Count Occurrences as a Dictionary

Question

Given the following list:

l:`a`a`a`a`a`a`b`c`c`d`e`e`e`e

Count the number of occurrences of each element in l. Return the result as a dictionary.

Expected result:

a| 6
b| 1
c| 2
d| 1
e| 4

Solution

count each group l

Alternative Solution

(distinct l)!sum each (distinct l)=\:l

Alternative Solution

exec count i by l from ([] l)

Explanation

group l returns a dictionary mapping each distinct value to the list of indexes where it occurs:

q)group l
a| 0 1 2 3 4 5
b| ,6
c| 7 8
d| ,9
e| 10 11 12 13

Applying the each iterator:

'

to count runs count independently against every value of that dictionary, turning each list of indexes into its length — giving the occurrence count per key, while keeping the original keys untouched.

The first alternative builds the same result manually: (distinct l)=\:l pairs each distinct value against the whole list using each-left, producing one boolean mask per distinct value, and sum each counts the 1s in each mask.

The second alternative uses qSQL's by clause to group and count directly.