Steven,

You can do it with assign() if you keep the names when you put the items in
the list:

Dlist <- list(D1=D1, D2=D2) # put the names of the objects in the list
Newlist <- lapply(Dlist, function(x) x[, columns]) # create a new list with
the output

for(i in seq(length(Newlist))) {
  assign(names(Newlist)[i],Newlist[[i]]) # assign the new objects to the
original names
}

You may want to keep in mind, though:

> fortune(236)

The only people who should use the assign function are those who fully
understand why you should never use the assign function.
   -- Gregory L. Snow
      R-help (July 2009)

You might want to ask yourself whether this is really the best way to
achieve what you want to do.

Jonathan


On Thu, Jul 15, 2010 at 7:18 PM, Steven Kang <stochastickang@gmail.com>wrote:

> Hi all,
>
> There are matrices with same column names but arranged in different orders
> and I desire columns of these matrices to have same order.
>
> For example, below are 2 arbitrary data sets with columns arranged in
> different order. I require columns of these to have same order as specified
> in "columns" object and the results stored in the original object names.
> I know this can be done simply by:
> D1 <- D1[, columns]
>
> But if there are hundreds of matrices, then more efficient method is
> required.
>
> columns <- c("A", "B", "C")
> D1 <- matrix(rnorm(6), nrow = 2, dimnames = list(c("R1", "R2"), c("C", "A",
> "B")))
> D2 <- matrix(rnorm(6), nrow = 2, dimnames = list(c("R1", "R2"), c("C", "B",
> "A")))
>
> > D1
>                    C                            A                        B
> R1 -0.653978178594122 -0.15910510749630 0.90507729153852
> R2  0.015557641181675 -0.73944224596032 0.23484927168787
> > D2
>                  C                           B                            A
> R1 0.18843559757623 0.207589297797905 -0.018884844424975
> R2 1.87387725184456 0.050349118287824 -1.796404635019739
>
> Dlist <- list(D1, D2)
> lapply(Dlist, function(x) x[, columns])
>
> [[1]]
>                   A                           B                          C
> R1 -0.15910510749630 0.90507729153852 -0.653978178594122
> R2 -0.73944224596032 0.23484927168787  0.015557641181675
> [[2]]
>                    A                          B                          C
> R1 -0.018884844424975 0.207589297797905 0.18843559757623
> R2 -1.796404635019739 0.050349118287824 1.87387725184456
> How can the results from the lapply function be stored back into the
> original object names?
>
> Many thanks.
>
>
>
> --
> Steven
>
>        [[alternative HTML version deleted]]
>
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>

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