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Top Forums UNIX for Advanced & Expert Users Sort words based on word count on each line Post 302945614 by martinsmith on Monday 1st of June 2015 03:52:48 AM
Old 06-01-2015
Question Sort words based on word count on each line

Hi Folks Smilie

I have a .txt file with thousands of words. I'm trying to sort the lines in order based on number of words per line.

Example

from:
Code:
word
word word word
word word
word word word word
word
word word word
word word


to desired output:
Code:
word
word
word word
word word
word word word
word word word
word word word word


Any help would be greatly appreciated.
Thanks for your help!

ps. Hope i posted this in the correct forum board. My apologies if it's in the wrong section

Last edited by Don Cragun; 06-01-2015 at 05:08 AM.. Reason: Add CODE tags.
 

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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
/sys/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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