Awk to print lines in file with multiple delimiters

Question

I have a file that looks like:

chr1    1197592 .   C   A   .   .   DP=67;ECNT=1;NLOD=8.12  GT:AD:AF:F1R2   0/1:37,2:0.063:13,0
chr1    1355707 .   G   T   .   .   DP=69;ECNT=1;NLOD=4.51  GT:AD:AF:F1R2   0/1:50,3:0.059:20,3
chr1    1641723 .   TC  T   .   .   DP=59;ECNT=1;NLOD=2.40  GT:AD:AF:F1R2   0/1:30,2:0.089:12,4
chr1    2550056 .   TC  CT  .   .   DP=99;ECNT=1;NLOD=9.03  GT:AD:AF:F1R2   0/1:63,2:0.053:33,2

I would like to print the lines for which the 3rd value in the 10th column is greater than 0.06.

cat file.txt | cut -f 10 | cut -f 3 | awk -F':' '$3>0.06'

Will only give me:

0/1:37,2:0.063:13,0
0/1:30,2:0.089:12,4

And I would like the entire line. Can that be done with awk?

Answer 1

As soon as you start discarding data in a pipeline (which is what cut does), you can't get it back in a later stage of the same pipeline.

Instead,

$ awk 'split($10,a,":") && a[3] > 0.06' file
chr1    1197592 .   C   A   .   .   DP=67;ECNT=1;NLOD=8.12  GT:AD:AF:F1R2   0/1:37,2:0.063:13,0
chr1    1641723 .   TC  T   .   .   DP=59;ECNT=1;NLOD=2.40  GT:AD:AF:F1R2   0/1:30,2:0.089:12,4

This takes the 10th whitespace-delimited field and splits it on colons into the array a. It prints the original line if the split() generated any elements in a and if the third element of a is greater than 0.06.

If the fields in the file are tab-delimited and if any field contains spaces, then use -F '\t' with awk to make sure that it finds the right fields correctly (the given example data does not have any such issues as far as I can see).

---

Addressing the follow-up question in comments:

awk 'split($10,a,":") && a[3] > 0.06 && split($11,b,":") && b[3] > 0.01' file

Answer 2

awk -F':' '$(NF-1) > 0.06 {print $0}' file

Just use ':' as separator and evaluate the penultimate column

Answer 3

Having read the comments it's clear that your sample data doesn't match your actual data (and it's using spaces instead of TABs to begin with, which means that even the command output of your own device won't work), but having clobbered together a different input file like so:

cat file.txt
chr1    1197592 .       C       A       .       .       DP=67;ECNT=1;NLOD=8.12  GT:AD:AF:F1R2   0/1:37,2:0.063:13,0     0/1:37,2:0.0005:13,0
chr1    1355707 .       G       T       .       .       DP=69;ECNT=1;NLOD=4.51  GT:AD:AF:F1R2   0/1:50,3:0.059:20,3     0/1:50,3:0.0005:20,3
chr1    1641723 .       TC      T       .       .       DP=59;ECNT=1;NLOD=2.40  GT:AD:AF:F1R2   0/1:30,2:0.089:12,4     0/1:30,2:0.0015:12,4
chr1    2550056 .       TC      CT      .       .       DP=99;ECNT=1;NLOD=9.03  GT:AD:AF:F1R2   0/1:63,2:0.053:33,2     0/1:63,2:0.0005:33,2

This will then output all lines matching both conditions:

awk -F"[  :]" '$15>0.06 && $19>0.001' file.txt
chr1    1641723 .   TC  T   .   .   DP=59;ECNT=1;NLOD=2.40  GT:AD:AF:F1R2   0/1:30,2:0.089:12,4 0/1:30,2:0.0015:12,4

Answer 4

I have used below method to achieve the result

for  i in `awk  '{print $NF}' file.txt | awk -F ":" '$3>"0.06"{print $0}'`; do awk -v i="$i" '$NF ==i{print $0}' file.txt ;done

output

chr1    1197592 .   C   A   .   .   DP=67;ECNT=1;NLOD=8.12  GT:AD:AF:F1R2   0/1:37,2:0.063:13,0
chr1    1641723 .   TC  T   .   .   DP=59;ECNT=1;NLOD=2.40  GT:AD:AF:F1R2   0/1:30,2:0.089:12,4
[root@praveen_linux_example ~]#