CODEWITHSUNDEEP
awk
pawamoy
pawamoy

Reputation: 3806

Awk how to negate a condition

I'm trying to compute some stuff in awk, and at the end print the result in the order of the input. For each line, I check if it has not been already seen. If not, I add it to the array and also store it in an order array.

{
    if (! $0 in seen) {
        seen[$0] = 1
        order[o++] = $0
    }
} END {
    for (i=0; i<o; i++)
        printf "%s\n", order[i]
}

You can try it with 

printf 'a\nb\na\nc\nb\na\n' | awk script_above

It prints nothing. If I print the variable o at the end, it shows that its value is still 0. What am I doing wrong?

Upvotes: 1

Views: 3837

Answers (2)

jas
jas

Reputation: 10865

You just need to add parens to get the right operator precedence*:

# a.awk

{
    if (!($0 in seen)) {
        seen[$0] = 1
        order[o++] = $0
    }
}

END {
    for (i=0; i<o; i++)
        printf "%s\n", order[i]
}

Test:

$ awk -f a.awk file
a
b
c

* (The unary ! binds more tightly than the in operator: https://www.gnu.org/software/gawk/manual/html_node/Precedence.html)

Upvotes: 3

RavinderSingh13
RavinderSingh13

Reputation: 133610

What you are trying to do is in Shell way, awk has a way where you could keep checking if an element is part of an array or not, try following once.

printf 'a\nb\na\nc\nb\na\n' | awk '
    !seen[$0]++ {
        order[o++] = $0
    }
END {
    for (i=0; i<o; i++)
        printf "%s\n", order[i]
}'

Here !seen[$0]++ means it is checking condition if an element is NOT a part of indexes of array named a then go inside the BLOCK(where your next statements are provided) then it does ++ which makes sure that this element(which was NOT there in array before checking condition)'s counter incremented by 1 so that next time this !seen[$0]++` condition is NOT TRUE for the already passed element.

Upvotes: 3

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