I have a tab-delimited text file in the following format.
Col1 | Col2 | Col3 123.0 | 534.2 | Blah0 2031/23/12 23.00 | 786.2 | Blah1 2033/01/01 12.40 | 343.0 | Blah2 2031/27/11 I need to remove all the characters after the space from the last column. So my output would be
Col1 | Col2 | Col3 123.0 | 534.2 | Blah0 23.00 | 786.2 | Blah1 12.40 | 343.0 | Blah2 How should I go about this using Awk or something similar?
With awk:
awk -F '\t' 'BEGIN { OFS = FS } NR != 1 { sub(/ [^ ]*$/, "", $NF) } 1' filename That is:
BEGIN { OFS = FS } # the output should be separated the same way as # the input NR != 1 { # in all lines except the header line: sub(/ [^ ]*$/, "", $NF) # replace the last space and everything after it } # in the last field ($NF) with the empty string # (i.e., remove it) 1 # in all lines: print. If there can be several spaces in the last field and you want to remove everything after the first space, use sub(/ .*/, "", $NF) instead. It wasn't entirely clear in the question what should happen in such a case.
awk code will trim Col3 from the header.
(rev | cut -d ' ' -f 2-7 | rev) < file?-f 2- instead of -f 2-7, but what if there's a line in which the last field doesn't contain a space at all? It's tabular data; awk was pretty much made for this. (Also, I disagree that awk is not unix-like, considering that it is part of POSIX :P)