Fastest way to determine if a character index is a carriage return (\n)?

Hi all.

1.15 seconds per 100,000 iterations:

if regexp(text(index),'[\n]','once')
    y = 1;
else
    y = 0;
end

1.25 seconds per 100,000 iterations:

ret =  regexp(text(index),'[\n]');
if any(ret==index)
    y = 1;
else
    y = 0;
end

I would think there is a faster way, but I have yet to find it. Any Suggestions??

Thanks so Much!

Will

EDIT: I acknowledge there are more than 1 type of carriage return, for simplicity lets assume the text only has one type.

[Stephen23]

"there are more than 1 type of carriage return,"

There exactly one carriage return character, in MATLAB represented by the escaped character \r.

Here is a table of the tab and line control characters defined in the ASCII standard:

 8  \b  Backspace
 9  \t  Horizontal Tab
 10  \n  Line Feed
 11  \v  Vertical Tab
 12  \f  Form Feed
 13  \r  Carriage Return

Several of these (and combinations thereof) have been used to indicate a newline in a text file, but a newline standard is not the same thing as a carriage return character.

[Guillaume]

AS per dpb answer, for finding a single character, direct comparison is going to be A LOT faster than involving a regular expression engine. I just wanted to comment on the regular expression.

[] is used in regular expressions to group several characters together (i.e. match any of the characters within the brackets). When there's only one character to match, the brackets serve no purpose, so

regexp(text, '\n', once)

would have worked just as well and would avoid wondering if the fact that there's only one character in the class is a bug or not.

I doubt it would make any difference to speed.

Answer 1

[dpb]

y=any(text(index)==char(10));

[Will Kinsman]

Ah! thanks! was actually looking for char(15) for my solution but this is much faster!! Thanks! Ultimately using:

function y = is_CR(text,index)
%return if an index is a carriage return
if text(index)==char(15)
    y = 1;
else
    y = 0;
end

[dpb]

I'd still suggest

function y = is_CR(text,index)
% return TRUE if an index is a carriage return, else FALSE

    y=(text(index)==15);
end

to return a logical instead of double for a logical function, particularly one you named is something. Plus, if you're looking for performance, eliminate an unnecessary if...end clause--it's totally superfluous; you've already got a 0|1 result from the test; it won't get any "one-er"...

On a stylistic note, char(15) is NOT CR but SI (Shift-In or X-Off) in common usage nomenclature which could be confusing for other users...see <http://www.ascii-code.com/>

[Will Kinsman]

hahahaaa done! Super appreciate the help. This massively sped up my program!

[dpb]

No problem, glad to help where can... :)

Another note is that if performance is the issue, as written passing the index into the function it probably actually would make more sense to simply do the test inline instead of using the function.

Answer 2

[Image Analyst]

I'm not sure what 15 is (it's a "shift in" character) but, like Stephen says carriage return is 13. And you don't need char(). So it would be

crIndexes = yourText == 13;

Here's a full demo showing lots of possible ways that new lines show up in the ASCII bytes:

yourText = sprintf('abc\n def \r hij \r\n  klm \n\r theEnd')
crIndexes = yourText == 13 % Find all CR
lfIndexes = yourText == 10 % Find all LF
% Find pairings with strfind():
crLF_location = strfind(yourText, [13, 10])
lfCR_location = strfind(yourText, [10, 13])

[dpb]

..."you don't need char()..."

Good catch, IA; I'll make the correction. Normally I don't do that, not sure why did this go-'round... :(