How to do uniq -d without presorting (or something similar)

问题

I am aware that I can remove duplicated lines without presorting, for example:

awk '!x[$0]++' file

However, my goal is to only print lines which are duplicated and only once. If it were not for the presorting problem

sort | uniq -d

would be perfect. BUT the order is of great importance to me. Is there a way to do this with awk, grep or something similar?

I am looking for a one liner which does not require writing a script if possible.

回答1:

Just check the value of x[$0]:

awk 'x[$0]++ == 1' file.txt

The above will print a line when it's seen for the second time.

Or with prefixed ++:

awk '++x[$0] == 2' file.txt

回答2:

Try simply removing ! from your awk command.

bash-4.1$ cat /tmp/inp.txt 
abc
def
abc
xyz
def
hello

bash-4.1$ awk 'x[$0]++ == 1' /tmp/inp.txt 
abc
def

Hope this is what you expect!!

来源：https://stackoverflow.com/questions/38203260/how-to-do-uniq-d-without-presorting-or-something-similar