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Top Forums UNIX for Advanced & Expert Users Sort words based on word count on each line Post 302945623 by martinsmith on Monday 1st of June 2015 05:44:16 AM
Old 06-01-2015
Quote:
Originally Posted by Don Cragun
Assuming you don't have more than 99,999 words on a line and that words are separated by sequences of one or more blanks, the following should work:
Code:
#!/bin/ksh
awk '{printf("%05d%s\n", NF, $0)}' file.txt|sort|awk '{print substr($0, 6)}'

Thank you Don! It works perfectly. Just what I needed. Many Thanks
 

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LOOK(1) 						      General Commands Manual							   LOOK(1)

NAME
look - find lines in a sorted list SYNOPSIS
look [ -dfnixtc ] [ string ] [ file ] DESCRIPTION
Look consults a sorted file and prints all lines that begin with string. It uses binary search. The following options are recognized. Options dfnt affect comparisons as in sort(1). -i Interactive. There is no string argument; instead look takes lines from the standard input as strings to be looked up. -x Exact. Print only lines of the file whose key matches string exactly. -d `Directory' order: only letters, digits, tabs and blanks participate in comparisons. -f Fold. Upper case letters compare equal to lower case. -n Numeric comparison with initial string of digits, optional minus sign, and optional decimal point. -t[c] Character c terminates the sort key in the file. By default, tab terminates the key. If c is missing the entire line comprises the key. If no file is specified, /lib/words is assumed, with collating sequence df. FILES
/lib/words SOURCE
/src/cmd/look.c SEE ALSO
sort(1), grep(1) DIAGNOSTICS
The exit status is ``not found'' if no match is found, and ``no dictionary'' if file or the default dictionary cannot be opened. LOOK(1)
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