### estimate methylation level
EstiMu <- function(counts, sf){
### function to estimate mu
### counts: expression count, input1, ip1,
### input2, ip2, input3, ip3, ...
x = counts[, seq(1, ncol(counts), 2)] # input
y = counts[, seq(2, ncol(counts), 2)] # ip
n = ncol(y)
if(is.null(sf)){
sf = colSums(counts, na.rm = TRUE)/median(
colSums(counts, na.rm = TRUE))
}
sf.x = sf[seq(1, ncol(counts), 2)]
sf.y = sf[seq(2, ncol(counts), 2)]
lambda_x.hat = sweep(x, 2, sf.x, FUN = "/")
lambda_y.hat = sweep(y, 2, sf.y, FUN = "/")
#### version 1
# p.hat = lambda_y.hat/(lambda_x.hat + lambda_y.hat)
# mu.hat = rowMeans(p.hat)
#
#### version 2
mu = rowMeans(lambda_y.hat)/rowMeans(
lambda_x.hat + lambda_y.hat)
mu[mu ==0] = 0.01
mu[mu==1] = 0.99
return(mu)
}
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