How to cut first n and last n columns

Question

How can I cut off the first n and the last n columns from a tab delimited file   I tried this to cut first n column  But I have no idea to combine first and last n column  cut -f 1-10 -d   lt CTR gt v  lt TAB gt   filename

User · Accepted Answer

Cut can take several ranges in -f   Columns up to 4 and from 7 onwards     cut -f -4 7-   or for fields 1 2 5 6 and from 10 onwards   cut -f 1 2 5 6 10-   etc

User · Answer

You can cut using following   -d   delimiter  -f for fields  t used for tab separated fields  cut -d   t  -f 1-3 7-

User · Answer

you can use awk  for example  cut off 1st 2nd and last 3 columns  awk   for i 3 i lt  NF-3 i    print  i   file   if you have a programing language such as Ruby  1 9      ruby -F  t  -ane  print  F 2  -3  join   t    file

User · Answer

You can use Bash for that   while read -a cols  do echo   cols    0 1    cols    1 -1   done  lt  file txt

User · Answer

Try the following   echo a b c   awk -F       1        NF       print   OFS

User · Answer

Use   cut -b COLUMN N BEGINS-COLUMN N UNTIL INPUT TXT  gt  OUTPUT TXT   -f doesn t work if you have  tabs  in the text file

User · Answer

To use AWK to cut off the first and last fields   awk    1        NF       print   inputfile   Unfortunately  that leaves the field separators  so  aaa bbb ccc   becomes   space bbb space    To do this using kurumi s answer which won t leave extra spaces  but in a way that s specific to your requirements   awk   delim       for  i 2 i lt  NF-1 i     printf delim   s    i  delim   OFS   printf   n    inputfile   This also fixes a couple of problems in that answer   To generalize that   awk -v skipstart 1 -v skipend 1   delim       for  i skipstart 1 i lt  NF-skipend i     printf delim   s    i  delim   OFS   printf   n    inputfile   Then you can change the number of fields to skip at the beginning or end by changing the variable assignments at the beginning of the command

User · Answer

The first part of your question is easy  As already pointed out  cut accepts omission of either the starting or the ending index of a column range  interpreting this as meaning either    from the start to column n  inclusive     or    from column n  inclusive  to the end     respectively     printf  this is a test    cut -d  -f-2 this is   printf  this is a test    cut -d  -f3- a test   It also supports combining ranges  If you want  e g   the first 3 and the last 2 columns in a row of 7 columns     printf  foo bar baz qux quz quux quuz    cut -d  -f-3 6- foo bar baz quux quuz   However  the second part of your question can be a bit trickier depending on what kind of input you   re expecting  If by    last n columns    you mean    last n columns  regardless of their indices in the overall row      i e  because you don   t necessarily know how many columns you   re going to find in advance  then sadly this is not possible to accomplish using cut alone  In order to effectively use cut to pull out    the last n columns    in each line  the total number of columns present in each line must be known beforehand  and each line must be consistent in the number of columns it contains   If you do not know how many    columns    may be present in each line  e g  because you   re working with input that is not strictly tabular   then you   ll have to use something like awk instead  E g   to use awk to pull out the last 2    columns     awk calls them fields  the number of which can vary per line  from each line of input     printf   a n a b n a b c n a b c d n    awk -F    print   NF-1  FS   NF     a a b b c c d

[linux] How to cut first n and last n columns?

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