Is the following more like what you want:
a=c(1,1,1,1,1) # generate the first list
b=c(2,2,2)# generate a second list
d=list(a,b)# make a list of a and b
(
e=c(d,a)
)
[[1]]
[1] 1 1 1 1 1
[[2]]
[1] 2 2 2
[[3]]
[1] 1
[[4]]
[1] 1
[[5]]
[1] 1
[[6]]
[1] 1
[[7]]
[1] 1
If no, have you read sec. 6 in An Introduction to R [the first
option available from help.start()]?
In this example, note that e[1] is a list with only one
attributes, namely the vector 1 1 1 1 1; e[[1]] is not a list but that
vector itself.
hope this helps. spencer graves
Karla Sartor wrote:
Hello all,
As a general programming question I can't seem to figure out how to
make a list of lists in R.
As matrices won't work as they have to be rectangular.
I am sure that there is an easy solution but...
the specific situation is this:
- I have created a Tukey confidence interval table and have listed the
means that are not significantly different
- then using these not significantly different pairs I have created
the groups of means that are not significantly different from each other
the issue then is that many of these lists are subsets of other lists
and I need to check for this.
Below is a little program is illustrate the issue
a=c(1,1,1,1,1) # generate the first list
b=c(2,2,2)# generate a second list
c=c(a,b) #combine them
cat(c, \n)# and print
1 1 1 1 1 2 2 2 # this is 1-D!!! ahh
d=list(a,b)# make a list of a and b
d # and print
[[1]] #this is exactly what I
want, but continue
[1] 1 1 1 1 1
[[2]]
[1] 2 2 2
e=list(d,a) # now on the next iteration I
need to add another list to this list of lists
e # and print
[[1]] # ahh all hell has broken
loose and this is not what I want
[[1]][[1]] # desired result below
[1] 1 1 1 1 1
[[1]][[2]]
[1] 2 2 2
[[2]]
[1] 1 1 1 1 1
-
desired result
#wrong code but this is what I want to happen
a=c(1,1,1,1,1,1,1)
for(i in 1:5) {
a=list(a,1:5)
}
output I want is (something like)
[1] 1 1 1 1 1 1 1
[2] 1 2 3 4 5
[3] 1 2 3 4 5
[4] 1 2 3 4 5
[5] 1 2 3 4 5
[6] 1 2 3 4 5
so then I could call cat(a[1]) and get 1 1 1 1 1 1 1 1 and cat(a[2])
and get 1 2 3 4 5
Anyone know the answer (hopefully simple)
Cheers,
Karla Sartor
--
Karla Sartor
Montana State University - LRES
[EMAIL PROTECTED]
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Spencer Graves, PhD, Senior Development Engineer
O: (408)938-4420; mobile: (408)655-4567
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