When computer time is cheap, use this routine to get a best Pedigree alignment. This routine will try all possible founder orders, and return the one with the least stress.
Usage
# S4 method for class 'Pedigree'
best_hint(obj, wt = c(1000, 10, 1), tolerance = 0)Arguments
- obj
A Pedigree object
- wt
A vector of three weights for the three error measures. Default is
c(1000, 10, 1).The number of duplicate individuals in the plot
The sum of the absolute values of the differences in the positions of duplicate individuals
The sum of the absolute values of the differences between the center of the children and the parents.
- tolerance
The maximum stress level to accept. Default is
0
Details
The auto_hint() routine will rearrange sibling order,
but not founder order.
This calls auto_hint() with every possible founder
order, and finds that plot with the least "stress".
The stress is computed as a weighted sum of three error measures:
nbArcs The number of duplicate individuals in the plot
lgArcs The sum of the absolute values of the differences in the positions of duplicate individuals
lgParentsChilds The sum of the absolute values of the differences between the center of the children and the parents
$$stress = wt[1] * nbArcs + wt[2] * lgArcs + wt[3] * lgParentsChilds $$
If during the search, a plot is found with a stress level less than tolerance, the search is terminated.
Examples
data(sampleped)
ped <- Pedigree(sampleped[sampleped$famid == 1,])
best_hint(ped)
#> An object of class "Hints"
#> Slot "horder":
#> 1_101 1_102 1_103 1_104 1_105 1_106 1_107 1_108 1_109 1_110 1_111 1_112 1_113
#> 1 1 3 4 5 3 7 4 9 1 2 3 13
#> 1_114 1_115 1_116 1_117 1_118 1_119 1_120 1_121 1_122 1_123 1_124 1_125 1_126
#> 4 1 3 17 2 4 20 21 22 23 24 25 26
#> 1_127 1_128 1_129 1_130 1_131 1_132 1_133 1_134 1_135 1_136 1_137 1_138 1_139
#> 27 28 29 30 31 32 33 34 35 2 37 38 39
#> 1_140 1_141
#> 40 41
#>
#> Slot "spouse":
#> idl idr anchor
#> 1 1_112 1_118 right
#> 2 1_114 1_115 right
#> 3 1_109 1_110 left
#>