# Vocabulary/at

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`[x] u@v y`Atop Conjunction

Rank *-- depends on the rank of v --*
WHY IS THIS IMPORTANT?

Forms the *composition* (`u@v`) of verbs `u` and `v` .

The resulting verb is applied **independently to each cell of the argument(s)**.

- 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| +-------+

The result of `u@v` is a tacit verb equivalent to ` [x] (u@:v)"v y`

`[x]` is used to signify that the phrase holds good whether `x` is present or not.

See: **More Information** for a visual comparison of At (`@:`), **Atop** (`@`), Compose (`&`) and Appose (`&:`).

### 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.

### 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` .

**Note:** unlike At (`@:`), the *rank* of **Atop** (`u@v`) depends on the ranks of `u` and `v`.

2. The difference between At (`u@:v`) and **Atop** (`u@v`) is shown in the **first two** columns of the diagram below.

The above diagram also shows Appose (`&:`) and Compose (`&`) for overall comparison.

3. 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 thedyadsare 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.