Hi Andrea,
I'm also a newbie here. Let me try to help with your questions.
> 1 - Counting elements line by line.
>
> I'm looking for the number of prime factors of a list of
> numbers If I was going for a single number i'd do:
>
> > #q:8
> > 3
>
> which is perfect. But I can't do it for a list of numbers,
> as q:1+i.10 returns a matrix and # returns the number of
> rows. How can I get the number of prime factor line by line?
q:6 returns a list of rank 1, q:6 7 returns a matrix of rank 2.
# applied to the matrix count the rows; you have to apply # to
each row separately. # as monad has rank _, so you can change it:
#"1 q:1+i.10
3 3 3 3 3 3 3 3 3 3
So, # counts not only prime factors, but also fillings, which
were created when results were joined into the matrix. You have to
apply # before the results are joined. You can do it this way:
#...@q:1+i.10
0 1 1 2 1 2 1 3 2 2
This actually makes a single verb, #...@q: , which first does q: and
then does # with the result of q: . The rank of this verb is same
as the rank of q: .
> 2 - Zeroes matter when counting.
>
> How can I eliminate items with value 0 from a list?
> i.e. given 1 0 2 0 3 0 0 4 5 0 0
> 6 i'd like to obtain 1 2 3 4 5 6
>
This problem appears rather often in my experience. I've found the
following way to handle it: first I find out which elements are
not zeros and then fetch those elements from original array. Like
this:
t1 =. -. 0 = 1 0 2 0 3 0 0 4 5 0 0
t1 # 1 0 2 0 3 0 0 4 5 0 0
1 2 3 4 5
Or it can be expressed in tacit form:
(#~ -...@=&0) 1 0 2 0 3 0 0 4 5 0 0
1 2 3 4 5
Here =&0 - monad, which checks for equality to 0, -...@=&0 does the
same and then reverses the results - 0 makes 1, 1 makes 0, and #~
applies the # after swapping x and y arguments.
Alex
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