"
The pattern you're searching for is trivial. So is the replacement. There's simply no need to use a regex at all. Under the circumstances, I'd just use std::string::find in a loop, skipping replacement of the first instance you find.
std::string s = "a = b; // ~ first line // ~ second line // ~ third line";
std::string pat = "// ~";
std::string rep = "\n// ~";
auto pos = s.find(pat); // first one, which we skip
while ((pos=s.find(pat, pos+pat.size())) != std::string::npos) {
s.replace(pos, pat.size(), rep);
pos += rep.size() - pat.size();
}
std::cout << s << "\n";
When you're doing a replacement like this one, where the replacement string includes a copy of the string you're searching for, you have to be a little careful about where you start second (and subsequent) searches, to assure that you don't repeatedly find the same string you just replaced. That's why we keep track of the current position, and each subsequent search we make, we move the starting point far enough along that we won't find the same instance of the pattern again and again.
I must browse a collection of strings to replace a pattern and save the changes.
The saving operation is (very) expensive and out of my hands, so I would like to know beforehand if the replacement did anything.
I can use std::regex_search to gain knowledge on the pattern's presence in my input, and use capture groups to store details in a std::smatch. std::regex_replace does not seem to explicitely tell me wether it did anything.
The patterns and strings are arbitrarily long and complicated; running regex_replace after a regex_search seems wasteful.
I can directly compare the input and output to search for a discrepancy but that too is uncomfortable.
Is there either a simple way to observe regex_replace to determine its impact, or to use a smatch filled by the regex_search to do a faster replacement operation ?
Thanks in advance.
No regex_replace doesn't provide this info and yes you can do it with a regex_search loop.
For example like this:
std::regex pattern("...");
std::string replacement_format = "...";
std::string input = "......"; // a very, very long string
std::string output, replacement;
std::smatch match;
auto begin = input.cbegin();
int replacements = 0;
while (std::regex_search(begin, input.cend(), match, pattern)) {
output += match.prefix();
replacement = match.format(replacement_format);
if (match[0] != replacement) {
replacements++;
}
output += replacement;
begin = match.suffix().first;
}
output.append(begin, input.cend());
if (replacements > 0) {
// process output ...
}
Live demo
As regex_replace creates a copy of your string you could simply compare the replaced string with the original one and only "store" the new one if they differ.
For C++14 it seems that regex_replace returns a pointer to the last place it has written to:
https://www.cplusplus.com/reference/regex/regex_replace/ Versions 5
and 6 return an iterator that points to the element past the last
character written to the sequence pointed by out.
I could have a string like:
During this time , Bond meets a stunning IRS agent , whom he seduces .
I need to remove the extra spaces before the comma and before the period in my whole string. I tried throwing this into a char vector and only not push_back if the current char was " " and the following char was a "." or "," but it did not work. I know there is a simple way to do it maybe using trim(), find(), or erase() or some kind of regex but I am not the most familiar with regex.
A solution could be (using regex library):
std::string fix_string(const std::string& str) {
static const std::regex rgx_pattern("\\s+(?=[\\.,])");
std::string rtn;
rtn.reserve(str.size());
std::regex_replace(std::back_insert_iterator<std::string>(rtn),
str.cbegin(),
str.cend(),
rgx_pattern,
"");
return rtn;
}
This function takes in input a string and "fixes the spaces problem".
Here a demo
On a loop search for string " ," and if you find one replace that to ",":
std::string str = "...";
while( true ) {
auto pos = str.find( " ," );
if( pos == std::string::npos )
break;
str.replace( pos, 2, "," );
}
Do the same for " .". If you need to process different space symbols like tab use regex and proper group.
I don't know how to use regex for C++, also not sure if C++ supports PCRE regex, anyway I post this answer for the regex (I could delete it if it doesn't work for C++).
You can use this regex:
\s+(?=[,.])
Regex demo
First, there is no need to use a vector of char: you could very well do the same by using an std::string.
Then, your approach can't work because your copy is independent of the position of the space. Unfortunately you have to remove only spaces around the punctuation, and not those between words.
Modifying your code slightly you could delay copy of spaces waiting to the value of the first non-space: if it's not a punctuation you'd copy a space before the character, otherwise you just copy the non-space char (thus getting rid of spaces.
Similarly, once you've copied a punctuation just loop and ignore the following spaces until the first non-space char.
I could have written code. It would have been shorter. But i prefer letting you finish your homework with full understanding of the approach.
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 did a program to remove a group of Characters From a String. I have given below that coding here.
void removeCharFromString(string &str,const string &rStr)
{
std::size_t found = str.find_first_of(rStr);
while (found!=std::string::npos)
{
str[found]=' ';
found=str.find_first_of(rStr,found+1);
}
str=trim(str);
}
std::string str ("scott<=tiger");
removeCharFromString(str,"<=");
as for as my program, I got my output Correctly. Ok. Fine. If I give a value for str as "scott=tiger" , Then the searchable characters "<=" not found in the variable str. But my program also removes '=' character from the value 'scott=tiger'. But I don't want to remove the characters individually. I want to remove the characters , if i only found the group of characters '<=' found. How can i do this ?
The method find_first_of looks for any character in the input, in your case, any of '<' or '='. In your case, you want to use find.
std::size_t found = str.find(rStr);
This answer works on the assumption that you only want to find the set of characters in the exact sequence e.g. If you want to remove <= but not remove =<:
find_first_of will locate any of the characters in the given string, where you want to find the whole string.
You need something to the effect of:
std::size_t found = str.find(rStr);
while (found!=std::string::npos)
{
str.replace(found, rStr.length(), " ");
found=str.find(rStr,found+1);
}
The problem with str[found]=' '; is that it'll simply replace the first character of the string you are searching for, so if you used that, your result would be
scott =tiger
whereas with the changes I've given you, you'll get
scott tiger