Extract value until regexp case match

3 visualizaciones (últimos 30 días)
Dyel
Dyel el 23 de Sept. de 2022
Comentada: Dyel el 26 de Sept. de 2022
Hi,
I have this string. I was hoping to use regexp in order to extract the time and date embedded in this string. I was going to use regexp '.*?<=216', as prior to 216 is all the values I was hoping to extract. I was hoping that there would be an easy way to extract the date and time to a separate cell array. Thanks for any and all help.
"9/12/2022 7:38:51 PM
216 Indicator"
  2 comentarios
Rik
Rik el 23 de Sept. de 2022
While this could work, is there a reason why you can't match the date itself directly?
Dyel
Dyel el 23 de Sept. de 2022
Is there a way using regexp to extract the datetime value, or something similar from a char array?

Iniciar sesión para comentar.

Respuesta aceptada

Stephen23
Stephen23 el 23 de Sept. de 2022
Does the data you want to match always occur at the start of the string?
A = "9/12/2022 7:38:51 PM 216 Indicator";
B = regexp(A,'^\S+\s+\S+\s[AP]M','match','once')
B = "9/12/2022 7:38:51 PM"
T = datetime(B,'inputFormat','d/M/u h:m:s aa')
T = datetime
09-Dec-2022 19:38:51
  1 comentario
Dyel
Dyel el 26 de Sept. de 2022
This did the trick. Thanks Stephen.
How I did it was to split it up using the SplitString fuction, and then convert the string with the date to datetime, in case anyone may find it useful.
for j = 1:length(cf)
loopname = cf(j).name;
filetext = fileread(loopname);
txtStr = convertCharsToStrings(filetext);
indVal{i,j} = cell2mat(regexp(filetext, '(?<=Indicator[^0-9]*)[+-]?[0-9]*\.?[0-9]+', 'match'));
splitStr = regexp(txtStr,'\n','split');
dateStr = splitStr{1};
date = datetime(dateStr);
dateVal{i,j} = date;
end

Iniciar sesión para comentar.

Más respuestas (0)

Categorías

Más información sobre Characters and Strings en Help Center y File Exchange.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by