Sent: Thursday, July 23, 2009 12:00 PM
To: r-help@r-project.org
Subject: [R] Random # generator accuracy
Dan Nordlund wrote:
It would be necessary to see the code for your 'brief test' before
anyone
could meaningfully comment on your results. But your results for a
single
test could have been
Dan Nordlund wrote:
It would be necessary to see the code for your 'brief test' before anyone
could meaningfully comment on your results. But your results for a single
test could have been a valid random result.
I've re-created what I did below. The problem appears to be with the
weighting
[mailto:r-help-boun...@r-
project.org] On Behalf Of Jim Bouldin
Sent: Thursday, July 23, 2009 12:00 PM
To: r-help@r-project.org
Subject: [R] Random # generator accuracy
Dan Nordlund wrote:
It would be necessary to see the code for your 'brief test' before
anyone
could meaningfully comment
-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
project.org] On Behalf Of Jim Bouldin
Sent: Thursday, July 23, 2009 12:00 PM
To: r-help@r-project.org
Subject: [R] Random # generator accuracy
Dan Nordlund wrote:
It would be necessary to see the code for your
-
From: Jim Bouldin [mailto:jrboul...@ucdavis.edu]
Sent: Thursday, July 23, 2009 12:49 PM
To: Greg Snow; r-help@r-project.org
Subject: RE: [R] Random # generator accuracy
Thanks Greg, that most definitely was it. So apparently the default is
sampling without replacement. Fine
On 23-Jul-09 17:59:56, Jim Bouldin wrote:
Dan Nordlund wrote:
It would be necessary to see the code for your 'brief test'
before anyone could meaningfully comment on your results.
But your results for a single test could have been a valid
random result.
I've re-created what I did below.
OOPS! The result of a calculation below somehow got omitted!
(325820+326140+325289+325098+325475+325916)/
(174873+175398+174196+174445+173240+174110)
# [1] 1.867351
to be compared (as at the end) with the ratio 1.867471 of the
expected number of weight=2 to expected number of weight=1.
You are absolutely correct Ted. When no weights are applied it doesn't
matter if you sample with or without replacement, because the probability
of choosing any particular value is equally distributed among all such.
But when they're weighted unequally that's not the case.
It is also
Indeed, Jim! And that's why I said to read carefully what is said about
prob in '?sample':
If 'replace' is false, these probabilities are applied
sequentially, that is the probability of choosing the
next item is proportional to the weights amongst the
remaining items.
Whereas, if you
On Thu, 23 Jul 2009 ted.hard...@manchester.ac.uk wrote:
The general problem, of sampling without replacement in such a way
that for each item the probability that it is included in the sample
is proportional to a pre-assigned weight (sampling with probability
proportional to size) is quite
Perfectly explained Ted. One might, at first reflection, consider that
simply repeating the values 7 through 12 and sampling (w/o replacement)
from among the 18 resulting values, would be similar to just doubling the
selection probabilities for 7 through 12 and then sampling. That would
clearly
On 23-Jul-09 22:16:39, Thomas Lumley wrote:
On Thu, 23 Jul 2009 ted.hard...@manchester.ac.uk wrote:
The general problem, of sampling without replacement in such a way
that for each item the probability that it is included in the sample
is proportional to a pre-assigned weight (sampling with
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