How to group a list by name with a newline after each group in shell? -

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i have list group name. should done newline after each group. here example file:

$ cat file 2015-07-09 07:03:46    7.5 gib apple-foo.txt.gpg 2015-07-22 11:36:36    6.9 gib apple-bar.txt.gpg 2015-07-27 04:40:34   31.0 gib banana-here.txt.gpg 2015-07-07 20:28:17   30.6 gib banana-even.txt.gpg 2015-07-19 15:02:20   30.8 gib banana-more.txt.gpg 2015-07-26 00:05:11    1.9 gib coconut-something.txt.gpg 2015-07-23 03:34:41    2.1 gib coconut-else.txt.gpg 2015-07-24 03:34:40   12.1 gib date-yougetit.txt.gpg

and output trying get:

2015-07-09 07:03:46    7.5 gib apple-foo.txt.gpg 2015-07-22 11:36:36    6.9 gib apple-bar.txt.gpg  2015-07-27 04:40:34   31.0 gib banana-here.txt.gpg 2015-07-07 20:28:17   30.6 gib banana-even.txt.gpg 2015-07-19 15:02:20   30.8 gib banana-more.txt.gpg  2015-07-26 00:05:11    1.9 gib coconut-something.txt.gpg 2015-07-23 03:34:41    2.1 gib coconut-else.txt.gpg  2015-07-24 03:34:40   12.1 gib date-yougetit.txt.gpg

i manage extract unique names (apple, banana, coconut, date) failing add new line after last occurrence of each unique name. able me out? awk , sed welcome.

an awk solution:

awk -f\- 'nr>1&&$1!=last{print ""}{last=$1}1' infile

explanation

-f\- :set field separator –.

nr>1 :omit first line checking.

last=$1 :always save last occurrence of group key.

1 :print current line.

1!=last{print ""} :if key $1 not equal last print separator.

update

for current source use:

awk 'split($nf,a,"-"){current=a[1]}nr>1&&current!=last{print ""}{last=current}1' infile

explanation 2

split($nf,a,"-"){current=a[1] :to key a[1] last field of line $nf splitting @ - char.


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