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. 2020 Jan 6;6:e243. doi: 10.7717/peerj-cs.243
## Set parameters for hypothetical species ##
> N<-100
> Hstar<-10
> probs<-c(rep(0.30, 3), rep(0.10/7, 7)) # three dominant haplotypes each with 30% frequency
### Run simulations ###
> HACSObj<-HACHypothetical(N = N, Hstar = Hstar, probs = probs)
> HAC.simrep(HACSObj)
Simulating haplotype accumulation...
=========================================================— 100%
—Measures of Sampling Closeness —
Mean number of haplotypes sampled: 8.3291
Mean number of haplotypes not sampled: 1.6709
Proportion of haplotypes sampled: 0.83291
Proportion of haplotypes not sampled: 0.16709
Mean value of N*: 120.061
Mean number of specimens not sampled: 20.06099
Desired level of haplotype recovery has not yet been reached
====================================================— 100%
—Measures of Sampling Closeness —
Mean number of haplotypes sampled: 9.2999
Mean number of haplotypes not sampled: 0.7001
Proportion of haplotypes sampled: 0.92999
Proportion of haplotypes not sampled: 0.07001
Mean value of N*: 179.5718
Mean number of specimens not sampled: 12.57182
Desired level of haplotype recovery has not yet been reached
======================================================— 100%
—Measures of Sampling Closeness —
Mean number of haplotypes sampled: 9.5358
Mean number of haplotypes not sampled: 0.4642
Proportion of haplotypes sampled: 0.95358
Proportion of haplotypes not sampled: 0.04642
Mean value of N*: 188.7623
Mean number of specimens not sampled: 8.762348
Desired level of haplotype recovery has been reached
———- Finished. ———-
The initial guess for sampling sufficiency was N = 100 individuals
The algorithm converged after 6 iterations and took 33.215 s
The estimate of sampling sufficiency for p = 95% haplotype recovery is N∗ = 180 individuals
( 95% CI [178.278–181.722])
The number of additional specimens required to be sampled for p = 95% haplotype recovery is
N* - N = 80 individuals