> On 24 Aug 2017, at 11:58 , peter dalgaard wrote:
>
> M <- as.matrix(do.call(expand.grid, rep(list(0:1),5)))
> M
> # This is just 0:31 encoded as (little-endian) binary
> apply(M, 1, function(i) sum((2^(0:4))[i]))
>
> # now, use rows of M for logical indexing into "A"-"E"
> mode(M) <- "logical
> On 24 Aug 2017, at 11:58 , peter dalgaard wrote:
>
> apply(M, 1, function(i) sum((2^(0:4))[i]))
--
Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Office: A 4.23
Email: pd@cbs.dk Priv: pda...
> On 24 Aug 2017, at 01:25 , Duncan Murdoch wrote:
>
> On 23/08/2017 6:25 PM, Bert Gunter wrote:
>> Doesn't sort by size of subgroup. I interpret the phrase I asterisked as:
>
> You were fooled by Peter's tricky single negative.
>
...
Let's do this more carefully, then:
M <- as.matrix(do.c
On 23/08/2017 6:25 PM, Bert Gunter wrote:
Doesn't sort by size of subgroup. I interpret the phrase I asterisked as:
You were fooled by Peter's tricky single negative.
Duncan Murdoch
Your code does the following:
First subsets of size 1 are given.
Then all subsets of size 2.
Then all subset
Doesn't sort by size of subgroup. I interpret the phrase I asterisked as:
Your code does the following:
First subsets of size 1 are given.
Then all subsets of size 2.
Then all subsets of size 3.
etc.
Your code does not do this (quite).
If you meant something else, then please clarify.
Cheers,
> On 23 Aug 2017, at 23:12 , Bert Gunter wrote:
>
>> This points to a different algorithm where you write 0:(2^n-1) as n-digit
>> binary numbers and chose items corresponding to the 1s. That won't give the
>> combinations **sorted by size of selected subgroup** though. Something like
>> this:
Inline.
-- Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Wed, Aug 23, 2017 at 1:58 PM, peter dalgaard wrote:
>
>> On 23 Aug 2017, at 20:51 , Ista Zahn
> On 23 Aug 2017, at 20:51 , Ista Zahn wrote:
>
> On Wed, Aug 23, 2017 at 12:35 PM, Bert Gunter wrote:
>> ummm, Ista, it's 2^n.
>
> ummm yes ug.
>
You didn't really say otherwise: sum(choose(n,0:n)) == 2^n by the binomial
expansion of (1+1)^n (but you knew that)
This points to a differ
On Wed, Aug 23, 2017 at 12:35 PM, Bert Gunter wrote:
> ummm, Ista, it's 2^n.
ummm yes ug.
My point is, if the number of groups is large, check it before hand.
If you can check it without embarrassing yourself in public like I did
that's even better.
Best,
Ista
>
> Cheers,
> Bert
>
>
> Bert
On 2017-08-23 11:35 AM, Bert Gunter wrote:
ummm, Ista, it's 2^n.
or (2^n-1) if the empty set is not considered as a "combination"
;-) spencer
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (ak
ummm, Ista, it's 2^n.
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Wed, Aug 23, 2017 at 8:52 AM, Ista Zahn wrote:
> On Wed, Aug 23, 2017 at 1
On Wed, Aug 23, 2017 at 11:33 AM, Christofer Bogaso
wrote:
> Hi again,
>
> I am exploring if R can help me to get all possible combinations of
> members in a group.
>
> Let say I have a group with 5 members : A, B, C, D, E
>
> Now I want to generate all possible unique combinations with all
> poss
Please at least do a basic internet search before posting here.
Searching on "R combinations" brought up the combn() function, which
you can use in a simple loop to get your answer:
> lapply(seq_len(5), FUN = function(x)combn(LETTERS[1:5], x))
## You can use a for() loop if you prefer
[[1]]
lapply(1:5, function(x) combn(groups, x))
or perhaps
unlist(lapply(1:5, function(x) combn(groups, x, FUN = paste, collapse = ", ")))
Best,
Ista
On Wed, Aug 23, 2017 at 11:33 AM, Christofer Bogaso
wrote:
> Hi again,
>
> I am exploring if R can help me to get all possible combinations of
> membe
Hi again,
I am exploring if R can help me to get all possible combinations of
members in a group.
Let say I have a group with 5 members : A, B, C, D, E
Now I want to generate all possible unique combinations with all
possible lengths from that group e.g.
1st combination : A
2nd combination : B
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