familycheck | R Documentation |
Given a family id vector, also compute the familial grouping from first principles using the parenting data, and compare the results.
familycheck(famid, id, father.id, mother.id, newfam)
famid |
a vector of family identifiers |
id |
a vector of unique subject identifiers |
father.id |
vector containing the id of the biological father |
mother.id |
vector containing the id of the biological mother |
newfam |
the result of a call to |
The makefamid
function is used to create a
de novo family id from the parentage data, and this is compared to the
family id given in the data.
a data frame with one row for each unique family id in the
famid
argument.
Components of the output are
famid |
the family id, as entered into the data set |
n |
number of subjects in the family |
unrelated |
number of them that appear to be unrelated to anyone else in the entire pedigree set. This is usually marry-ins with no children (in the pedigree), and if so are not a problem. |
split |
number of unique "new" family ids. If this is 0, it means that no one in this "family" is related to anyone else (not good); 1 = everythings is fine; 2+= the family appears to be a set of disjoint trees. Are you missing some of the people? |
join |
number of other families that had a unique famid, but are actually joined to this one. 0 is the hope. If there are any joins, then an attribute "join" is attached. It will be a matrix with famid as row labels, new-family-id as the columns, and the number of subjects as entries. |
makefamid
, makekinship
# use 2 sample peds
data(sample.ped)
pedAll <- with(sample.ped, pedigree(id, father, mother, sex,
affected=cbind(affected, avail), famid=ped))
## check them giving separate ped ids
fcheck.sep <- with(sample.ped, familycheck(ped, id, father, mother))
fcheck.sep
## check assigning them same ped id
fcheck.combined <- with(sample.ped, familycheck(rep(1,nrow(sample.ped)), id, father, mother))
fcheck.combined
#make person 120's father be her son.
sample.ped[20,3] <- 131
fcheck1.bad <- try({with(sample.ped, familycheck(ped, id, father, mother))}, silent=FALSE)
## fcheck1.bad is a try-error
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