[R] apply one list to another one
David Winsemius
dwinsemius at comcast.net
Fri May 4 04:51:24 CEST 2012
On May 3, 2012, at 7:39 PM, Hui Du wrote:
>
> Hi All,
>
> Suppose I have the following codes:
>
> x = data.frame(A = rnorm(20), B = rnorm(20), C = rnorm(20))
> a = list()
> a[["1.1"]] = x
> a[["1.2"]] = x
>
> b = list()
> b[["1.1"]] = c("A", "B")
> b[["1.2"]] = c("B", "C")
>
>
> Now I want to apply b to a like this, for each element of 'a', only
> select the corresponding columns listed in 'b'. For example, after
> that operation,
>
> a[["1,1"]] become subset(a[["1.1"]], select = c("A", "B"))
>
> a[["1,2"]] become subset(a[["1.1"]], select = c("B", "C"))
I'm assuming the is a typo and you wanted
a[["1,2"]] become subset(a[["1.2"]], select = c("B", "C"))
>
> Do you know how to do it without looping?
using lapply is a loop construct but maybe it won't feel that way:
> lapply( names(a), function(x) subset(a[[x]], select=b[[x]]) )
[[1]]
A B
1 -0.660594340 -0.44402746
2 0.926022768 0.50906870
3 -0.357588460 -0.10657540
4 0.226316328 1.96415323
5 0.143330486 -0.89655890
6 -0.307193955 -1.10708473
7 -0.904927213 2.01289010
8 -1.922708953 0.74302970
9 1.008833376 0.90442422
10 1.095004836 1.23202474
11 -0.568905764 0.20946675
12 0.621113819 0.25753637
13 0.001997838 0.41691655
14 -0.206525640 0.94883919
15 -0.665738673 1.29442907
16 -1.225075753 0.49222447
17 0.224294111 -0.51241903
18 -0.170286569 0.09245769
19 0.396839258 0.23253796
20 0.645889199 0.52546631
[[2]]
B C
1 -0.44402746 -0.05013952
2 0.50906870 0.32247682
3 -0.10657540 -0.54740895
4 1.96415323 0.63823531
5 -0.89655890 0.54520142
6 -1.10708473 -0.14625567
7 2.01289010 -0.97021834
8 0.74302970 0.49252268
9 0.90442422 0.30610560
10 1.23202474 -0.52001742
11 0.20946675 -0.96469545
12 0.25753637 0.22606727
13 0.41691655 0.48728976
14 0.94883919 -0.99520018
15 1.29442907 0.60930763
16 0.49222447 -0.23321555
17 -0.51241903 -0.72746372
18 0.09245769 -1.96009906
19 0.23253796 -0.14303924
20 0.52546631 2.12944010
--
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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