I have a text file containing :
A 25 27 50
B 35 75
C 75 78
D 99 88 76
I wanted to delete the line that does not have the fourth field(the fourth pair of digits). Expected output :
A 25 27 50
D 99 88 76
I know that awk command would be the best option for such task, but i'm wondering what's the problem with my sed command since it should work as you can see below :
sed -E '/^[ABCD] ([0-9][0-9]) \1$/d' text.txt
Using POSIX ERE with back-referencing (\1) to refer to the previous pattern surrounded with parenthesis.
I have tried this command instead :
sed -E '/^[ABCD] ([0-9][0-9]) [0-9][0-9]$/d' text.txt
But it seems to delete only the first occurrence of what i want. I would appreciate further explanation of,
Santiago Trujillo
Much much easier with awk:
awk 'NF == 4' file
A 25 27 50
D 99 88 76
This awk command uses default field separator of space or tab and checks a condition NF == 4 to make sure we print lines with 4 fields only.
With sed it would be (assuming no leading+trailing spaces in each line):
sed -nE '/^[^[:blank:]]+([[:blank:]]+[^[:blank:]]+){3}$/p' file
A 25 27 50
D 99 88 76
With your shown samples in sed program you could try following. Written and tested in GNU sed.
sed -nE '/^([^[:space:]]+[[:space:]]+){3}[^[:space:]]+$/p' Input_file
Explanation: Simply stopping the printing for lines by sed's -n option. Then using -E for using ERE in program. In main program using regex to match from starting non-space(1 or more occurrences) followed by spaces(1 or more occurrences) and this combo 3 times(to match 3 fields basically) which is followed by non spaces 1 or more occurrences till end of line's value, if this regex matched then print that line.
This might work for you (GNU sed):
sed -En 's/\S+/&/4p' file
Turn off implicit printing -n and on extended regexp -E.
Substitute the 4th field with itself and print the result.