Clash Royale CLAN TAG#URR8PPP

Distribute elements of one list over elements of another list

I have two lists:

l1:`a`b`c;
l2: til 20;


I am trying to create a dictionary 'd' that contains the elements of 'l1' as key and the elements of 'l2' evenly distributed over it. So like this:

d:(`a`b`c)!(0j, 3j, 6j, 9j, 12j, 15j, 18j;1j, 4j, 7j, 10j, 13j, 16j, 19j;2j, 5j, 8j, 11j, 14j, 17j)


The order of the elements is not relevant, I just need them balanced. I was able to achieve that in an iterative way (happy to add the code, if that's considered helpful), but there must be a more elegant way (potentially with adverbs?).

2 Answers
2

It can be done using the group :

group

q)group (count[l2]#l1)
(`a`b`c)!(0j, 3j, 6j, 9j, 12j, 15j, 18j;1j, 4j, 7j, 10j, 13j, 16j, 19j;2j, 5j, 8j, 11j, 14j, 17j)


If your l2 is something else instead of til 20 , then you have to lookup the items back after grouping :

l2

til 20

q)l2: 20#.Q.a
q)l2
"abcdefghijklmnopqrst"
q)l2 group (count[l2]#l1) // lookup the items back from l2 after grouping
(`a`b`c)!("adgjmps";"behknqt";"cfilor")


You can use the reshape functionality of the take operator #. It takes two arguments: a LHS of at least 2 dimensions and the list to reshape.

#

For example (3;4)#til 12 will reshape the list 0 1 ... 12 into a 3 by 4 matrix

(3;4)#til 12

0 1 ... 12

In our case, the number of the number of elements in l1 will will not necessary divide exactly into the number of elements in l2 (we don't want a rectangular matrix). Instead we can supply a null as the second dimension which will take care of distributing the remainders.

l1

l2

q) l1!(count[l1];0N)#l2
a| 0 1 2 3 4 5
b| 6 7 8 9 10 11 12
c| 13 14 15 16 17 18 19


This method performs very well for larger input lists.

As a side note, when using .Q.fc to split a vector argument over n slaves for multi-threading, kdb uses the # operator to reshape the vector into n vectors, one for each slave.

.Q.fc

n

#

n

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