I don't think I understand you entirely, but here are a couple
things.
You need to shorten the loop so you don't run off the end.
You basically just write the code as you've stated it except that
"and" is && for single values and & for vectors
You don't say what should happen if the test is not true.
You can probably use "ifelse" instead of a loop.
iseq <- 2:(length(x.dif) - 2)
test <- x.dif[iseq-1] <= 0 & x.dif[iseq] > 0 & x.dif[iseq + 2] > 0
ifelse(test, iseq + 2, NA)
I suspect that the last term in test should be i+1 and not i+2 ?
You can see S Poetry for details.
Patrick Burns
Burns Statistics
[EMAIL PROTECTED]
+44 (0)20 8525 0696
http://www.burns-stat.com
(home of S Poetry and "A Guide for the Unwilling S User")
ebashi wrote:
I'll appreciate if some one can help me with the
following loop. This is the logic of the loop,
if we have the following data;
x.df
x.dif
. .
. .
102 0.00
103 0.42
104 0.08
105 0.00
106 0.00
107 0.00
108 -0.16
109 -0.34
110 0.00
111 -0.17
112 -0.33
113 0.00
114 0.00
115 0.00
116 0.33
117 0.17
118 0.00
. .
. .
I'm trying to find i's where
for (i in 2:length(x.dif))
if (x.dif[i-1]<=0 and x.dif[i]>0 and x.dif[i+2]>0)
it would return i+2 to me,
How can I turn this to a right format as a loop.(I
can't figure out the syntax)
Cheers,
Sean
_______________________________________________
[EMAIL PROTECTED] mailing list
https://stat.ethz.ch/mailman/listinfo/r-sig-finance
______________________________________________
[EMAIL PROTECTED] mailing list
https://stat.ethz.ch/mailman/listinfo/r-devel
