#!/bin/ksh
# Final component of script name.
IAm=${0##*/}

# Absolute pathname of control file.
CF='/some/dir/control.status'

# Absolute pathname of directory containing the *.pnt files to be procssed.
DataDir='/some/same_or_other/directory'

if ! cd "$DataDir"
then	exit 1
fi
if ! read BaseYear BaseMonth < "$CF"
then	exit 2
fi
BaseFile="$BaseYear$BaseMonth.pnt"
if [ ! -r "$BaseFile" ]
then	printf "%s: Can't read base file (%s).\n" "$IAm" "$DataDir/$BaseFile" >&2
	exit 3
fi

sort -bn -k1,1 -k2,2 -k3,3 *"$BaseMonth.pnt" | awk '
# Set output field separator to <tab>.
BEGIN {	OFS = "\t"
}

# Function to print a group of elements that all have identical values in the
# first and second input fields.
function print_group() {
	# Check to see if we have data to process...
	if(cnt) {
		# Look for the 1st change in values after the mid-point for
		# this set group.
		for(i = int((cnt + 1) / 2) + 1; i <= cnt; i++)
			if(d[i] != d[i - 1])
				break
		# For each set of duplicate values after the midpoint, reset
		# the rank for those points to the end of the set instead of
		# the start of the set.
		while(i < cnt) {
			if(c[i] > 1)
				for(j = i; j <= i + c[i] - 1; j++)
					r[j] += c[i] - 1
			i += c[i]
		}
		# Print the data and rank for each element of the set in the
		# base file.
		for(i = 1; i <= cnt; i++)
			if(d[i] in P) {
				print d[i], r[i]
				delete P[d[i]]
			}
	}
	# Reset variables for next group.
	cnt = con = 0
}

# Gather points to process from 1st input file...
FNR == NR {
	# Gather data from the base file (given as first file operand)...
	# Gather list of point pairs to be processed.
	L[$1 OFS $2]

	# Gather list of points and value triples to be printed.
	P[$1 OFS $2 OFS $3]
	next
}

# Skip points not found in the 1st input file...
!(($1 OFS $2) in L) {
	next
}

# Look for a change in the first two input fields...
$1 != l1 || $2 != l2 || NR == 1 {
	# We have found a change in values.  Print the results from the
	# previous group, if there was one.
	print_group()

	# Note first two field values so we notice the next change.
	l1 = $1
	l2 = $2

	# Clear the remembered 3rd field value to prevent contamination from
	# the previous group.
	l3 = ""
}

# Gather data for this group...
{	# Save the data for this line.
	d[++cnt] = $1 OFS $2 OFS $3

	# Calculate the rank for this line.  (At this point, we do not know
	# what the midpoint will be for this group, so all of these are saved
	# with the rank being the lowest rank for the set of lines with
	# identical third field values.  The group_print() function wll make
	# adjustments for sets of ranks after the midpoint in the group.)
	if($3 != l3 || cnt == 1) {
		# A change in field 3 values has been found.  Save the value
		# and rank for this set.
		l3 = $3
		lr = cnt
		# Clear the count of the consecutive number of lines with the
		# same value.
		con = 0
	} 

	# Set the rank for this line.
	r[cnt] = lr

	# Set number of consecutive lines that have this third field value.
	for(i = cnt - con++; i <= cnt; i++)
		c[i] = con
}

# We have found EOF.
END {	# Print the data for the last group.
	print_group()
}' "$BaseFile" -

				delete P[d[i]]

				break

24.5625	-81.8125	39.16	1
24.5625	-81.7708	40.81	1
24.5625	-81.7292	46.73	1
24.5625	-81.6875	52.67	1
24.6042	-81.6458	62.22	2
24.6458	-81.5625	66.18	2
24.6458	-81.4792	68.19	2
24.6875	-81.5625	67.32	2
24.6875	-81.3958	71.72	2
24.7292	-81.3958	73.26	2
24.7708	-80.9375	90.29	2
25.1458	-81.1042	116.34	1
25.1458	-81.0625	117.04	1
25.1458	-81.0208	119.01	1
25.1458	-80.9792	118.53	1
25.1458	-80.9375	118.07	1
25.1458	-80.7708	142.98	1
25.1458	-80.7292	149.23	2
25.1458	-80.4375	171.91	2
25.1458	-80.3958	172.67	2
25.1875	-81.1042	122.42	1
25.1875	-81.0625	125.46	1
25.1875	-81.0208	125.53	1
25.1875	-80.9792	125.67	1
25.1875	-80.9375	127.46	1
25.1875	-80.8958	130.94	1