Check if an awk array contains a value

Question

With Perl you can check if an array contains a value

$ perl -e '@foo=(444,555,666); print 555 ~~ @foo ? "T" : "F"'
T

However with awk, this similar command is checking the array indexes rather than values

$ awk 'BEGIN {split("444 555 666", foo); print 555 in foo ? "T" : "F"}'
F

How can I check if an array contains a particular value with awk?

Like Etan says, you need a for-loop. You can achieve a somewhat flexible result by creating a new array (hash) with the values as keys, e.g.: `awk 'BEGIN { split("444 555 666", foo); for(i=1; i<=length(foo); i++) bar[foo[i]]; print 555 in bar ? "T" : "F" }'` — Thor, Nov 04 '14 at 23:19
@Thor - Can't you just use `for( i in foo ) bar[foo[i]]` instead? — n0741337, Nov 04 '14 at 23:24
@n0741337: You are right, shorter and cleaner, so it becomes: `awk 'BEGIN { split("444 555 666", foo); for(i in foo) bar[foo[i]]; print 555 in bar ? "T" : "F" }'`. — Thor, Nov 04 '14 at 23:27

score 28 · Accepted Answer · edited Nov 12 '17 at 09:39

Awk noob here. I digested Steven's answer and ended up with this hopefully easier to understand snippet below. There are 2 more subtle problems:

An Awk array is actually a dictionary. It's not ["value1", "value2"], it's more like {0: "value1", 1: "value2"}.
in checks for keys, and there is no built-in way to check for values.

So you have to convert your array (which is actually a dictionary) to a dictionary with the values as keys.

BEGIN {

    split("value1 value2", valuesAsValues)
    # valuesAsValues = {0: "value1", 1: "value2"}

    for (i in valuesAsValues) valuesAsKeys[valuesAsValues[i]] = ""
    # valuesAsKeys = {"value1": "", "value2": ""}
}

# Now you can use `in`
($1 in valuesAsKeys) {print}

For one-liners:

echo "A:B:C:D:E:F" | tr ':' '\n' | \
awk 'BEGIN{ split("A D F", parts); for (i in parts) dict[parts[i]]=""}  $1 in dict'

score 0 · Answer 2 · answered Nov 19 '21 at 19:25

what I do is structure my data to take advantage of what the language provides.

So instead of saying :

BEGIN {split("444 555 666", foo); .... }

say :

BEGIN { bar[ "444" ] ; ... }

or ala another suggestion:

BEGIN { 
  split("444 555 666", foo);
  for( i in FOO ) bar[foo[i]];
}

remember arrays in awk are really associative arrays.

score 0 · Answer 3 · answered May 18 '22 at 08:18

0

gawk 'BEGIN { split("a b c d",parts); for (i in parts) dict[parts[i]]=1 } $1 in dict {print}' some.log

answered May 18 '22 at 08:18

Ryan Liu

499
4
5

While this code may solve the question, [including an explanation](//meta.stackexchange.com/q/114762) of how and why this solves the problem would really help to improve the quality of your post, and probably result in more up-votes. Remember that you are answering the question for readers in the future, not just the person asking now. Please [edit] your answer to add explanations and give an indication of what limitations and assumptions apply. – Yunnosch Sep 08 '22 at 21:07

score 0 · Answer 4 · answered Jan 25 '23 at 21:28

BEGIN {
  split("value1 value2 value3", valueArray)
  # valueArray is {1: "value1", 2: "value2", 3: "value2"}
}
{
  # given $1 is the value to test for
  for (i in valueArray) {
    if ( $1 == valueArray[i] ) {
      print $1
      # OR 
      print "True"
      # ...
    }
  }
}

score 0 · Answer 5 · answered Jan 28 '23 at 20:37

0

In the spirit of a one-liner (see Thamme's answer) that sticks more to the OP's perl reference (ternary operator):

echo -e "a\nb\nc" | awk '{t[$1]}END{print (("a" in t) ? "yes":"no")}' # prints yes, try with ("m" in t)

answered Jan 28 '23 at 20:37

ExpertNoob1

900
1
8
17

Check if an awk array contains a value

5 Answers5

For one-liners:

Linked