Delete every line if occurence found - awk

Question

I have a file with this format content:
1 6 8
1 6 9
1 12 20
1 6
2 8
2 9
2 12
2 20
2 35
I want to delete all the lines if the number (from 2nd or 3rd column but not from 1st) is found in the next lines whether it is in the 2nd or 3rd column inluding the line where the initial number is found.
I should have this as an output:
2 35
I've tried using:
awk '{for(i=2;i<=NF;i++){if($i in a){next};a[$i]}} 1'
but it doesn't seem to work.
What is wrong ?

Answer 1

One-pass awk that hashes all the records to r[NR] and keeps another array a[$i] for the values seen in fields $2,...NF.
awk ' {
for(i=2;i<=NF;i++) # iterate fields starting from the second
if($i in a) { # if field value was seen before
delete r[a[$i]] # delete related record
a[$i]="" # clear a
f=1 # flag up
} else { # if it was not seen before
a[$i]=NR # add record number to a
r[NR]=$0
}
if(f!=1) # if flag was not raised
r[NR]=$0 # store record on record number
else # if it was raised
f="" # flag down
}
END {
for(i=1;i<=NR;++i)
if(i in r)
print r[i] # output remaining
}' file
Output:
2 35

Answer 2

The simplest way is a double-pass algorithm where you read your file twice.
The idea is to store all values in an array a and count how many times they appear. If the value appears 2 or more times, it means you have found more then a single entry and you should not print the line.
awk '(NR==FNR){a[$2]++; if(NF>2) a[$3]++; next}
(NF==2) && (a[$2]==1);
(NF==3) && (a[$2]==1 && a[$3]==1)' <file> <file>
In practice, you should avoid things such as a[var]==1 if you are not sure whether var is in the array as it will create that array element. However, since we never increase it any more, it is fine to proceed.
If you want to achieve the same thing with more then three fields you can do:
awk '(NR==FNR){for(i=2;i<=NF;++i) a[$i]++; next }
{for(i=2;i<=NF;++i) if(a[$i]>1) next }
{print}' <file> <file>
While both these solutions read the file twice, you can also store the full file in memory and read the file only a single time. This, however, is exactly the same algorithm:
awk '{for(i=2;i<=NF;++i) a[$i]++; b[NR]=$0}
END{ for(j=1;j<=NR;++j) {
$0=b[j];
for(i=2;i<=NF;++i) if(a[$i]>1) continue
print $0
}
}' <file>
comment: this single-pass solution is very simple and stores the full file in memory. The solution of James Brown is very clever. It removes stuff from memory when they are not needed anymore. A bit shorter version is:
awk '{ for(i=2;i<=NF;++i) if ($i in a) delete b[a[$i]]; else { a[$i]=NR; b[NR]=$0 }}
END { for(n=1;n<=NR;++n) if(n in b) print b[n] }' <file>
note: you should never thrive for the shortest solution, but the most readable one!

Answer 3

Could you please try following.
awk '
FNR==NR{
for(i=2;i<=NF;i++){
a[$i]++
}
next
}
(NF==2 && a[$2]==1) || (NF==3 && a[$2]==1 && a[$3]==1)
' Input_file Input_file
Output will be as follows.
2 35

Answer 4

$ cat tst.awk
NR==FNR {
cnt[$2]++
cnt[$3]++
next
}
cnt[$2]<2 && cnt[$NF]<2
$ awk -f tst.awk file file
2 35

Answer 5

This might work for you (GNU sed):
sed -r 'H;s/^[0-9]+ +//;G;s/\n(.*\n)/\1/;h;$!d;s/^([^\n]*)\n(.*)/\2\n \1/;:a;/^[0-9]+ +([0-9]+)\n(.*\n)*[^\n]*\1[^\n]*\1[^\n]*$/bb;/^[0-9]+ +[0-9]+ +([0-9]+)\n(.*\n)*[^\n]*\1[^\n]*\1[^\n]*$/bb;/\n/P;:b;s/^[^\n]*\n//;ta;d' file
This is not a serious solution however it demonstrates what can be achieved using only matching and substitution.
The solution makes a copy of the original file and whilst doing so accumulates all numbers in the second and possible third fields of each record in a separate line which it maintains at the head of the copy.
At the end of the file, the first line of the copy contains all the pertinent keys and if there are duplicate keys then any line in the file that contains such a key is deleted. This is achieved by moving the keys (the first line) to the end of the file and matching the second (and possibly third) fields of each record on those keys.