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I've used the below command to count the number of words in a file:
tr ' ' '\n' < Filename | grep -c "WORD"
This returns the word list with counter. Now I want to count the number of words of a particular length. For example, given a file with these contents:
employee paresh emp employee jayesh hitesh
When I run the shell script with argument 6, it should display/count words with 6 characters (paresh, jayesh, and hitesh, with count 3). How can I do that?
If you grep for the regular expression ^.{6}$ it will return lines with exactly six characters:
$ tr ' ' '\n' < Filename | grep '^.\{6\}$'
paresh
jayesh
hitesh
$ tr ' ' '\n' < Filename | grep -c '^.\{6\}$'
3
grep will look for { and } symbols, but not 6 repetitions of preceeding pattern.
X{Y} means "find the X character Y times"); without the escapes it searches for actual braces
Jan 4, 2011 at 15:28
Also,
awk -v n=6 '{for (i=1; i<=NF; i++) if (length($i) == n) print $i}' file
or, with tr
tr ' ' '\n' file | awk -v n=6 'length($0)==n'
NF, you don't need to iterate to find the number of fields.
Dec 11, 2012 at 16:24
awk '{ print length($NF) }' filename | sort | uniq -c was useful to count word sizes in last column of a file. thanks