Print only first and last matching patterns - linux

Question

I am new to scripting and am learning as I go, I appreciate all and any help you can provide. I have a file with the following data:
0252 Fri 03 Jul 2015 84082679
0252 Fri 10 Jul 2015 81473945
0252 Fri 17 Jul 2015 87405062
0252 Fri 24 Jul 2015 89400396
0253 Fri 03 Jul 2015 29038894
0253 Fri 10 Jul 2015 29392107
0253 Fri 17 Jul 2015 31271055
0253 Fri 24 Jul 2015 31367348
071 Fri 03 Jul 2015 18594024
071 Fri 10 Jul 2015 18568430
071 Fri 17 Jul 2015 18648903
071 Fri 24 Jul 2015 18887643
072 Fri 03 Jul 2015 20141235
072 Fri 10 Jul 2015 19563727
072 Fri 17 Jul 2015 19573266
My desired output would look like:
0252 Fri 03 Jul 2015 84082679
0252 Fri 24 Jul 2015 89400396
0253 Fri 03 Jul 2015 29038894
0253 Fri 24 Jul 2015 31367348
071 Fri 03 Jul 2015 18594024
071 Fri 24 Jul 2015 18887643
072 Fri 03 Jul 2015 20141235
072 Fri 17 Jul 2015 19573266
The first column in the input data defines the "groups". From each group I want to print exactly two lines: the first line and the last line.
I would like to use awk to get my desired result, as i am trying to sort this information as final output. Any help is greatly appreciated, thank you.

Answer 1

Perl to the rescue!
perl -lane '
if ($F[0] eq $id) {
$keep = $_
} else {
$id = $F[0];
print $keep if defined $keep;
print
}
}{ print $keep
' < input.txt > output.txt
-n reads the input line by line
-a splits each line into the #F array
-l adds newline to print
$id is used to keep the value from the first column
$keep remembers the last line. When the $id changes, $keep and the current line are printed.
after the Eskimo greeting operator }{, the last line is printed once the whole file has been processed.

Answer 2

$ awk -v h=99 'h>$3{if (last) print last;print;} {h=$3;last=$0;} END{print last}' file
0252 Fri 03 Jul 2015 84082679
0252 Fri 24 Jul 2015 89400396
0253 Fri 03 Jul 2015 29038894
0253 Fri 24 Jul 2015 31367348
071 Fri 03 Jul 2015 18594024
071 Fri 24 Jul 2015 18887643
072 Fri 03 Jul 2015 20141235
072 Fri 17 Jul 2015 19573266
How it works
The script uses two variables: h and last. h is the value of the third field on the previous line and last is the text of the last line. Any decrease in h triggers printing.
-v h=99
Set initial value of h to a large number.
h>$3{if (last) print last;print;}
If h is larger than field 3, then print both the previous line (if there is one) and the current line.
h=$3;last=$0;
Update h and last.
END{print last}
Print the last line.

Answer 3

This might work for you (GNU sed):
sed -r '1p;N;/^(\S+\s+).*\n\1/D;2s/.*\n//' file
Always print the first line. Append the next line to the current line and compare the first field of the first with the first field of the second. It they are the same delete the first and repeat. Otherwise, print both lines but only the second if on line 2.

Answer 4

$ cat tst.awk
$1 != p1 { print p0 $0 }
{ p1 = $1; p0 = $0 ORS }
END { printf "%s", p0 }
$ awk -f tst.awk file
0252 Fri 03 Jul 2015 84082679
0252 Fri 24 Jul 2015 89400396
0253 Fri 03 Jul 2015 29038894
0253 Fri 24 Jul 2015 31367348
071 Fri 03 Jul 2015 18594024
071 Fri 24 Jul 2015 18887643
072 Fri 03 Jul 2015 20141235
072 Fri 17 Jul 2015 19573266