I have a comma',' FS filename as csv with n number of columns. I need to extract the unique value from colm.#1 with only corresponding values in colm.#10. So basically the column 10 is the date which is always unique for colm.#1 despite the other columns.
echo filename
colm.#1 colm.#2 colm.#3 colm.#4 colm.#5 colm.#6 colm.#7 colm.#8 colm.#9 colm.#10 colm.#11
a 231 412 30.84873962 3 1 1 2013 5/28/2013 6/6/2006 299
c 12 41 66.80690765 3 1 1 2014 5/25/2014 4/4/2004 351
d 35 6 25.91622925 3 1 2 2013 6/27/2013 3/3/2003 303
d 352 55 33.91288757 3 1 2 2014 6/26/2014 3/3/2003 355
a 86 3 30.58783722 3 1 3 2013 7/24/2013 6/6/2006 307
c 15 3242 26.6435585 3 1 3 2014 7/24/2014 4/4/2004 359
e 67 1 22.95526123 3 1 4 2013 8/21/2013 5/5/2005 311
a 464 64 4.804824352 3 1 4 2014 8/20/2014 6/6/2006 363
b 66 42 29.42435265 3 1 5 2014 9/18/2014 7/7/2007 367
m 24 2 66.10663319 3 1 6 2014 10/13/2014 9/9/2009 371
I tried the following command but it is only for colm#1 and I do not know how to get the corresponding value of the colm#10.
cut -d',' -f1 filename |uniq
O resultado esperado seria:
a 6/6/2006
b 7/7/2007
c 4/4/2004
d 3/3/2003
e 5/5/2005
m 9/9/2009
awk '{if ( ! ( $1 in Peers)) { Peers[$1]=$1 " " $10; print Peers[$1]} }' YourFile
isto leva a ordem como ela vem, se você precisar ordenar o resultado, um tipo em shel (entrada ou saída) ou (com o GNU awk)
awk '{if ( ! ( $1 in Peers)) Peers[$1]=$1 " " $10 } END{asort(Peers);for (Peer in Peers) print Peers[ Peer]}' YourFile