I'm having a condition as the particular text must be in the particular format where in I'm getting the text from the scanner and need to check if it is in the right format.
The format comes like
The starting letter must start with e or E, next letter might be any letter from a to z or A to Z alphabets, next 9 characteres must be numbers and the last two characters must be from anything within a to z or A to Z alphabets
I tried something like
if (_scannedCode.startsWith('e|E') && _scannedCode[1].startsWith('a-zA-Z') && _scannedCode.substring(2, 10))
but got struck.
Seeing the answers did get the condtions correctly but was struck up with one, so just wanted to get a clarification if its right or not
RegExp emo = new RegExp(r'[0-9]{6}(EM|em|eM|Em){2}[0-9]{10}$');
As i needed the first 6 characters to be numbers the next two characters be alphabet(em) and the remaining 10 characters be numbers.
final regex = RegExp(r'(e|E)[a-zA-z]\d{9}[a-zA-z]{2}');
if (_scannedCode.length == 13 && regex.hasMatch(_scannedCode) ) {
// your code
}
Related
I have a text file with 1 million decimal digits of "e" number with 80 digits on each line excluding the first and the last line which have 76 and 4 digits and the file has 12501 lines. I want to convert it into a vector in matlab with each digit on each row. I tried num2str function, but the problem is that it gets converted like for example '7.1828e79' (13 characters). What can I do?
P.S.1: The first two lines of the text file (76 and 80 digits) are:
7182818284590452353602874713526624977572470936999595749669676277240766303535 47594571382178525166427427466391932003059921817413596629043572900334295260595630
P.S.2: I used "dlmread" and got a 12501x1 vector, with the first and second row of 7.18281828459045e+75 and 4.75945713821785e+79 and the problem is that when I use num2str for example for the first row value, I get: '7.182818284590453e+75' as a string and not the whole 76 digits. My aim was to do something like this:
e1=dlmread('e.txt');
es1=num2str(e1);
for i=1:12501
for j=1:length(es1(1,:))
a1((i-1)*length(es1(1,:))+j)=es1(i,j);
end
end
e_digits=a1.';
but I get a string like this:
a1='7.182818284590453e+754.759457138217852e+797.381323286279435e+799.244761460668082e+796.133138458300076e+791.416928368190255e+79 5...'
with 262521 characters instead of 1 million digits.
P.S.3: I think the problem might be solved if I can manipulate the text file in a way that I have one digit on each line and simply use dlmread.
Well, this is not hard, there are many ways to do it.
So first you want to load in your file as a Char Array using something simple like (you want a Char Array so that you can easily manipulate it to forget about the lines breaks) :
C = fileread('yourfile.txt'); %loads file as Char Array
D = C(~isspace(C)); %Removes SPACES which are line-breaks
Next, you want to actually append a SPACE between each char (this is because you want to use the num2str transform - and matlab needs to see the space), you can do this using a RESHAPE, a STRTRIM or simply a REGEX:
E = strtrim(regexprep(D, '.{1}', '$0 ')); %Add a SPACE after each Numeric Digit
Now you can transform it using str2num into a Vector:
str2num(E)'; %Converts the Char Array back to Vector
which will give you a single digit each row.
Using your example, I get a vector of 156 x 1, with 1 digit each row as you required.
You can get a digit per row like this
fid = fopen('e.txt','r');
c = textscan(fid,'%s');
c=cat(1,c{:});
c = cellfun(#(x) str2num(reshape(x,[],1)),c,'un',0);
c=cat(1,c{:});
And it is not the only possible way.
Could you please tell what is the final task, how do you plan using the array of e digits?
I have an .nc file I'm reading in matlab, and getting info out of the time variable.
the code looks like this
>> ncreadatt(model_list{3},'T','units')
ans =
'months since 1850-01-01'
what I want to do is get just the '1850' out of the answer.
Regular expression is a very powerful tool to parse and manipulate strings.
Matlab has regexp command:
line = 'months since 1850-01-01';
res = regexp( line, '\s(\d+)-', 'tokens', 'once');
year = str2double(res{1})
And the results is:
year =
1850
The regular expression used '\s(\d+)-' means:
\s - look for a single white space character (the space before 1850).
'(\d+)' - look for one or more digit ('\d+'), the parentheses means that all charcters matching here will be saved as a "token".
'-' - look for a single '-' after the digits.
You can play with it on ideone.
I need to search a text file that is about 30 lines. I need it to search row by row and grab numbers based on their position in the text file that will remain constant throughout the text file. Currently I need to get the first 2 numbers and then the last 4 numbers of each row. My code now:
FileToOpen = fopen(textfile.txt)
if FileToOpen == -1
disp('Error')
return;
end
while true
msg = fgetl(FileToOpen)
if msg == -1
break;
end
end
I would like to use the fgetl command if possible as I somewhat know that command, but if their is an easier way that will be more than welcome.
