Hi Jim,
Dennis Murphy solved my problem by the following code.
Thank you for you suggestion. Will check out listBuilder function too.
best,
Zhongyi
# (2) A little more general: prespecify the number of list components
# and run a one-line loop to populate the list
> l <- vector('list', 6)
> for(i in seq_along(l)) l[[i]] <- v
> l
[[1]]
[[1]]$para1
[1] 1 2 3 4 5
[[1]]$para2
[1] 5 6 7 8 9
[[2]]
[[2]]$para1
[1] 1 2 3 4 5
[[2]]$para2
[1] 5 6 7 8 9
[[3]]
[[3]]$para1
[1] 1 2 3 4 5
[[3]]$para2
[1] 5 6 7 8 9
[[4]]
[[4]]$para1
[1] 1 2 3 4 5
[[4]]$para2
[1] 5 6 7 8 9
[[5]]
[[5]]$para1
[1] 1 2 3 4 5
[[5]]$para2
[1] 5 6 7 8 9
[[6]]
[[6]]$para1
[1] 1 2 3 4 5
[[6]]$para2
[1] 5 6 7 8 9
On Fri, Mar 12, 2010 at 3:06 AM, Jim Lemon <j...@bitwrit.com.au> wrote:
> On 03/12/2010 05:13 PM, Zhongyi Yuan wrote:
>
>> Dear R users:
>>
>> I am hoping that someone can help with constructing a list that consists
>> of
>> list with the number of lists variable.
>> i.e. to find a convenient express(or loop sentences) to realize the
>> following:
>> list( list(para1=p1, para2=p2), list(para1=p1, para2=p2), ....,
>> list(para1=p1,para2=p2) )
>>
>> Hi Zhongyi,
> Have a look at the listBuilder function in the crank package.
>
> Jim
>
>
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