count occurrence of a regex in a very long line

[Mike_Ca Li]

I have a very long line which has about 10 thousand words and many words with "regexpression" how to use a script command to get the count for "regexpression"? Note that one way is break the long line, by using space as separator, into lots of shorter line but that is not efficient. Thank you.

eg: quick brown fox jumps on the lazy dog quick brown fox jumps on the lazy dog quick brown fox jumps on the lazy dog quick brown fox jumps on the lazy dog quick brown fox jumps on the lazy dog quick brown fox jumps on the lazy dog quick brown fox jumps on the lazy dog quick brown fox jumps on the lazy dog quick brown fox jumps on the lazy dog
I need the count for "jumps" for instance.

[Steven E. Protter]

Shalom Mike,

Perhaps process the string with awk?

echo $string | awk -F '{print $1 ... }'

Maybe someting with xargs, not sure.

SEP

[Steven E. Protter]

Mike,

Ah, forget that stuff..... in my first post, unless it unexpected helpful.

store the data in a file:

grep jumps filename | wc -l

Gets you count of jumps.

SEP

[Sandman!]

How about a short one-line awk construct like...

# awk -F"jumps" '{cnt+=(NF-1)} END{print cnt}'

regards!

[Mike_Ca Li]

Thanks for reply SEP.
I tried
awk -F" " '{print NF}' filename
awk: record ... too long. Any other ideas?

[Vincent Fleming · Accepted solution]

Steven,

Having a bad day? (this isn't like you!)

grep jumps filename | wc -l

will tell you how many lines "jumps" appears on, not how many times it appears in one line.

I can't think of an elegent way of doing this, given the regular expression requirement... UNIX shell tools are all pretty much line-based.

You could use 'sed' to change spaces to \r (ie: so that each word is on it's own line), then pipe it through grep|wc or awk... or something similar, but that'll work only if the regex you're looking for does not contain spaces.

Of course, there's always a C program - that's ALWAYS elegent!

Regards,

Vince

[James R. Ferguson]

Hi Mike:

# perl -lne '$i++ while m/jumps/g;END{print $i}'

Regards!

...JRF...

[Vincent Fleming]

Better idea, use "tr" instead of "sed"...

oh - and I also meant \n, not \r...

echo "whatever" | tr ' ' '\n' | grep jumps | wc -l

"tr" is smaller and more efficient than sed since it does so little...

-Vince

[James R. Ferguson]

Hi (again) Mike:

I might add (of course) that you can pipe input to the perl code or specify the filename as an argument:

# perl -lne '$i++ while m/jumps/g;END{print $i}' filename

# echo ... | # perl -lne '$i++ while m/jumps/g;END{print $i}'

Regards!

...JRF...

[Hein van den Heuvel]

An other perl way readily allowing for a regexpr:

perl -ne 'print scalar split (/jumps/)."\n"' filename

Of course this prints one too many.

So fix to:

perl -ne '$x=scalar split(/jumps/) -1; print "$x\n"' filename


Or for single count of many words on many lines:

perl -ne '$x+=scalar split(/jumps/) -1;END{print "$x\n"}' filename


Hein.

[RAC_1]

No perl, no awk, just plain Os built in commands.

Are the words on line seperated by a space?? If yes, do as follows.

tr " " "\n" < your_file | grep -ic "your_reg_exp"

[Mike_Ca Li]

Thanks a lot for all the commands and suggestions.
I tested all the perl and OS commands. Plain OS command run faster