How to filter a list in J?

Question

I'm currently learning the fascinating J programming language, but one thing I have not been able to figure out is how to filter a list.

Suppose I have the arbitrary list 3 2 2 7 7 2 9 and I want to remove the 2s but leave everything else unchanged, i.e., my result would be 3 7 7 9. How on earth do I do this?

Forkey

Answer 1

The short answer

   2 (~: # ]) 3 2 2 7 7 2 9
3 7 7 9

The long answer

I have the answer for you, but before you should get familiar with some details. Here we go.

Monads, dyads

There are two types of verbs in J: monads and dyads. The former accept only one parameter, the latter accept two parameters.

For example passing a sole argument to a monadic verb #, called tally, counts the number of elements in the list:

   # 3 2 2 7 7 2 9
7

A verb #, which accepts two arguments (left and right), is called copy, it is dyadic and is used to copy elements from the right list as many times as specified by the respective elements in the left list (there may be a sole element in the list also):

   0 0 0 3 0 0 0 # 3 2 2 7 7 2 9
7 7 7

Fork

There's a notion of fork in J, which is a series of 3 verbs applied to their arguments, dyadically or monadically.

Here's the diagram of a kind of fork I used in the first snippet:

 x (F G H) y

      G
    /   \
   F     H
  / \   / \
 x   y x   y

It describes the order in which verbs are applied to their arguments. Thus these applications occur:

   2 ~: 3 2 2 7 7 2 9
1 0 0 1 1 0 1

The ~: (not equal) is dyadic in this example and results in a list of boolean values which are true when an argument doesn't equal 2. This was the F application according to diagram.

The next application is H:

   2 ] 3 2 2 7 7 2 9
3 2 2 7 7 2 9

] (identity) can be a monad or a dyad, but it always returns the right argument passed to a verb (there's an opposite verb, [ which returns.. Yes, the left argument! :)

So far, so good. F and H after application returned these values accordingly:

1 0 0 1 1 0 1
3 2 2 7 7 2 9

The only step to perform is the G verb application.

As I noted earlier, the verb #, which is dyadic (accepts two arguments), allows us to duplicate the items from the right argument as many times as specified in the respective positions in the left argument. Hence:

   1 0 0 1 1 0 1 # 3 2 2 7 7 2 9
3 7 7 9

We've just got the list filtered out of 2s.

Reference

Slightly different kind of fork, hook and other primitves (including abovementioned ones) are described in these two documents:

• A Brief J Reference (175 KiB)
• Easy-J. An Introduction to the World's most Remarkable Programming Language (302 KiB)

Other useful sources of information are the Jsoftware site with their wiki and a few mail list archives in internets.

Gony

J is truly an ingenious language. – Forkey

I took your example and have been playing around with it. For instance, to find all the numbers between 1 and 1000 that are evenly divisible by 3, I'd say 3 ((|=0:)#]) 1+i.1000 (with two powerful dyadic forks) and so on. Very useful! – Forkey

@yasir - I understand your desire for more rep, but this is really the wrong way to go about it. If you were the only editor, I'd suggest a flag to moderator or email to the team to reset the community wiki status, rather than adding a new answer to the question with the old answer's contents. – Surrey

@tv: ok ok, I'm newbie here, thanks, getting everything back, just didn't know about 8 max editions. :) – Lea

I don't have an un-wiki button, I'm afraid. It doesn't exist. – Fibrilla

Nice explanation! +1. Though (#~ ~:&2)3 2 2 7 7 2 9 looks easier for me. – Missie

Answer 2

Just to be sure it's clear, the direct way - to answer the original question - is this:

   3 2 2 7 7 2 9 -. 2

This returns

3 7 7 9

The more elaborate method - generating the boolean and using it to compress the vector - is more APLish.

To answer the other question in the very long post, to return the first element and the number of times it occurs, is simply this:

      ({. , {. +/ .= ]) 1 4 1 4 2 1 3 5
1 3

This is a fork using "{." to get the first item, "{. +/ . = ]" to add up the number of times the first item equals each element, and "," as the middle verb to concatenate these two parts.

Sokil

Answer 3

Also:

   2 ( -. ~ ]) 3 2 2 7 7 2 9
3 7 7 9

Yesseniayester

Actually, I think you can just say 2 -.~ 3 2 2 7 7 2 9. – Forkey

Answer 4

There are a million ways to do this - it bothers me, vaguely, that these these things don't evaluate strictly right to left, I'm an old APL programmer and I think of things as right to left even when they ain't.

If it were a thing that I was going to put into a program where I wanted to pull out some number and the number was a constant, I would do the following:

(#~ 2&~:)  1 3 2 4 2 5
1 3 4 5 

This is a hook sort of thing, I think. The right half of the expression generates the truth vector regarding which are not 2, and then the octothorpe on the left has its arguments swapped so that the truth vector is the left argument to copy and the vector is the right argument. I am not sure that a hook is faster or slower than a fork with an argument copy.

  +/3<+/"1(=2&{"1)/:~S:_1{;/5 6$1+i.6

156

This above program answers the question, "For all possible combinations of Yatzee dice, how many have 4 or 5 matching numbers in one roll?" It generates all the permutations, in boxes, sorts each box individually, unboxing them as a side effect, and extracts column 2, comparing the box against their own column 2, in the only successful fork or hook I've ever managed to write. The theory is that if there is a number that appears in a list of 5, three or more times, if you sort the list the middle number will be the number that appears with the greatest frequency. I have attempted a number of other hooks and/or forks and every one has failed because there is something I just do not get. Anyway that truth table is reduced to a vector, and now we know exactly how many times each group of 5 dice matched the median number. Finally, that number is compared to 3, and the number of successful compares (greater than 3, that is, 4 or 5) are counted.

This program answers the question, "For all possible 8 digit numbers made from the symbols 1 through 5, with repetition, how many are divisible by 4?"

I know that you need only determine how many within the first 25 are divisible by 4 and multiply, but the program runs more or less instantly. At one point I had a much more complex version of this program that generated the numbers in base 5 so that individual digits were between 0 and 4, added 1 to the numbers thus generated, and then put them into base 10. That was something like 1+(8$5)#:i.5^8 +/0=4|,(8$10)#. >{ ;/ 8 5$1+i.5 78125 As long as I have solely verb trains and selection, I don't have a problem. When I start having to repeat my argument within the verb so that I'm forced to use forks and hooks I start to get lost.

For example, here is something I can't get to work.

((1&{~+/)*./\(=1&{))1 1 1 3 2 4 1

I always get Index Error.

The point is to output two numbers, one that is the same as the first number in the list, the second which is the same as the number of times that number is repeated.

So this much works:

*./\(=1&{)1 1 1 3 2 4 1
1 1 1 0 0 0 0

I compare the first number against the rest of the list. Then I do an insertion of an and compression - and this gives me a 1 so long as I have an unbroken string of 1's, once it breaks the and fails and the zeros come forth.

I thought that I could then add another set of parens, get the lead element from the list again, and somehow record those numbers, the eventual idea would be to have another stage where I apply the inverse of the vector to the original list, and then use $: to get back for a recursive application of the same verb. Sort of like the quicksort example, which I thought I sort of understood, but I guess I don't.

But I can't even get close. I will ask this as a separate question so that people get proper credit for answering.

Dixie