Sample Input:
apple_ig
rabbit_cat_ig
dog_ig
bird_duck_ig
orange_ig
goat_ig
Expected output:
apple_ig
dog_ig
orange_ig
goat_ig
I need to ignore all data that contains underscore _ more than once in the entire string.
bash, ksh , sed, grep, awk
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5 Answers
$ grep -v '_.*_' file
apple_ig
dog_ig
orange_ig
goat_ig
With awk, to print only lines having maximum one underscore.
$ awk -F_ 'NF<=2' file
apple_ig
dog_ig
orange_ig
goat_ig
Using sed
$ sed -n '/^[^_]*_[[:alpha:]]\+$/p' input_file
apple_ig
dog_ig
orange_ig
goat_ig
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2
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I'm not sure why you'd use
[[:alpha:]]in the second instance though, it seems overly restrictive if the idea is to just count the underscores. Similarly, there's no need to require at least one character after the underscore -- and then there's the thing that\+is undefined in standard BRE, so that command might not work on all systems. Also that would miss any lines that have no underscores at all. I think you'd have to do something likesed -n '/^[^_]*\(_[^_]*\)\{0,1\}$/p'to do that positive match on lines that contain zero or one underscores.– ilkkachuMay 10 at 20:24
GNU sed we try to replace the second occurrence of _ and breakout on failureT, otherwise deleting the pattern space d.
sed 's/_/&/2;T;d' file
POSIX sed same as above but inverting the logic, since T command is a GNU add-on::
sed -n -e 's/_/&/2;t' -e 'p'file
awk
awk '!/(.*_){2}/' file
Perl in a scalar / void contexts the tr builtin emits a count of the number of transformations it does.
perl -pe '$_=$\ if tr/_// > 1' file
sed -n '/.*_.*_.*/!p' file.txt
output
apple_ig
dog_ig
orange_ig
goat_ig
answered May 11 at 10:04
