familycheck {kinship2}

Error check for a family classification

Description

Given a family id vector, also compute the familial grouping from first principles using the parenting data, and compare the results.

Usage

familycheck(famid, id, father.id, mother.id, newfam)

Arguments

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 makefamid. If this has allready been computed by the user, adding it as an argument shortens the running time somewhat.

Details

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.

Value

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.

See Also

makefamid, makekinship

Examples

# 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