David Winsemius wrote:
On Jul 5, 2009, at 12:19 PM, Mark Knecht wrote:
On Sun, Jul 5, 2009 at 8:18 AM, David
Winsemius<dwinsem...@comcast.net> wrote:
On Jul 5, 2009, at 10:50 AM, Uwe Ligges wrote:
David Winsemius wrote:
So if your values are calculated from other values then consider using
all.equal()
And repeated applications of the testing criteria process are
effective:
test[3,][which(names(test)=="C1"):(which(test[3,] == 0.0)-1)]
C1 C2 C3
3 0.52 0.66 0.51
(and a warning that does not seem accurate to me.)
In which(names(test) == "C1"):(which(test[3, ] == 0) - 1) :
numerical expression has 3 elements: only the first used
David,
# which(test[3,] == 0.0)
[1] 6 7 8
and in a:b a and b must be length 1 vectors (scalars) otherwise just
the
first element (in this case 6) is used.
That leads us to the conclusion that writing the line above is not
really
the cleanest way or you intended something different ....
Thanks, Uwe. I see my confusion. I did want 6 to be used and it
looks as
though I would not be getting in truouble this way, but a cleaner method
would be to access only the first element of which(test[3, ] == 0):
test[3,][ which(names(test) == "C1") : (which(test[3,] == 0.0)[1]-1) ]
David
Seems to me that all of the element were used. I cannot explain that
warning but am pretty sure it can be ignored.
David
OK - making lots more headway. Thanks for your help.
QUESTION: How do I handle the case where I'm testing for 0 and don't
find it? In this case I need to all of the row from C1:C6.
test <- data.frame(A=1:10, B=100, C1=runif(10), C2=runif(10),
C3=runif(10), C4=runif(10), C5=runif(10), C6=runif(10))
test<-round(test,2)
#Make array ragged
test$C3[2]<-0;test$C4[2]<-0;test$C5[2]<-0;test$C6[2]<-0
test$C4[3]<-0;test$C5[3]<-0;test$C6[3]<-0
test$C6[7]<-0
test$C4[8]<-0;test$C5[8]<-0;test$C6[8]<-0
test
#C1 always the same so calculate it only once
StartCol <- which(names(test)=="C1")
#Print row 3 explicitly
test[3,][StartCol :(which(test[3,] == 0.0)[1]-1)]
#Row 6 fails because 0 is not found
test[6,][StartCol :(which(test[6,] == 0.0)[1]-1)]
EndCol <- which(test[6,] == 0.0)[1]-1
EndCol
It's getting a bit Baroque, but here is a solution that handles an NA:
test[6,][StartCol :ifelse(is.na( which(test[6,] == 0.0)[1]) ,
ncol(test), which(test[6,] == 0.0)[1]-1 )
]
#####-----
C1 C2 C3 C4 C5 C6
6 0.33 0.84 0.51 0.86 0.84 0.15
Maybe an R-meister can offer something more compact?
So let's wait for some R-meister, I'd write even more ....
Reason: testing for exactly zero after possible calculations is a bit
dangerous and ifelse() is designed for vectorized operations but is not
efficient for scalar operations, particularly since both expressions are
evaluated, so if() else would be preferable, but we could use min()
instead. Finally, a:b could end up in 5:3 without a warning and I'd use
seq() instead.
Hence I'd prefer:
temp <- which(sapply(test[6,], function(x, y) isTRUE(all.equal(x,y)), 0))[1]
test[6, seq(from = StartCol, to = min(c(temp - 1, ncol(test)), na.rm =
TRUE), by = 1)]
Best,
Uwe Ligges
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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