Vocabulary/at
>> << Back to: Vocabulary Thru to: Dictionary
[x] u@v y [x] u@n y Atop Conjunction
Rank -- depends on the rank of v -- WHY IS THIS IMPORTANT?
u@v is equivalent to (u@:v)"v. That is,
u@v y <--> {{u (v y)}}"mv y
x u@v y <--> x {{u (x v y)}}"(lv,rv) y
where mv, lv, and rv are the monadic, left, and right ranks of v, respectively. You should understand Rank (") and At (@:) before trying to understand Atop (@) .
Forms the composition of verbs u and v; the resulting verb is applied independently to each v-cell of the argument(s).
This contrasts with u@:v which applies v to the entire argument(s) and then applies u on the entire filled and assembled result of v .
- Operand u is always executed monadically on the result of each application of v
- Operand v is executed either monadically or dyadically depending whether u@v has been called monadically or dyadically.
#@> 'Newton';'Einstein' 6 8 2 3 <@, 4 5 +-------+ |2 3 4 5| +-------+
See: More Information for a visual comparison of At (@:), Atop (@), Compose (&) and Appose (&:).
Noun right operand (u@n)
If the right operand of @ is a noun n, the resulting verb u@n is equivalent to u@(n"_) , where the constant verb (n"_) ignores its argument(s) and produces the result n , which in turn becomes the argument to u.
The resulting verb ignores its argument(s) and is customarily called with an empty array argument, e.g. ''.
greet=: echo@'hello' greet'' hello 100 i.@4 'apple' 0 1 2 3
".@'name' is a good way to take the value of the noun name at the point of execution rather than the point of parsing.
Common uses
Implement: f(g(x)) -- the mathematical composition of the two functions: f and g.
mean=: +/ % # cat=: ,&1"1 ]z=: i.2 3 NB. sample noun 0 1 2 3 4 5 cat z NB. appends 1 to each row of z 0 1 2 1 3 4 5 1 mean@cat z NB. mean of the ROWS of cat z 1 3.25 mean@cat b.0 NB. rank of (mean@cat) 1 1 1
You can also implement: f(g(x)) with: Compose (&) or Cap ([:) or At (@:).
But see Rank in a hurry: an insidious rank problem for how and when these different methods give different results.
Should I use @ or @:?
Suppose your y is a list of boxes and you want to create a list of the number of items in the contents of each box. You want to apply # individually on the contents of each box, so you use @ to get
#@> 4 5 6 7;1 2;3 4 2 1
If you used @:, the interpreter would open all the boxes and join them together, then applying # just once on the assembled result:
#@:> 4 5 6 7;1 2;3 3
Suppose your y is a list of numbers and you want to total of the squares of all the numbers. You want to apply +/ to all the squares, so you use @: to get
+/@:*: 3 4 5 50
If you used @, +/ would be applied on each result item of *: - and because the rank of *: is 0, you would be applying +/ on individual atoms:
+/@* 3 4 5 9 16 25
Related Primitives
At (@:), Compose (&), Appose (&:), Hook ((u v)), Fork ((f g h))
More Information
1. Phrase (u@:v)"v means "apply v followed by u on each cell of the operand(s) independently". The rank of a cell is given by the rank of the verb v . The results from the cells are collected to produce the overall result of u@v .
An illustration of this explanation follows, extracted from the series of diagrams referenced below:
Atop (u@v) , At (u@:v) , Appose (&:) , and Compose (&) are also visualized along with verb application and rank here: a series of flow diagrams
2. So what's the difference between Atop (@) and Compose (&) ?
None at all, for the monads (u@v) and (u&v)
u&v y ↔ u v y u@v y ↔ u v y
But the dyads are different:
x u&v y ↔ (v x) u (v y) x u@v y ↔ u x v y
According to the J Dictionary -- &: is equivalent to & except that the ranks of the resulting function are infinite; the relation is similar to that between @: and @
Oddities
The J Dictionary's definition of [x] u@v y as equivalent to u [x] v y has caused much confusion, because the statement must be read in the context of the rank of v .
The precise definition is as above.