awk在列中打印特定数量的字符

Question

I have a file with many columns and rows and I want to remove the rows that are more than one character in the fourth and fifth columns.

Input:

--- 22:16050115:G:A 16050115 GGG A
--- 22:16050213:C:T 16050213 C T
--- 22:16050319:C:T 16050319 C T
--- 22:16050527:C:A 16050527 C AAA
--- 22:16050568:C:A 16050568 CC A
--- 22:16050607:G:A 16050607 G A
--- 22:16050627:G:T 16050627 G TGG
--- 22:16050646:G:T 16050646 G T
--- 22:16050655:G:A 16050655 GTAA A
...

Desired output:

--- 22:16050213:C:T 16050213 C T
--- 22:16050319:C:T 16050319 C T
--- 22:16050607:G:A 16050607 G A
--- 22:16050646:G:T 16050646 G T
...

Thank you very much.

Answer 1

awk 'length($4)==1 && length($5)==1' inputfile
--- 22:16050213:C:T 16050213 C T
--- 22:16050319:C:T 16050319 C T
--- 22:16050607:G:A 16050607 G A
--- 22:16050646:G:T 16050646 G T

This will check the length of $4 and $5 using length() function of awk . This is using comparison operator == . You can modify it to < , > , <= etc. So the above command will print the lines which have only one character in their 4th and 5th column.