Re: [R] R lappy, sapply or mapply question

[Frank Schwidom]

On 2015-10-09 20:15:16, liqunhan--- via R-help wrote:

> for (k in 1 : 5) {
>   xlist <- list(x5 = dailyrecord$a[k], x6 = dailyrecord$e[k], x7 = 
> dailyrecord$f[k])
>   result_forloop[k] <- fun3(list1, list2, xlist)
> }

result_forloop <- lapply( 1 : 5, function( k) {

 tmpRow <- dailyrecord[ k, ]

 xlist <- with( df, list( x5= a, x6= e, x7= f))

 fun3(list1, list2, xlist)

})

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Re: [R] R lappy, sapply or mapply question

[Frank Schwidom]

correction:

On 2015-10-10 16:08:39, Frank Schwidom wrote:
> On 2015-10-09 20:15:16, liqunhan--- via R-help wrote:
> 
> > for (k in 1 : 5) {
> >   xlist <- list(x5 = dailyrecord$a[k], x6 = dailyrecord$e[k], x7 = 
> > dailyrecord$f[k])
> >   result_forloop[k] <- fun3(list1, list2, xlist)
> > }
> 
> result_forloop <- lapply( 1 : 5, function( k) {
> 
>  tmpRow <- dailyrecord[ k, ]
> 
>  xlist <- with( tmpRow, list( x5= a, x6= e, x7= f))
> 
>  fun3(list1, list2, xlist)
> 
> })
> 
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

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Re: [R] R lappy, sapply or mapply question

[Adams, Jean]

You were very close.  Try this.

df <- data.frame(x5=dailyrecord$a, x6 = dailyrecord$e, x7 = dailyrecord$f)
apply(df, 1, function(row) fun3(list1, list2, as.list(row)))

Jean

On Fri, Oct 9, 2015 at 3:15 PM, liqunhan--- via R-help  wrote:

>
>
> Hello, R-experts,
> In R-program, I have a question about the apply-family.
> I want to use apply-family to replace a for-loop in my R-code,But, lapply
> returns a list of 3 (each component is the same), sapply returns a matrix,
> and mapply with error message.
> how to use apply-family function so that it returns a vector, as it does
> when using for-loop in my R-codes below?
> Hope to hear back soon!
> Thank you very much!
>
>
>
> #-
> # Below is my R-codes:
> #---  begin of  R code --
> # suppose list1 returned by fun1()
> # fun1() is a R-script with about 200 lines
> list1 <- list(u=3.8, v=53.42)# suppose list2 returned by fun2()
> # fun2() is a R-script with about 5000 lines
> list2 <- list(x=3.8, y=-9,3)# demo fun3(), the actual function is much
> more complicated
> fun3 <- function(xlist1, xlist2, xlist3) {
>
>   x1 <- xlist1$u
>   x2 <- xlist1$v
>
>   x3 <- xlist2$x
>   x4 <- xlist2$y
>
>   x5 <- xlist3$x5
>   x6 <- xlist3$x6
>   x7 <- xlist3$x7
>
>   w <- x1^2 + sqrt(x2+x3) - 0.75*x4 + exp(x5)
>   z <- sin(x2)/x7 + x6/(x3+x4)
>   return(w+z)
> }dailyrecord <- data.frame(a = rnorm(5), b = rnorm(5), c =
> rnorm(5),
>   d = rnorm(5), e = rnorm(5), f =
> rnorm(5),
>   g = rnorm(5))
> result_forloop <- rep(0, 5)
> # use for - loop ## how to avoid the for-loop ??for (k in 1 : 5) {
>   xlist <- list(x5 = dailyrecord$a[k], x6 = dailyrecord$e[k], x7 =
> dailyrecord$f[k])
>   result_forloop[k] <- fun3(list1, list2, xlist)
> }# use lapply  #--- return a list of 3 
> xlst <- list(x5=dailyrecord$a, x6 = dailyrecord$e, x7 = dailyrecord$f)
> result_lapply <- lapply(xlst, function(s) fun3(list1, list2, xlst))
>
> # use sapply  #--- return a matrix  ---
> result_sapply <- sapply(xlst, function(s) fun3(list1, list2, xlst))
>
> # use mapply  #--- error  ---
> result_mapply <- mapply(fun3, xlist1 = list1, xlist2 = list2, xlist3 =
> xlst)
>
> #---  end of R code --
>
>
> [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.

[[alternative HTML version deleted]]

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