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edited Nov 6, 2015 at 1:38

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Here is another awk solution, similar to @Thor's answer, less concise but more efficient:

awk '!NF {print;next}; !($0 in a) {a[$0];print}' file

With this, we only check a[$0] has existed or not. If not, initializing it then print. In this case, we don't have any reference, assignment to a[$0] if it existed.

Here is another awk solution, similar to @Thor's answer, less concise but more efficient:

awk '!NF {print;next} !($0 in a) {a[$0];print}' file

With this, we only check a[$0] has existed or not. If not, initializing it then print. In this case, we don't have any reference, assignment to a[$0] if it existed.

Here is another awk solution, similar to @Thor's answer, less concise but more efficient:

awk '!NF {print;next}; !($0 in a) {a[$0];print}' file

With this, we only check a[$0] has existed or not. If not, initializing it then print. In this case, we don't have any reference, assignment to a[$0] if it existed.

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answered May 21, 2014 at 6:37

cuonglm

158.2k
41
342
420

Here is another awk solution, similar to @Thor's answer, less concise but more efficient:

awk '!NF {print;next} !($0 in a) {a[$0];print}' file

With this, we only check a[$0] has existed or not. If not, initializing it then print. In this case, we don't have any reference, assignment to a[$0] if it existed.