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Bash - remove duplicates without sort


 
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# 1  
Old 08-31-2013
Bash - remove duplicates without sort

I need to use bash to remove duplicates without using sort first.

I can not use:
Code:
cat file | sort | uniq

But when I use only

Code:
cat file | uniq

some duplicates are not removed.
# 2  
Old 08-31-2013
Obviously, the simple way to do this is:
Code:
sort -u file

but you tell us we can't do that without saying why. Is there a requirement to output lines in the same order they were in in the input file? If so, is it important to keep a particular one of the duplicated lines in the output? Or, do you want every line that had one or more duplicates removed from the output?

In the 2.5 years you've been a member of this forum, there have seen dozens of examples using awk to do this where the 1st duplicated input line is kept or the last duplicated input line is kept, or all duplicated lines are removed. If keeping the same order is important, it is more difficult to keep the last duplicate than it is to keep the 1st duplicate.

So, what are the real requirements?
# 3  
Old 09-01-2013
Try
Code:
awk '!a[$0]++' file

# 4  
Old 09-01-2013
Question

Quote:
Originally Posted by Jotne
Try
Code:
awk '!a[$0]++' file

Hi.
I tried:
Code:
echo -e "prova\012zappa\012prova\012quadro\012cesto\012zappa" | awk '!a[$0]++'

and it works Smilie.

Please explain me what does it means:
Code:
'!a[$0]++'

# 5  
Old 09-01-2013
This User Gave Thanks to Jotne For This Post:

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