在R中创建一元运算符

时间:2018-01-11 13:48:28

标签: r

是否可以在R中创建一元运算符?我知道可以像这样创建二元运算符:

setGeneric("%power%", function(x, y) x ^ y)
2 %power% 4

但是可以创建像-这样的一元运算符。我试过像:

setGeneric("%-%", function(x) -x)
%-% 3

但它不起作用

2 个答案:

答案 0 :(得分:5)

R解析器不支持自定义一元运算符。

R language definition支持的运算符列表的副本:

-   Minus, can be unary or binary
+   Plus, can be unary or binary
!   Unary not
~   Tilde, used for model formulae, can be either unary or binary
?   Help
:   Sequence, binary (in model formulae: interaction)
*   Multiplication, binary
/   Division, binary
^   Exponentiation, binary
%x% Special binary operators, x can be replaced by any valid name
%%  Modulus, binary
%/% Integer divide, binary
%*% Matrix product, binary
%o% Outer product, binary
%x% Kronecker product, binary
%in%    Matching operator, binary (in model formulae: nesting)
<   Less than, binary
>   Greater than, binary
==  Equal to, binary
>=  Greater than or equal to, binary
<=  Less than or equal to, binary
&   And, binary, vectorized
&&  And, binary, not vectorized
|   Or, binary, vectorized
||  Or, binary, not vectorized
<-  Left assignment, binary
->  Right assignment, binary
$   List subset, binary

(解析器也支持二进制运算符:=,这里没有记录,因为它不被基数R使用。)

请注意,唯一的自定义运算符(&#34; %x%特殊二元运算符 x 可以替换为任何有效名称&#34;)是二进制。

因此,您唯一的选择是重载现有的一元运算符,分别为它编写方法。

答案 1 :(得分:1)

虽然我不熟悉setGeneric,但我可以回答这个问题

<块引用>

是否可以在 R 中创建一元运算符?

是的,有点,但不是真的。你可以伪造它:

# LET'S USE A BINARY OPERATOR TO DEFINE A UNARY OPERATOR:
# THE SYMBOL /*/<- IS SUPPOSED TO LOOK LIKE PERCENT-WILDCARD-PERCENT--ASSIGNMENT-ARROW
`%/*/<-%` <- function ( name , FUN , safe = TRUE ) {
    `%//%` <- paste0
    NAME <- "%" %//% name %//% "%"
    PARENT <- parent.frame ()
    if ( safe && exists ( NAME , PARENT ) )
            stop ( NAME %//% " exists." )
    assign (
        x = NAME ,
        value = function ( x , ignored ) FUN ( x ) ,
        envir = PARENT ) }

.. <- NULL            # THIS IS WHAT I MEAN BY FAKING IT...
"-"    %/*/<-%    `-` # ... `%-%` IS ACTUALLY A BINARY OPERATOR....
1 %-%..               # ... IN THIS CALL, `..` IS THE SECOND ARGUMENT.
# [1] -1
1 %-%.. %-%..
# [1] 1

"t"   %/*/<-%   `t` 
m <- matrix(1:4, 2)
m
#      [,1] [,2]
# [1,]    1    3
# [2,]    2    4
m %t%..
#      [,1] [,2]
# [1,]    1    2
# [2,]    3    4

"-"   %/*/<-%   `-` 
# Error in "-" %/*/<-% `-` : %-% exists.

i <- floor ( runif ( 9 , min = 1 , max = 10 ) )
i
# [1] 2 3 2 1 7 3 9 5 9
unique(i)
# [1] 2 3 1 7 9 5
"u"   %/*/<-%   function ( x ) sort ( unique ( x ) )
i %u%..
# [1] 1 2 3 5 7 9