I'm trying to check if a string contains one of four sub strings in a simpler way than this:
if (imageUrl.contains('.jpg') ||
imageUrl.contains('.png') ||
imageUrl.contains('.tif') ||
imageUrl.contains('.gif')) {
}
Is there a way to do this? For example checking against a list?
You can use a regex pattern instead of a simple string:
imageUrl.contains(new RegExp("\.(jpg|png|tif|gif)"))
Might be somewhat simpler.
RegularExpression can solve your problem. RegEx are used to search patterns in strings.
RegEx example:
^The matches any string that starts with The
end$ matches a string that ends with end
^The end$ exact string match (starts and ends with The end)
abc* matches a string that has ab followed by zero or more c
Related
I need to count the number of leading tabs in a Swift string. I know there are fairly simple solutions (e.g. looping over the string until a non-tab character is encountered) but I am looking for a more elegant solution.
I have attempted to use a regex such as ^\\t* along with the .numberOfMatches method but this detects all the tab characters as one match. For example, if the string has three leading tabs then that method just returns 1. Is there a way to write a regex that treats each individual tab character as a single match?
Also open to other ways of approaching this without using a regex.
Here is a non-regex solution
let count = someString.prefix(while: {$0 == "\t"}).count
You may use
\G\t
See the regex demo.
Here,
\G - matches a string start position or end of the previous match position, and
\t - matches a single tab.
Swift test:
let string = "\t\t123"
let regex = try! NSRegularExpression(pattern: "\\G\t", options: [])
let numberOfOccurrences = regex.numberOfMatches(in: string, range: NSRange(string.startIndex..., in: string))
print(numberOfOccurrences) // => 2
I have the following string:
{\"Id\":\"135\",\"Type\":0}
The number in the Id field will vary, but will always be an integer with no comma separator. I'm not sure how to get just that value from that string given that it's string data type and not real "XML". I was toying with the replace() function, but the special characters are making it more complex than it seems it needs to be.
is there a way to convert that to XML or something that I can reference the Id value directly?
Maybe use a regular expression, e.g.
import re
txt = "{\"Id\":\"135\",\"Type\":0}"
x = re.search('"Id":"([0-9]+)"', txt)
if x:
print(x.group(1))
gives
135
It is assumed here that the ids are numeric and consist of at least one digit.
Non-regex answer as you asked
\" is an escape sequence in python.
So if {\"Id\":\"135\",\"Type\":0} is a raw string and if you put it into a python variable like
a = '{\"Id\":\"135\",\"Type\":0}'
gives
>>> a
'{"Id":"135","Type":0}'
OR
If the above string is python string which has \" which is already escaped, then do a.replace("\\","") which will give you the string without \.
Now just load this string into a dict and access element Id like below.
import json
d = json.loads(a)
d['Id']
Output :
135
diary_file = tempname();
diary(diary_file);
myFun();
diary('off');
output = fileread(diary_file);
I would like to search a string from output, but also to ignore spaces and upper/lower cases. Here is an example for what's in output:
the test : passed
number : 4
found = 'thetest:passed'
a = strfind(output,found )
How could I ignore spaces and cases from output?
Assuming you are not too worried about accidentally matching something like: 'thetEst:passed' here is what you can do:
Remove all spaces and only compare lower case
found = 'With spaces'
found = lower(found(found ~= ' '))
This will return
found =
withspaces
Of course you would also need to do this with each line of output.
Another way:
regexpi(output(~isspace(output)), found, 'match')
if output is a single string, or
regexpi(regexprep(output,'\s',''), found, 'match')
for the more general case (either class(output) == 'cell' or 'char').
