Re: [R] Help with pedigree() function in kinship2

[Therneau, Terry M., Ph.D.]

Your problem is that PatientID, FatherID, MotherID are factors.  The authors of kinship2 
(myself and Jason) simply never thought of someone doing this.  Yes, that is an oversight. 
 We will correct it by adding some more checks and balances.  For now, turn your id 
variables into character or numeric.

Terry Therneau

On 05/13/2015 05:00 AM, r-help-requ...@r-project.org wrote:
Dear R-help,

I am interested in plotting some pedigrees and came across the kinship2
package. What follows is an example of the pedigrees I am working with
Now, when running

## check package availability
if(!require(kinship2)) install.packages('kinship2')
require(kinship2)


## data to plot
d <- structure(list(FamilyID = c("1", "1", "1", "1", "1", "1", "1",
"1", "1"), PatientID = structure(c(2L, 3L, 5L, 11L, 12L, 15L,
16L, 17L, 6L), .Label = c(" 1", " 2", " 3", " 4", " 5", " 6",
" 7", " 9", "10", "11", "13", "14", "18", "20", "23", "24", "25",
"27", "28", "29", "30", "31", "33", "34", "35", "37", "38", "39",
"41", "43", "45", "50", "62", "63", "64", "65", "66", "67", "85",
"88"), class = "factor"), FatherID = structure(c(1L, 1L, 6L,
1L, 5L, 6L, 1L, 7L, 6L), .Label = c("0", "1", "10", "11", "13",
"2", "23", "27", "28", "3", "33", "34", "35", "38", "5", "62",
"64", "66", "9"), class = "factor"), MotherID = structure(c(1L,
1L, 7L, 1L, 14L, 7L, 1L, 5L, 7L), .Label = c("0", "10", "18",
"2", "24", "29", "3", "30", "33", "34", "39", "4", "43", "5",
"6", "63", "65", "9"), class = "factor"), Sex = structure(c(2L,
1L, 1L, 2L, 2L, 2L, 1L, 2L, 2L), .Label = c("Female", "Male"), class =
"factor"),
     AffectionStatus = structure(c(1L, 1L, 2L, 1L, 2L, 2L, 1L,
     2L, 2L), .Label = c("1", "2"), class = "factor")), .Names =
c("FamilyID",
"PatientID", "FatherID", "MotherID", "Sex", "AffectionStatus"
), row.names = c(NA, 9L), class = "data.frame")

## plotting
ped <- with(d, pedigree(PatientID, FatherID, MotherID, Sex,  affected =
AffectionStatus, famid = FamilyID))

## Error in pedigree(PatientID, FatherID, MotherID, Sex, affected =
AffectionStatus,  :
##                    Value of 'dadid' not found in the id list 1/0 1/0 1/2
1/0 1/2

I get an error.  My sessionInfo() is at the end.  I was wondering if
someone could help me to dissect what the cause of this error is and how it
can be fixed.

Thank you very much for your help.

Best regards,
Jorge Velez.-

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