`multidog()` does not work with odd ploidy species when `model = "f1"`

[kostasgalexiou]

Hi,

I am running multidog on my work's cluster, using the following script

updog.R

library(updog)
library(future)
library(ggplot2)

sessionInfo()

refmat <- as.matrix(read.table("ref_matrix.tab", header = TRUE))
sizemat <- as.matrix(read.table("size_matrix.tab", header = TRUE))
ploidy <- 5

#plot_geno(refvec = refmat, sizevec = sizemat, ploidy = ploidy)
mout <- multidog(refmat = refmat, sizemat = sizemat, ploidy = ploidy, 
                 model = "f1", nc=4)

and getting the following error:

Working on it...Error in { : 
  task 1 failed - "assert_that: missing values present in assertion"
Calls: multidog -> %dorng% -> do.call -> %dopar% -> <Anonymous>
Además: There were 50 or more warnings (use warnings() to see the first 50)
Ejecución interrumpida

I would appreciate any help. Many thanks

Below I have added the sessionInfo() output:

R version 3.4.0 (2017-04-21)
Platform: x86_64-pc-linux-gnu (64-bit)
Running under: CentOS Linux 7 (Core)

Matrix products: default
BLAS: /home/kalexiou/local/software/R-3.4.0/lib/libRblas.so
LAPACK: /home/kalexiou/local/software/R-3.4.0/lib/libRlapack.so

locale:
 [1] LC_CTYPE=es_ES.UTF-8       LC_NUMERIC=C              
 [3] LC_TIME=es_ES.UTF-8        LC_COLLATE=es_ES.UTF-8    
 [5] LC_MONETARY=es_ES.UTF-8    LC_MESSAGES=es_ES.UTF-8   
 [7] LC_PAPER=es_ES.UTF-8       LC_NAME=C                 
 [9] LC_ADDRESS=C               LC_TELEPHONE=C            
[11] LC_MEASUREMENT=es_ES.UTF-8 LC_IDENTIFICATION=C       

attached base packages:
[1] stats     graphics  grDevices utils     datasets  base     

other attached packages:
[1] ggplot2_3.1.0 future_1.23.0 updog_2.1.2  

loaded via a namespace (and not attached):
 [1] Rcpp_1.0.0               parallelly_1.30.0        magrittr_1.5            
 [4] tidyselect_0.2.5         munsell_0.5.0            colorspace_1.4-0        
 [7] R6_2.4.0                 RcppArmadillo_0.10.8.1.0 doRNG_1.8.2             
[10] rlang_0.3.1              foreach_1.5.1            stringr_1.4.0           
[13] dplyr_0.8.0.1            plyr_1.8.4               globals_0.14.0          
[16] rngtools_1.5.2           tools_3.4.0              parallel_3.4.0          
[19] grid_3.4.0               gtable_0.2.0             withr_2.1.2             
[22] ggthemes_4.0.1           iterators_1.0.10         lazyeval_0.2.1          
[25] assertthat_0.2.0         digest_0.6.18            tibble_2.0.1            
[28] crayon_1.3.4             purrr_0.3.0              codetools_0.2-15        
[31] glue_1.3.0               stringi_1.3.1            compiler_3.4.0          
[34] pillar_1.3.1             methods_3.4.0            scales_1.0.0            
[37] doFuture_0.12.0          listenv_0.8.0            pkgconfig_2.0.2

[dcgerard]

Hey @kostasgalexiou,

Thanks for trying out {updog}!

Right now, {updog} does not support odd ploidy levels when model = "f1". I think modeling meiosis in odd ploidy species is non-trivial, and likely dependent on the species, right? So I would recommend using model = "norm" for now.

Side question: do you actually have data from an organism with all pentaploid bisexual reproduction? I would love to learn more.

- David

[kostasgalexiou]

Hey @kostasgalexiou,

Thanks for trying out {updog}!

Right now, {updog} does not support odd ploidy levels when model = "f1". I think modeling meiosis in odd ploidy species is non-trivial, and likely dependent on the species, right? So I would recommend using model = "norm" for now.

Side question: do you actually have data from an organism with all pentaploid bisexual reproduction? I would love to learn more.

• David

Hi David,

Thanks for your fast reply! I have tried it and works as it should be.

I have a blueberry segregating population, where one parent is supposed to be a tetraploid and the other a hexaploid. I was told that the progeny should be 5ploid...I cannot say more because I am not very familiar with the data. I am responsible for a specific analysis and the data interpretation will be done by another person.

Thanks again, Kostas