Advantages:
Fast.
robust (ALL whitespace (not just spaces) is removed)
more flexible (you can return starting/ending indices of the match, tokenize, etc.)
will return original case of the match in output
Disadvantages:
more typing
less obvious (more documentation required)
will return original case of the match in output (yes, there's two sides to that coin)
That last point in both lists is easily forced to lower or uppercase using lower() or upper(), but if you want same-case, it's a bit more involved:
C = regexpi(output(~isspace(output)), found, 'match');
if ~isempty(C)
C = found; end
for single string, or
C = regexpi(regexprep(output, '\s', ''), found, 'match')
C(~cellfun('isempty', C)) = {found}
for the more general case.
You can use lower to convert everything to lowercase to solve your case problem. However ignoring whitespace like you want is a little trickier. It looks like you want to keep some spaces but not all, in which case you should split the string by whitespace and compare substrings piecemeal.
I'd advertise using regex, e.g. like this:
a = regexpi(output, 'the\s*test\s*:\s*passed');
If you don't care about the position where the match occurs but only if there's a match at all, removing all whitespaces would be a brute force, and somewhat nasty, possibility:
a = strfind(strrrep(output, ' ',''), found);
This may be a very simple task for many but I could not find anything appropriate for me.
I have a file name: filenm_A006.2011.269.10.47.G25_2010
I want to separate all its parts (separated by . and _) to use them separately. How can I do it with simple matlab commands?
Kind Regards,
Mushi
I recommend regexp:
fname = 'filenm_A006.2011.269.10.47.G25_2010';
parts = regexp(fname, '[^_.]+', 'match');
parts =
'filenm' 'A006' '2011' '269' '10' '47' 'G25' '2010'
You can now refer to parts{1} through parts{8} for the pieces. Explanation: the regexp pattern [^_.] means all characters not equal to _ or ., and the + means you want groups of at least 1 character. Then 'match' asks the regexp function to return a cell array of the strings of all the matches of that pattern. There are other regexp modes; for example, the indices of each piece of the file.
Use the command
strsplit.
cellArrayOfParts = strsplit(fileName,{'.' '_'});
You can use strsplit to split it:
strsplit('filenm_A006.2011.269.10.47.G25_2010',{'_','.'})
ans =
'filenm' 'A006' '2011' '269' '10' '47' 'G25' '2010'
Another option is to use regexp, like Peter suggested.
Hi need help in using regexp for condition matching.
ex.my file has the following content
{hello.program='function'`;
bye.program='script'; }
I am trying to use regexp to match the string that has .program='function' in them:
pattern = '[.program]+\=(function)'
also tried pattern='[^\n]*(.hello=function)[^\n]*';
pattern_match = regexp(myfilename,pattern , 'match')
but this returns me pattern_match={} while i expect the result to be hello.program='function'`;
If 'function' comes with string-markers, you need to include these in the match. Also, you need to escape the dot (otherwise, it's considered "any character"). [.program]+ looks for one or several letters contained in the square brackets - but you can just look for program instead. Also, you don't need to escape the =-sign (which is probably what messed up the match).
cst = {'hello.program=''function''';'bye.program=''script'''; };
pat = 'hello\.program=''function''';
out = regexp(cst,pat,'match');
out{1}{1} %# first string from list, first match
hello.program='function'
EDIT
In response to the comment
my file contains
m2 = S.Parameter;
m2.Value = matlabstandard;
m2.Volatility = 'Tunable';
m2.Configurability = 'None';
m2.ReferenceInterfaceFile ='';
m2.DataType = 'auto';
my objective is to find all the lines that match, .DataType='auto'
Here's how you find the matching lines with regexp
%# read the file with textscan into a variable txt
fid = fopen('myFile.m');
txt = textscan(fid,'%s');
fclose(fid);
txt = txt{1};
%# find the match. Allow spaces and equal signs between DataType and 'auto'
match = regexp(txt,'\.DataType[ =]+''auto''','match')
%# match is not empty only if a match was found. Identify the non-empty match
matchedLine = find(~cellfun(#isempty,match));
Try this as it matches .program='function' exactly:
(\.)program='function'
I think this did not work:
'[.program]+\=(function)'
because of how the []'s work. Here is a link explaining why I say that: http://www.regular-expressions.info/charclass.html