This looks like a good start. You should be able to use msg - '0' to get the value of the numbers. For ascii code the digits are placed next to each other in the right order (0,1,2,3,4,5,6,7,8,9). What you do when you subtract with '0' is that you subtract msg with the ascii code of '0'. You will then get the digits as
tmp = msg - '0';
idx = find(tmp>=0 & tmp < 10); % Get the position in the row
val = tmp(idx); % Or tmp(tmp>=0 & tmp < 10) with logical indexing.
I agree that fgetl is probably the best to use for text without specific structure. However, in case you have a special structure of the text you can use that and thus be able to use more effective algorithms.
In case you was actually after finding the absolute position of the digits in the text, you can save the msgLength = msgLength + length(msg) for every iteration and use that to calculate the absolute position of the digits.
I have a string and I need two characters to be returned.
I tried with strsplit but the delimiter must be a string and I don't have any delimiters in my string. Instead, I always want to get the second number in my string. The number is always 2 digits.
Example: 001a02.jpg I use the fileparts function to delete the extension of the image (jpg), so I get this string: 001a02
The expected return value is 02
Another example: 001A43a . Return values: 43
Another one: 002A12. Return values: 12
All the filenames are in a matrix 1002x1. Maybe I can use textscan but in the second example, it gives "43a" as a result.
(Just so this question doesn't remain unanswered, here's a possible approach: )
One way to go about this uses splitting with regular expressions (MATLAB's strsplit which you mentioned):
str = '001a02.jpg';
C = strsplit(str,'[a-zA-Z.]','DelimiterType','RegularExpression');
Results in:
C =
'001' '02' ''
In older versions of MATLAB, before strsplit was introduced, similar functionality was achieved using regexp(...,'split').
If you want to learn more about regular expressions (abbreviated as "regex" or "regexp"), there are many online resources (JGI..)
In your case, if you only need to take the 5th and 6th characters from the string you could use:
D = str(5:6);
... and if you want to convert those into numbers you could use:
E = str2double(str(5:6));
If your number is always at a certain position in the string, you can simply index this position.
In the examples you gave, the number is always the 5th and 6th characters in the string.
filename = '002A12';
num = str2num(filename(5:6));
Otherwise, if the formating is more complex, you may want to use a regular expression. There is a similar question matlab - extracting numbers from (odd) string. Modifying the code found there you can do the following
all_num = regexp(filename, '\d+', 'match'); %Find all numbers in the filename
num = str2num(all_num{2}) %Convert second number from str
I would like to modify a string that will have make the first letter capitalized and all other letters lower cased, and anything else will be unchanged.
I tried this:
function new_string=switchCase(str1)
%str1 represents the given string containing word or phrase
str1Lower=lower(str1);
spaces=str1Lower==' ';
caps1=[true spaces];
%we want the first letter and the letters after space to be capital.
strNew1=str1Lower;
strNew1(caps1)=strNew1(caps1)-32;
end
This function works nicely if there is nothing other than a letter after space. If we have anything else for example:
str1='WOW ! my ~Code~ Works !!'
Then it gives
new_string =
'Wow My ^code~ Works !'
However, it has to give (according to the requirement),
new_string =
'Wow! My ~code~ Works !'
I found a code which has similarity with this problem. However, that is ambiguous. Here I can ask question if I don't understand.
Any help will be appreciated! Thanks.
Interesting question +1.
I think the following should fulfil your requirements. I've written it as an example sub-routine and broken down each step so it is obvious what I'm doing. It should be straightforward to condense it into a function from here.
Note, there is probably also a clever way to do this with a single regular expression, but I'm not very good with regular expressions :-) I doubt a regular expression based solution will run much faster than what I've provided (but am happy to be proven wrong).
%# Your example string
Str1 ='WOW ! my ~Code~ Works !!';
%# Convert case to lower
Str1 = lower(Str1);
%# Convert to ascii
Str1 = double(Str1);
%# Find an index of all locations after spaces
I1 = logical([0, (Str1(1:end-1) == 32)]);
%# Eliminate locations that don't contain lower-case characters
I1 = logical(I1 .* ((Str1 >= 97) & (Str1 <= 122)));
%# Check manually if the first location contains a lower-case character
if Str1(1) >= 97 && Str1(1) <= 122; I1(1) = true; end;
%# Adjust all appropriate characters in ascii form
Str1(I1) = Str1(I1) - 32;
%# Convert result back to a string
Str1 = char(Str1);