This is an R package for simulating pooled genetic screens and single cell phenotyping. The question is for a given library size and a cell-level classifier, on average how many cells of different genotypes and how many wells are needed to accurately classify a genotype as having the phenotype?
install.packages("remotes"
remotes::install_github("maomlab/spaCRPower")
To begin, simulate a screen
data <- spaCRPower::simulate_screen(
n_genes_per_library = 200,
gene_abundance_factor_mu = 1,
gene_abundance_factor_var = .01,
gene_hit_rate = 0.05,
n_wells_per_screen = 80,
well_abundance_factor_mu = 1,
well_abundance_factor_var = .01,
n_genes_per_well = 10,
n_cells_per_well_mu = 850,
n_cells_per_well_var = 1000,
class_pos_mu = 0.99,
class_pos_var = 0.0001,
class_neg_mu = 0.01,
class_neg_var = 0.0001)
investigate properties of the simulated screen
library(patchwork)
plot1 <- plot_genes_per_well(data)
plot2 <- plot_wells_per_gene(data)
plot3 <- plot_positivity_rate_by_well(data)
plot4 <- plot_well_gene_heatmap(data)
(plot1 + plot2) / (plot3 + plot4)
predict the which genotypes are hits:
model_data <- prepare_model_data(data)
model <- compile_model(model_data)
model_fit <- fit_model(model_data, model)
model_estimate <- gather_model_estimate(model_fit)
model_evaluation <- evaluate_model_fit(data, model_estimate)
Or sweep over parameters:
scores <- spaCRPower::scan_parameters(
n_genes_per_library = 200,
gene_abundance_factor_mu = 1,
gene_abundance_factor_var = .01,
gene_hit_rate = 0.05,
n_wells_per_screen = seq(5, 500, length.out = 100),
well_abundance_factor_mu = 1,
well_abundance_factor_var = .01,
n_genes_per_well = c(1, 3, 10, 30),
n_cells_per_well_mu = 850,
n_cells_per_well_var = 1000,
class_pos_mu = 0.99,
class_pos_var = 0.0001,
class_neg_mu = 0.01,
class_neg_var = 0.0001,
progress_file = "parameter_scan.tsv",
verbose = TRUE,
refresh = 0)
