I work in Qt Creator (Community) 5.5.1. For example, I have
string="44° 36' 14.2\" N, 33° 30' 58.6\" E, 0m"
of QString. I know, that I must parse it, but i don't know how, because I have never faced with the problem like it. From our string I want to get some other smaller strings:
cgt = "44"; cmt = "36"; cst = "14.2"
cgg = "33"; cmg = "30"; csg = "58.6"
What must I do for working my programm how I said?
I need real code. Thanks.
The simplest way to start would be string.split(' ') - that would yield the list of the string components that were separated by the space character (' '). If you're sure the string will always be formatted exactly like this, you can first remove all the special characters (° and so on).
Then analyze the resulting QStringList. Again, if the format is fixed, you can check that the number of list items matches the expected number, and then get degrees as list[0], minutes as ``list[1]` and so on.
Another alternative would be to use QRegExp for parsing the string (splitting it into substrings based on regex), but I find it too complicated for use cases where split works just as well.
"I need code" is not the kind of question you should be asking, SO is about "gimme knowledge" not about "do my work" questions. A good question should demonstrate your effort to solve the problem, so people can tell you what you are doing wrong. Not only does your question lack any such effort, but you didn't expend any even when Devopia did half of the work for you. Keep that in mind for your future questions.
struct G {
double cgt, cmt, cst, cgg, cmg, csg;
};
G parse(QString s) {
QStringList list = s.split(QRegExp("[^0-9.]"), QString::SkipEmptyParts);
G g;
g.cgt = list.at(0).toDouble();
g.cmt = list.at(1).toDouble();
g.cst = list.at(2).toDouble();
g.cgg = list.at(3).toDouble();
g.cmg = list.at(4).toDouble();
g.csg = list.at(5).toDouble();
return g;
}
Related
P.S: --> I know there is an easy solution to my needs, and I can do it that way but, -- I am looking for a "diff" solution for learning sake & challenge sake. So, this is just to solve an algorithm in a lesser traditional way.
I am working on solving an algorithm, and thought I had everything working well but one use case is failing. That is because I am building a regexp dynamically - now, my issue is this.
I need to match letters sequentially up until one doesn't match, then I just "match" what did match sequentially.
so... lets say I was matching this:
"zaazizz"
with this: /\bz[a]?[z]?/
"zizzi".match(/\bz[z]?[i]?/)
currently, that is matching with a : [zi], but the match should only be [z]
zzi only matches "z" from the front of "zizzi", in that order zzi - I now I am using [z]? etc... so it is optional.. but what I really need is match sequentially.. I'd only get "zi" IF from the front, it matched: zzi per my regex.... so, some sort of lookahead or ?. I tried ?= and != no luck.
I still think a non-regex-approach is best here. Have a look at the following JS-Code:
var match = "abcdef";
var input = "abcxdef";
var mArray = match.split("");
var inArray = input.split("");
var max = Math.min(mArray.length, inArray.length) - 1;
for (var i = 0; i < max; i++) {
if (mArray[i] != inArray[i]) { break; }
}
input.substring(0, i);
Where match is the string to be partially matched, input is the input and input.substring(0, i) is the result of the matching part. And you can change match as often as you like.
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 have to write the following as it is.
('trial1' = Ozone1, 'trial2' = Ozone2, trial3 = Ozone3,...........trial1000 = Ozone1000)
I want to write this with one command in R. How do I do it?
I tried it using paste0
Let us take only 5 as number of repetitions:
paste0("trial",1:5,"= Ozone", 1:5)
I get this as result.
"trial1= Ozone1" "trial2= Ozone2" "trial3= Ozone3" "trial4= Ozone4" "trial5= Ozone5"
But it is not the way I wanted it. I want the output to come out as it is like (not even in inverted commas):
('trial1' = Ozone1, 'trial2' = Ozone2, 'trial3' = Ozone3, 'trial4' = Ozone4, 'trial5 = Ozone5)
Also as you can see, it is not a string i.e. output should not come between inverted commas as "........". I want it as it is exactly.
How do i do it?
This will generate the string you want...
paste0('(',paste0("'trial",1:1000,"'= Ozone",1:1000,collapse=' ,'),')')
This will print the string without quotes...
print(paste0('(',paste0("'trial",1:10,"'= Ozone",1:10,collapse=' ,'),')'), quote=FALSE)
I hope it answered your question...
You need to escape the single quotes, ie \', and use the collapse argument of paste0:
paste0("(", paste0("\'trial",1:5,"\' = Ozone",1:5, collapse=", "), ")")
[1] "('trial1' = Ozone1, 'trial2' = Ozone2, 'trial3' = Ozone3, 'trial4' = Ozone4, 'trial5' = Ozone5)"
I have a list of several phrases in the following format
thisIsAnExampleSentance
hereIsAnotherExampleWithMoreWordsInIt
and I'm trying to end up with
This Is An Example Sentance
Here Is Another Example With More Words In It
Each phrase has the white space condensed and the first letter is forced to lowercase.
Can I use regex to add a space before each A-Z and have the first letter of the phrase be capitalized?
