在使用biomaRt库的函数上下文中调用变量时出现意外结果
我有一个一般性的问题,我认为可以归结为某种范围问题 下面是使用biomaRt的getSequence()函数的公式片段。用户在自定义函数中输入(1)基因名称,并可选地输入(2)上游要导入的碱基对数在使用biomaRt库的函数上下文中调用变量时出现意外结果,r,R,我有一个一般性的问题,我认为可以归结为某种范围问题 下面是使用biomaRt的getSequence()函数的公式片段。用户在自定义函数中输入(1)基因名称,并可选地输入(2)上游要导入的碱基对数 # Load libraries library(biomaRt) # Let's make a custom "getSequence" function getUpstream <- function(x, bp.upstream = 50){ bp.upstream <- b
# Load libraries
library(biomaRt)
# Let's make a custom "getSequence" function
getUpstream <- function(x, bp.upstream = 50){
bp.upstream <- bp.upstream
ensembl <- useMart("ensembl", dataset = "hsapiens_gene_ensembl")
upstream.master <- NULL
for(i in x){
upstream.i <- getSequence(id = i,
type = "hgnc_symbol",
seqType = "coding_gene_flank",
upstream = bp.upstream,
mart = ensembl
)
upstream.master <- rbind(upstream.master, upstream.i)
}
return(upstream.master)
}
意外的是,如果没有以下行,函数将无法工作:
bp.upstream <- bp.upstream
bp.upstream这里有一个避免范围问题的解决方法
# Load libraries
library(biomaRt)
# Let's make a custom "getSequence" function
getUpstream <- function(x, bp.upstream = 50){
ensembl <- useMart("ensembl", dataset = "hsapiens_gene_ensembl")
upstream.master <- lapply(x, function(i,stream)
getSequence(id = i,
type = "hgnc_symbol",
seqType = "coding_gene_flank",
upstream = stream,
mart = ensembl),stream=bp.upstream)
upstream.master
}
#加载库
图书馆(生物艺术)
#让我们创建一个自定义的“getSequence”函数
逆流而上
# Load libraries
library(biomaRt)
# Let's make a custom "getSequence" function
getUpstream <- function(x, bp.upstream = 50){
ensembl <- useMart("ensembl", dataset = "hsapiens_gene_ensembl")
upstream.master <- lapply(x, function(i,stream)
getSequence(id = i,
type = "hgnc_symbol",
seqType = "coding_gene_flank",
upstream = stream,
mart = ensembl),stream=bp.upstream)
upstream.master
}