- 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
- ...
How to identify a string and copy the value from the text file
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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
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per isakson
il 28 Apr 2015
Modificato: per isakson
il 5 Mag 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
 
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
This variant doesn't capture a string of spaces
cac = regexp(str, '(?<=ROTOR TIMECONST) +\d++ +(?=m)', 'match','once')
but is significantly slower.
4 Commenti
per isakson
il 5 Mag 2015
Modificato: per isakson
il 5 Mag 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.
Più risposte (1)
the cyclist
il 28 Apr 2015
Modificato: the cyclist
il 28 Apr 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.
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