I thought of doing something like
([a-z]+)([A-Z])([a-z]+)([A-Z])([a-z]+) // etc
$1 $2$3 $4$5 // etc
but on 50 records of varying length, my idea is a poor solution. Is there a way to regex in a way that will be more dynamic? Thanks
A Java fragment I use looks like this (now revised):
result = source.replaceAll("(?<=^|[a-z])([A-Z])|([A-Z])(?=[a-z])", " $1$2");
result = result.substring(0, 1).toUpperCase() + result.substring(1);
This, by the way, converts the string givenProductUPCSymbol into Given Product UPC Symbol - make sure this is fine with the way you use this type of thing
Finally, a single line version could be:
result = source.substring(0, 1).toUpperCase() + source(1).replaceAll("(?<=^|[a-z])([A-Z])|([A-Z])(?=[a-z])", " $1$2");
Also, in an Example similar to one given in the question comments, the string hiMyNameIsBobAndIWantAPuppy will be changed to Hi My Name Is Bob And I Want A Puppy
For the space problem it's easy if your language supports zero-width-look-behind
var result = Regex.Replace(#"thisIsAnExampleSentanceHereIsAnotherExampleWithMoreWordsInIt", "(?<=[a-z])([A-Z])", " $1");
or even if it doesn't support them
var result2 = Regex.Replace(#"thisIsAnExampleSentanceHereIsAnotherExampleWithMoreWordsInIt", "([a-z])([A-Z])", "$1 $2");
I'm using C#, but the regexes should be usable in any language that support the replace using the $1...$n .
But for the lower-to-upper case you can't do it directly in Regex. You can get the first character through a regex like: ^[a-z] but you can't convet it.
For example in C# you could do
var result4 = Regex.Replace(result, "^([a-z])", m =>
{
return m.ToString().ToUpperInvariant();
});
using a match evaluator to change the input string.
You could then even fuse the two together
var result4 = Regex.Replace(#"thisIsAnExampleSentanceHereIsAnotherExampleWithMoreWordsInIt", "^([a-z])|([a-z])([A-Z])", m =>
{
if (m.Groups[1].Success)
{
return m.ToString().ToUpperInvariant();
}
else
{
return m.Groups[2].ToString() + " " + m.Groups[3].ToString();
}
});
A Perl example with unicode character support:
s/\p{Lu}/ $&/g;
s/^./\U$&/;
I am using vb.net to parse my own basic scripting language, sample below. I am a bit stuck trying to deal with the 2 separate types of nested brackets.
Assuming name = Sam
Assuming timeFormat = hh:mm:ss
Assuming time() is a function that takes a format string but
has a default value and returns a string.
Hello [[name]], the time is [[time(hh:mm:ss)]].
Result: Hello Sam, the time is 19:54:32.
The full time is [[time()]].
Result: The full time is 05/06/2011 19:54:32.
The time in the format of your choice is [[time([[timeFormat]])]].
Result: The time in the format of your choice is 19:54:32.
I could in theory change the syntax of the script completely but I would rather not. It is designed like this to enable strings without quotes because it will be included in an XML file and quotes in that context were getting messy and very prone to errors and readability issues. If this fails I could redesign using something other than quotes to mark out strings but I would rather use this method.
Preferably, unless there is some other way I am not aware of, I would like to do this using regex. I am aware that the standard regex is not really capable of this but I believe this is possible using MatchEvaluators in vb.net and some form of recursion based replacing. However I have not been able to get my head around it for the last day or so, possibly because it is hugely difficult, possibly because I am ill, or possibly because I am plain thick.
I do have the following regex for parts of it.
Detecting the parentheses: (\w*?)\((.*?)\)(?=[^\(+\)]*(\(|$))
Detecting the square brackets: \[\[(.*?)\]\](?=[^\[+\]]*(\[\[|$))
I would really appreciate some help with this as it is holding the rest of my project back at the moment. And sorry if I have babbled on too much or not put enough detail, this is my first question on here.
Here's a little sample which might help you iterate through several matches/groups/captures. I realize that I am posting C# code, but it would be easy for you to convert that into VB.Net
//these two may be passed in as parameters:
string tosearch;//the string you are searching through
string regex;//your pattern to match
//...
Match m;
CaptureCollection cc;
GroupCollection gc;
Regex r = new Regex(regex, RegexOptions.IgnoreCase);
m = r.Match(tosearch);
gc = m.Groups;
Debug.WriteLine("Number of groups found = " + gc.Count.ToString());
// Loop through each group.
for (int i = 0; i < gc.Count; i++)
{
cc = gc[i].Captures;
counter = cc.Count;
int grpnum = i + 1;
Debug.WriteLine("Scanning group: " + grpnum.ToString() );
// Print number of captures in this group.
Debug.WriteLine(" Captures count = " + counter.ToString());
if (cc.Count >= 1)
{
foreach (Capture cap in cc)
{
Debug.WriteLine(string.format(" Capture found: {0}", cap.ToString()));
}
}
}
Here is a slightly simplified version of the code I wrote for this. Thanks for the help everyone and sorry I forgot to post this before. If you have any questions or anything feel free to ask.
Function processString(ByVal scriptString As String)
' Functions
Dim pattern As String = "\[\[((\w+?)\((.*?)\))(?=[^\(+\)]*(\(|$))\]\]"
scriptString = Regex.Replace(scriptString, pattern, New MatchEvaluator(Function(match) processFunction(match)))
' Variables
pattern = "\[\[([A-Za-z0-9+_]+)\]\]"
scriptString = Regex.Replace(scriptString, pattern, New MatchEvaluator(Function(match) processVariable(match)))
Return scriptString
End Function
Function processFunction(ByVal match As Match)
Dim nameString As String = match.Groups(2).Value
Dim paramString As String = match.Groups(3).Value
paramString = processString(paramString)
Select Case nameString
Case "time"
Return getLocalValueTime(paramString)
Case "math"
Return getLocalValueMath(paramString)
End Select
Return ""
End Function
Function processVariable(ByVal match As Match)
Try
Return moduleDictionary("properties")("vars")(match.Groups(1).Value)
Catch ex As Exception
End Try
End Function