bash print unique entries in specific column

# Basic syntax:
awk -F'delimiter' '!a[$column_number]++' input_file
# Where:
#	- The column_number is the column you want to use for checking
#		if a line is unique
#	- The delimiter is the delimiter used in the input_file. Note that
#		awk recognizes spaces, tabs, and new lines as delimiters by 
#		default, so this can be omitted if your file uses one of those.

# Example usage 1:
# Say you have a comma-delimited file with the following rows and only
#	want one copy of the row based on the third field
10,15-10-2014,abc
20,12-10-2014,bcd
10,09-10-2014,def
40,06-10-2014,ghi
10,15-10-2014,abc # Looking only at field 3, this row duplicates row 1

# Running:
awk -F',' '!a[$3]++' input_file 
# Would return:
10,15-10-2014,abc
20,12-10-2014,bcd
10,09-10-2014,def
40,06-10-2014,ghi
# The second instance of abc in the third field/column is ignored. 

# Note, this only returns the first line that has a duplicate entry in
#	your specified column.

# Example usage 2:
# Say you have a tab-delimited file with the following rows and only
#	want one copy of each row based on the first AND fourth fields
a	11	b	17
c	23	d	27
a	31	e	17 # Looking only at fields 1 and 4 this row duplicates row 1
a	17	f	42

# Running:
awk '!a[$1$4]++' input_file
# Would return:
a	11	b	17
c	23	d	27
a	17	f	42

# Short answer:
# An easy way to do this is with an awk array/dictionary. 
# Google: "bash print keys of awk array" for the Grepper answer on this