AWK count occurrences of column A based on uniqueness of column B - linux

Question

I have a file with several columns and I want to count the occurrence of one column based on a second columns value being unique to the first column
For example:
column 10 column 15
-------------------------------
orange New York
green New York
blue New York
gold New York
orange Amsterdam
blue New York
green New York
orange Sweden
blue Tokyo
gold New York
I am fairly new to using commands like awk and am looking to gain more practical knowledge.
I've tried some different variations of
awk '{A[$10 OFS $15]++} END {for (k in A) print k, A[k]}' myfile
but, not quite understanding the code, the output was not what I've expected.
I am expecting output of
orange 3
blue 2
green 1
gold 1

Answer 1

With GNU awk. I assume tab is your field separator.
awk '{count[$10 FS $15]++}END{for(j in count) print j}' FS='\t' file | cut -d $'\t' -f 1 | sort | uniq -c | sort -nr
Output:
3 orange
2 blue
1 green
1 gold
I suppose it could be more elegant.

Answer 2

Single GNU awk invocation version (Works with non-GNU awk too, just doesn't sort the output):
$ gawk 'BEGIN{ OFS=FS="\t" }
NR>1 { names[$2,$1]=$1 }
END { for (n in names) colors[names[n]]++;
PROCINFO["sorted_in"] = "#val_num_desc";
for (c in colors) print c, colors[c] }' input.tsv
orange 3
blue 2
gold 1
green 1
Adjust column numbers as needed to match real data.
Bonus solution that uses sqlite3:
$ sqlite3 -batch -noheader <<EOF
.mode tabs
.import input.tsv names
SELECT "column 10", count(DISTINCT "column 15") AS total
FROM names
GROUP BY "column 10"
ORDER BY total DESC, "column 10";
EOF
orange 3
blue 2
gold 1
green 1