How to identify a string and copy the value from the text file

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madhu T S
madhu T S le 28 Avr 2015
Commenté : madhu T S le 18 Mai 2015
Hii ppl... can anyone suggest how to identify a string in a txt file... im using some instructions and it is working well... if a string something like 151.04 LMD[q=1,d=2] 99.9972 % .. it has to identify the line of the string and give me the value as 99.9972.. if the string is like 150.04 ROTOR TIMECONST 940 m Im able to get the value as 940.. but the brackets [] are giving me problem.. how to solve this?? please suggest me

Réponse acceptée

per isakson
per isakson le 28 Avr 2015
Modifié(e) : per isakson le 5 Mai 2015
This script
str = fileread( 'cssm.txt' );
cac = regexp( str, '(?<=LMD\[q=1,d=2\][ ]+)[\d\.]++(?=[ ]+\%)', 'match','once' )
cac = regexp( str, '(?<=ROTOR TIMECONST[ ]+)\d++(?=[ ]+m)' , 'match','once' )
returns
cac =
99.9972
cac =
940
where cssm.txt contains
Hii ppl... can anyone suggest how to identify a string
in a txt file... im using some instructions and it is
working well... if a string something like
151.04 LMD[q=1,d=2] 99.9972 % .. it has to identify
the line of the string and give me the value as 99.9972..
if the string is like 150.04 ROTOR TIMECONST 940 m Im able
to get the value as 940.. but the brackets [] are giving me
problem.. how to solve this?? please suggest me
Mostly, regex is fast enough
&nbsp
ADDENDUM
The script below does the "same" job orders of magnitude faster. (I guess, I did a beginners mistake.)
str = fileread( 'cssm.txt' );
cac = regexp(str, '(?<=LMD\[q=1,d=2\])[ \d\.]++(?=\%)', 'match','once')
cac = regexp(str, '(?<=ROTOR TIMECONST)[ \d]++(?=m)' , 'match','once')
returns
cac =
99.9972
cac =
940
The differences are that in this latter case
  • the spaces, which surrounds the value, are captured together with the value.
  • if there is only spaces between "ROTOR TIMECONST" and "m", this script captures the string of spaces
  • ...
This variant doesn't capture a string of spaces
cac = regexp(str, '(?<=ROTOR TIMECONST) +\d++ +(?=m)', 'match','once')
but is significantly slower.
  4 commentaires
per isakson
per isakson le 5 Mai 2015
Modifié(e) : per isakson le 5 Mai 2015
Yes, but faster .... . What's the fastest way depends heavily on the size and content of the file. BTW: what do you mean by slow?
I've added a faster script to my answer.
madhu T S
madhu T S le 18 Mai 2015
Hi per isakson... actually slow in the sense... if I use the code to get the value once its working fine... i have 60 lines of data in which the braces [] present.. it taking long time to process these 60 lines and to give value for these 60 lines...

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Plus de réponses (1)

the cyclist
the cyclist le 28 Avr 2015
Modifié(e) : the cyclist le 28 Avr 2015
Here is one way, assuming that the value you are looking for is always between two spaces at the end of the string:
S = '150.04 ROTOR TIMECONST 940 m';
% Find where the spaces are
indexToSpaces = regexp(S,' ');
% Get the characters that are between that last two spaces
valueString = S(indexToSpaces(end-1)+1:indexToSpaces(end)-1);
% Convert those characters to numeric
value = str2double(valueString);
If that is not correct, then you need to think carefully about what rule will consistently identify where the value is. Then maybe we can help you translate that into MATLAB code.
  1 commentaire
madhu T S
madhu T S le 4 Mai 2015
thanks for your answer the cyclist, Im using the same logic but its working for others except for 151.04 LMD[q=1,d=2] 99.9972 %, the sqare brackets making problem for the above code... However I got code from per isakson and its working well... thanks again for your feed back the cyclist

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