data2=reduce(data,by="all")
fit<- openCR.fit(data2,mask=mask1,type="JSSAsecrb")
predict(fit)
#$lambda0
# session estimate SE.estimate lcl ucl
#1 1 0.003594342 0 0.003594342 0.003594342
$phi
session estimate SE.estimate lcl ucl
1 1 NA NA NA NA
$b
session estimate SE.estimate lcl ucl
1 1 1 NA 1 1
$superD
estimate SE.estimate lcl ucl
1 2.483975e-05 0 2.483975e-05 2.483975e-05
$sigma
session estimate SE.estimate lcl ucl
1 1 6620.637 0 6620.637 6620.637
(this happens to several datasets I am testing always in the same order of difference)
Interestingly, by mistake one time I was running openCR I forced the spacing when calling the mask to be half of it actually is (so in this case if I use the 1km spaced mask but when loading it I specify spacing=500) then I have a much similar result with secr (secr using the correct mask with correct spacing):
mask2<-read.mask(data=mask, spacing = 250, header=TRUE) # mask here is the same file as before just spacing =250 differs
fit2<- openCR.fit(data2,mask=mask2,type="JSSAsecrb")#Spatial
predict(fit2)
$lambda0
session estimate SE.estimate lcl ucl
1 1 0.003534561 0 0.003534561 0.003534561
$phi
session estimate SE.estimate lcl ucl
1 1 NA NA NA NA
$b
session estimate SE.estimate lcl ucl
1 1 1 NA 1 1
$superD
estimate SE.estimate lcl ucl
1 0.0001155695 0 0.0001155695 0.0001155695
$sigma
session estimate SE.estimate lcl ucl
1 1 7949.61 0 7949.61 7949.61
Plus, the non-spatial super N estimated with openCR gives 53 which divided by the correct sized area (283575ha) is roughly the secr estimated density (0.000187 ind/ha).
I am sorry if I am missing something obvious here but is there any difference in the units of D between openCR and secr?
Thank you so much!
Rafael