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Hello,

I know how to reduce a single-session caphist :

Code: Select all

bear08_reduce<-reduce(bear2008,newocc=list(1:2, 3:4))

But, I can't quite figure out how to reduce a multi-session caphist where the structure of the reduce is session-specific.

For example, I can reduce a large multi-session caphist in this way, but all sessions get reduced in the same way;

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test<-lapply(bearCH, function(x) reduce(x,newocc=list(1:2, 3:4)))

I've been trying head this route, but can't seem to figure it out:

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test<-mapply(reduce, bearCH_F,
       reduce=lapply(bearCH_F, function(x) reduce(x,newocc=list(1:2, 3:4))),
       MoreArgs = list(Session = 1:5)
       )

I have 9 years of data, 2006-2014

I want to reduce 2006-2010 in this fashion:

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bear_reduce<-reduce(bear,newocc=list(1:2, 3:4))

2011-2012,2014 I don't want to do anything to,

and I want to do this to 2013:

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bear_reduce<-reduce(bear,newocc=list(1:2, 3, 4 ))

thoughts?

Without looking at the detail = I think you are on the right track with mapply, but perhaps need SIMPLIFY = FALSE.
Murray

Thanks, Murray

When I do so, I get the following.

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> test<-mapply(reduce, bearCH_F,
+        reduce=lapply(bearCH_F, function(x) reduce(x,newocc=list(1:2, 3:4))),
+        MoreArgs = list(session = 1:5),
+        SIMPLIFY = FALSE
+        )

Session 2008
Conflicting number of occasions in usage matrix
capthist : 3 occasions
usage(traps(capthist)) : 4 occasions
Session 2009
Conflicting number of occasions in usage matrix
capthist : 3 occasions
usage(traps(capthist)) : 4 occasions
Session 2010
Conflicting number of occasions in usage matrix
capthist : 3 occasions
usage(traps(capthist)) : 4 occasions


> class(test)<-class(bearCH)
> summary(test)


Error in counts[8, ] <- apply(usage(traps), 2, function(x) sum(x > 0)) :
  number of items to replace is not a multiple of replacement length

And not sure how to implement the second argument for 2013 where I reduce only sessions 1:2.

You want to provide a list of values for newoccasions, one per input session. Without checking details, how about something like this (note need to restore the class of the resulting list):

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newocc <- list(c(1:2, 3:4), NULL, c(1:2,3,4))[c(1,1,1,1,1,2,3,2)]
newCH <- mapply(reduce, bearCH_F, newoccasions = newocc, SIMPLIFY = FALSE)
class(newCH) <- class(bearCH_F)


Ah I can see we are very close!!

The issue with the:

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Session 2008
Conflicting number of occasions in usage matrix
capthist : 3 occasions
usage(traps(capthist)) : 4 occasions
Session 2009
Conflicting number of occasions in usage matrix
capthist : 3 occasions
usage(traps(capthist)) : 4 occasions
Session 2010
Conflicting number of occasions in usage matrix
capthist : 3 occasions
usage(traps(capthist)) : 4 occasions


stemmed from the fact that a bunch of females had no captures in session 1 when I subsetted the data, so secr didn't seem to be able to combine. If I apply the reduce to the male + female dataset first, then subset by sex I think I'll be better off.

Now, the problem with the code below seems to be that c(1:2, 3:4) and c(1,2, 3:4) produce the same string in a list, 1,2,3,4.

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> newocc <- list(c(1:2, 3:4), NULL, c(1:2,3,4))[c(1,1,1,1,1,2,2,3,2)]
> newocc
[[1]]
[1] 1 2 3 4

[[2]]
[1] 1 2 3 4

[[3]]
[1] 1 2 3 4

[[4]]
[1] 1 2 3 4

[[5]]
[1] 1 2 3 4

[[6]]
NULL

[[7]]
NULL

[[8]]
[1] 1 2 3 4

[[9]]
NULL

I got it, this is how to reduce a multi-session caphist

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newocc <- list(
              list(1:2,3:4),
              NULL,
              list(1:2,3,4)
              )[c(1,1,1,1,1,2,2,3,2)]


newCH <- mapply(reduce, bearCH, newoccasions = newocc, SIMPLIFY = FALSE)
class(newCH) <- class(bearCH)