David: Thanks. I cannot believe that I had not tried the simple,
lwf[lwf$bushno==145 , "bout3"] <- 1
I will mess around with the runs suggestion too.
Thanks, Don
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Dennis. Tres cool. I will try it. regards, MN
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Hi:
A simple diagnostic is to check how many distinct run lengths exist
in a row - ideally, it should be one or two. If it's more than two,
something
is amiss. Hence, define f() as a function to determine the number of
distinct
runs in a given row and call the apply() function with it:
f <- funct
On May 31, 2010, at 5:29 PM, Mr. Natural wrote:
The data frame is lwf that records the survival of bushes over an 8
year
period. Years are called bouts. Dead bushes are recorded as zeros,
and live
bushes as "1."
str(lwf)
'data.frame': 638 obs. of 9 variables:
$ bushno: int 1 2 3 4 5 6
The data frame is lwf that records the survival of bushes over an 8 year
period. Years are called bouts. Dead bushes are recorded as zeros, and live
bushes as "1."
str(lwf)
'data.frame': 638 obs. of 9 variables:
$ bushno: int 1 2 3 4 5 6 7 8 9 10 ...
$ bout1 : int 0 1 0 1 1 1 0 1 0 1 ...
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