Hi,
On Jul 15, 2009, at 1:00 PM, Chyden Finance wrote:
Hello!
For the past three years, I have been using R extensively in my PhD
program in Finance for statistical work. Normally, I can figure
this kind of thing out on my own, but I am completely stumped as to
why the following code does not work.
I have two variables: sigs and p0_recent.
dim(sigs) = 296 3
dim(p0_recent) = 504 7
I want to compare each element in the first column of sigs with the
elements in the first column of p0_recent.
In other words, does sigs[i,1] == p0_recent[j,1], where i =
1:dim(sigs)[1] and j = 1:dim(p0_recent)[1].
I've been trying:
> for (j in 1:dim(p0_recent)[1]) {
+ for (i in 1:dim(sigs)[1]) {
+ if (sigs[i,1] == p0_recent[j,1]) {
+ print(sigs[i,1])
+ }}}
But, I get:
Error in Ops.factor(sigs[i, 1], p0_recent[j, 1]) :
level sets of factors are different
It seems that this particular problem is due to the fact that you are
comparing two sets of factors with different levels, which is what the
Ops.factor error is saying. But let's make it more clear:
R> f1 <- factor(c('a','b','c'))
R> f1
[1] a b c
> f2 <- factor(c('c','d','e'))
[1] c d e
Levels: c d e
> f1[3] == f2[1]
Error in Ops.factor(f1[3], f2[1]) : level sets of factors are different
Is there a better way than for loops to compare each element in one
column to each element in another column of different length? If
not, can anyone suggest a solution?
Probably better ways, but how about starting by first nuking the inner
loop? Something like this should work:
for (j in 1:nrow(p0_recent)) {
found <- match(p0_recent[j,1], sigs[,1])
...
}
In each iteration "found" will be NA if the p0_recent element does not
appear in sigs[,1], otherwise it will be a vector of indices into
sigs[,1] that equal the p0_recent element you're querying with.
(Assuming you fix your factor/level issue)
HTH,
-steve
--
Steve Lianoglou
Graduate Student: Physiology, Biophysics and Systems Biology
Weill Medical College of Cornell University
Contact Info: http://cbio.mskcc.org/~lianos/contact
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