I have a tab delimited table for which I want to print all lines where column 'x' is greater than 'Y'. I have attempted using the code below but am new to using awk so am unsure how to use it based on columns.
awk '$X >= Y {print} ' Table.txt | cat > Wanted_lines
Y are values from 1 to 100.
If the input were like below with column X were the second column.
1 30
2 50
3 100
4 100
5 80
6 79
7 90
The wanted output would be:
3 100
4 100
5 80
7 90
The first 2 lines of the file is:
1 OTU1 243622 208679 121420 265864 0 0 2 0 0 11 1 5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 839604 OTU1 - Archaea 100% Euryarchaeota 100% Methanobacteria 100% Methanobacteriales 100% Methanobacteriaceae 100% Methanobrevibacter 100%
2 OTU2 84366 120817 15834 74737 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 295755 OTU2 - Archaea 100% Euryarchaeota 100% Methanobacteria 100% Methanobacteriales 100% Methanobacteriaceae 100% Methanobrevibacter 100%
First
awk's default internal field separator (FS) will work on space or tab delimited files.
Secondly
awk '$x > FLOOR' Table.txt
Where $x is the target column, and FLOOR is the actual numeric floor (i.e. 5000, etc ...)
Example file: awktest
500 100
400 1100
1000 400
1200 500
awk '$1 > 1000' awktest
1200 500
awk '$1 >= 1000' awktest
1000 400
1200 500
Thus, you should be able to use a relational expression to print the lines where x > y, in the form:
awk '$x > $y' awktest
Where $x is a numeric column as in $1, or other.
Where $y is a numeric column as in $2, or other.
Example:
awk '$1 > $2' awktest
or ...
awk '$2 > $1' awktest
awk numbers are floating point numbers, so you can compare decimals, too.
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