On Nov 23, 2010, at 11:04 AM, Charles C. Berry wrote:
On Tue, 23 Nov 2010, Dennis Murphy wrote:
Interesting. Check this out:
u <- sample(c(TRUE, FALSE), 10, replace = TRUE)
u
[1] FALSE FALSE TRUE FALSE FALSE TRUE FALSE FALSE FALSE FALSE
class(u)
[1] "logical"
u + 0
[1] 0 0 1 0 0 1 0 0 0 0
0 + u
[1] 0 0 1 0 0 1 0 0 0 0
v <- rpois(10, 3)
!duplicated(v)
[1] TRUE FALSE TRUE TRUE TRUE TRUE FALSE TRUE FALSE TRUE
class(!duplicated(v))
[1] "logical"
!duplicated(v) + 0
[1] TRUE FALSE TRUE TRUE TRUE TRUE FALSE TRUE FALSE TRUE
0 + !duplicated(v)
[1] 1 0 1 1 1 1 0 1 0 1
# Now assign !duplicated(v) to an object:
w <- !duplicated(v)
class(w)
[1] "logical"
0 + w
[1] 1 0 1 1 1 1 0 1 0 1
w + 0
[1] 1 0 1 1 1 1 0 1 0 1
I can see *what* is going on, but what is the reason for it? I see
another
notebook entry coming :)
See
?Arithmetic
and read the paragraph under Details starting 'Logical vectors'
Chuck;
Compare these three, all of which are using binary operators on
logical vectors which is what is being discussed in ?Arithmetic:
> duplicated(c("a", "a", "b") ) + 0
[1] 0 1 0
> !duplicated(c("a", "a", "b") ) + 0
[1] TRUE FALSE TRUE
> 0 + !duplicated(c("a", "a", "b") )
[1] 1 0 1
I believe the proper place to go is ?Syntax where operator precedence
is discussed. I think the precendence rules implicitly do this in the
second instance, because "+" has higher precendence than negation:
! ( duplicated(c("a", "a", "b") ) + 0 )
--
David.
Chuck
Dennis
On Tue, Nov 23, 2010 at 6:12 AM, David Winsemius <dwinsem...@comcast.net
>wrote:
On Nov 23, 2010, at 8:33 AM, Joel wrote:
Is there any similar function in R to the first. in SAS?
What it dose is:
Lets say we have this table:
a b c
1 1 5
1 0 2
2 0 2
2 0 NA
2 9 2
3 1 3
and then I want do to do one thing the first time the number 1
appers in a
and something else the secund time 1 appers in a and so on.
so
something similar to:
if first.a {
a$d<-1
}else{
a$d<-0
}
The duplicated function which returns a logical vector with those
features
can easily be coerced to numeric.
df$d <- as.numeric(!duplicated(df$a))
I was a bit puzzled about my failure to get coercion by the method
which I
thought was supposed to work, namely adding 0.
df$e <- !duplicated(df$a)+0 # does not coerce
df$e <- 0 + !duplicated(df$a) # pre-adding 0 does coerce
Maybe the rules on coercion were amended.
--
David
This would give me
a b c b
1 1 5 1
1 0 2 0
2 0 2 1
2 0 NA 0
2 9 2 0
3 1 3 1
Is there such a function in R or anything similar?
thx
//Joel
--
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______________________________________________
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David Winsemius, MD
West Hartford, CT
______________________________________________
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______________________________________________
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Charles C. Berry Dept of Family/
Preventive Medicine
cbe...@tajo.ucsd.edu UC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego
92093-0901
David Winsemius, MD
West Hartford, CT
______________________________________________
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.