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R中的高级函数-是否有正式的复合运算符或curry函数?

higher-order-functions functional-programming r

Alex Brown · 技术社区 · 14 年前

我可以在r中创建compose操作符:

 `%c%` = function(x,y)function(...)x(y(...))

使用方法如下:

 > numericNull = is.null %c% numeric
 > numericNull(myVec)
 [2] TRUE FALSE

但是我想知道是否有一组正式的函数来做这类事情以及其他操作,比如在r中进行货币化,这在很大程度上是为了减少代码中括号、函数关键字等的数量。

我的咖喱功能:

> curry=function(...){
    z1=z0=substitute(...);z1[1]=call("list");
    function(...){do.call(as.character(z0[[1]]),
                          as.list(c(eval(z1),list(...))))}}
> p = curry(paste(collapse=""))
> p(letters[1:10])
[1] "abcdefghij"

这尤其适用于骨料:

> df = data.frame(l=sample(1:3,10,rep=TRUE), t=letters[1:10])
> aggregate(df$t,df["l"],curry(paste(collapse="")) %c% toupper)
  l    x
1 1  ADG
2 2  BCH
3 3 EFIJ

我发现它比:

> aggregate(df$t, df["l"], function(x)paste(collapse="",toupper(x)))
  l    x
1 1  ADG
2 2  BCH
3 3 EFIJ

基本上我想知道-这已经为R做了吗?

4 回复 | 直到 6 年前

elviejo79 9 年前

functional

   library(functional)
   newfunc <- Curry(oldfunc,x=5)

https://cran.r-project.org/web/packages/functional/index.html

ROxigen

Shane 14 年前

the roxygen package see the source code here Byron Ellis's solution on R-Help

Curry <- function(FUN,...) {
  .orig = list(...);
  function(...) do.call(FUN,c(.orig,list(...)))
}

Compose <- function(...) {
  fs <- list(...)
  function(...) Reduce(function(x, f) f(x),
                       fs,
                       ...)
}

Reduce Map Filter

> library(roxygen)
> p <- Curry(paste, collapse="")
> p(letters[1:10])
[1] "abcdefghij"

Compose

> Compose(function(x) x[length(x):1], Curry(paste, collapse=""), toupper)(letters)
[1] "ZYXWVUTSRQPONMLKJIHGFEDCBA"

> aggregate(df[,"t"], df["l"], Compose(Curry(paste, collapse=""), toupper))
  l    x
1 1  ABG
2 2 DEFH
3 3  CIJ

plyr by aggregate

> library(plyr)
> ddply(df, .(l), function(df) paste(toupper(df[,"t"]), collapse=""))
  l   V1
1 1  ABG
2 2 DEFH
3 3  CIJ

Richie Cotton Joris Meys 11 年前

roxygen
this conversation

cmc 6 年前

plot(rnorm(1000),rnorm(1000)) data.frame

> data.frame( rnorm(5), rnorm(5), first=rpois(5,1), second=rbinom(5,1,0.5) )
    rnorm.5. rnorm.5..1 first second
1  0.1964190 -0.2949770     0      0
2  0.4750665  0.8849750     1      0
3 -0.7829424  0.4174636     2      0
4  1.6551403  1.3547863     0      1
5  1.4044107 -0.4216046     0      0

Curry <- function(FUN, ...) {
    .orig = match.call()
    .orig[[1]] <- NULL # Remove first item, which matches Curry
    .orig[[1]] <- NULL # Remove another item, which matches FUN
    function(...) {
        .inner = match.call()
        .inner[[1]] <- NULL # Remove first item, which matches Curry
        do.call(FUN, c(.orig, .inner), envir=parent.frame())
    }
}

match.call ... FUN Curry

.orig .orig[[1]]<-NULL

Curry(data.frame, rnorm(5), rnorm(5) )( first=rpois(5,1) , second=rbinom(5,1,0.5) )

envir=parent.frame()