Hi Michael,
The days in your example do not look continuous (at least from my
thinking), so you may have extra requirements in mind, but take a look
at this code. My general thought was first to turn each column into a
logical vector (c1 >= 100 and c2 >= 8). Taking advantage of the fact
that R treats TRUE as 1 and FALSE as 0, compute a rolling mean. If
(and only if) 5 consecutive values are TRUE, the mean will be 1. Next
I added the rolling means for each column, and then tested whether any
were 2 (i.e., 1 + 1).
Cheers,
Josh
###################
#Load required package
library(zoo)
#Your data with ds converted to Date
#from dput()
dat <-
structure(list(ds = structure(c(14702, 14729, 14730, 14731, 14732,
14733, 14734, 14735, 14736, 14737, 14738, 14739, 14740, 14741,
14742, 14743, 14744), class = "Date"), c1 = c(100L, 11141L, 3L,
7615L, 6910L, 5035L, 3007L, 4L, 8335L, 2897L, 6377L, 3177L, 7946L,
8705L, 9030L, 8682L, 8440L), c2 = c(0L, 15L, 16L, 14L, 17L, 3L,
15L, 14L, 17L, 13L, 17L, 17L, 15L, 0L, 16L, 16L, 1L)), .Names = c("ds",
"c1", "c2"), row.names = c(NA, -17L), class = "data.frame")
#Order by ds
dat <- dat[order(dat$ds), ]
yourvar <- 0
#Test that 5 consecutive values from c1 AND c2 meet requirements
if(any(
c(rollmean(dat$c1 >= 100, 5) + rollmean(dat$c2 >= 8, 5)) == 2)
) {yourvar <- 1}
###################
On Sat, Jul 17, 2010 at 2:38 PM, Michael Hess <mlh...@med.umich.edu> wrote:
Sorry for not being clear.
In the dataset there are around 100 or so days of data (in the case also rows
of data)
I need to make sure that the person meets that c1 is at least 100 AND c2 is at
least 8 for 5 of 7 continuous days.
I will play with what I have and see if I can find out how to do this.
Thanks for the help!
Michael
Stephan Kolassa 07/17/10 4:50 PM >>>
Mike,
I am slightly unclear on what you want to do. Do you want to check rows
1 and 7 or 1 *to* 7? Should c1 be at least 100 for *any one* or *all*
rows you are looking at, and same for c2?
You can sort your data like this:
data <- data[order(data$ds),]
Type ?order for help. But also do this for added enlightenment...:
library(fortunes)
fortune("dog")
Next, your analysis on the sorted data frame. As I said, I am not
entirely clear on what you are looking at, but the following may solve
your problem with choices "1 to 7" and "any one" above.
foo <- 0
for ( ii in 1:(nrow(data)-8) ) {
if (any(data$c1[ii+seq(0,6)]>=100) & any(data$c2[ii+seq(0,6)]>=8)) {
foo <- 1
break
}
}
The variable "foo" should contain what you want it to. Look at ?any
(and, if this does not do what you want it to, at ?all) for further info.
No doubt this could be vectorized, but I think the loop is clear enough.
Good luck!
Stephan
Michael Hess schrieb:
Hello R users,
I am a researcher at the University of Michigan looking for a solution to an R
problem. I have loaded my data in from a mysql database and it looks like this
data
ds c1 c2
1 2010-04-03 100 0
2 2010-04-30 11141 15
3 2010-05-01 3 16
4 2010-05-02 7615 14
5 2010-05-03 6910 17
6 2010-05-04 5035 3
7 2010-05-05 3007 15
8 2010-05-06 4 14
9 2010-05-07 8335 17
10 2010-05-08 2897 13
11 2010-05-09 6377 17
12 2010-05-10 3177 17
13 2010-05-11 7946 15
14 2010-05-12 8705 0
15 2010-05-13 9030 16
16 2010-05-14 8682 16
17 2010-05-15 8440 15
What I am trying to do is sort by ds, and take rows 1,7, see if c1 is at least
100 AND c2 is at least 8. If it is not, start with check rows 2,8 and if not
there 3,9....until it loops over the entire file. If it finds a set that
matches, set a new variable equal to 1, if never finds a match, set it equal to
0.
I have done this in stata but on this project we are trying to use R. Is this
something that can be done in R, if so, could someone point me in the correct
direction.
Thanks,
Michael Hess
University of Michigan
Health System
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______________________________________________
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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.
