创建基因列表[重复]

Question

我需要创建一个列表，每个基因包含一个向量。向量应该是对每个基因使用 func 的结果。

Answer 1

您好，欢迎来到堆栈溢出。由于您没有提供任何数据进行测试，因此我制作了一些虚拟数据，如下所示。所以这些是虚拟基因的例子：

valid_codons <- c("aaa", "aac", "aag", "aat", "aca", "acc", "acg", "act",
                  "aga", "agc", "agg", "agt", "ata", "atc", "atg", "att", "caa", "cac", 
                  "cag", "cat", "cca", "ccc", "ccg", "cct", "cga", "cgc", "cgg", "cgt",
                  "cta", "ctc", "ctg", "ctt", "gaa", "gac", "gag", "gat", "gca", "gcc",
                  "gcg", "gct", "gga", "ggc", "ggg", "ggt", "gta", "gtc", "gtg", "gtt",
                  "taa", "tac", "tag", "tat", "tca", "tcc", "tcg", "tct", "tga", "tgc",
                  "tgg", "tgt", "tta", "ttc", "ttg", "ttt")



genes <- replicate(3800, {
  paste0(sample(valid_codons, sample(5:20, 1), replace = TRUE), collapse = "")
})
print(head(genes, 3))
#> [1] "gggtacaaagtgcat"                        
#> [2] "cggaaaaccggggcgtgtccg"                  
#> [3] "ggaccactattactctcctcgggtatagatacccgaggt"

我从函数中假设您正在使用的数据结构是字符向量，我是这样制作的：

genes_chars <- strsplit(genes, "")
print(head(genes_chars, 2))
#> [[1]]
#>  [1] "g" "g" "g" "t" "a" "c" "a" "a" "a" "g" "t" "g" "c" "a" "t"
#> 
#> [[2]]
#>  [1] "c" "g" "g" "a" "a" "a" "a" "c" "c" "g" "g" "g" "g" "c" "g" "t" "g" "t" "c"
#> [20] "c" "g"

现在谈到您的实际问题，我将您提供的 codon_count() 函数包装在 lapply 循环中以计算结果。

codon_count <- function(gene) {
  answer <- rep(0, 64)
  names(answer) <- valid_codons
  for(i in seq(from=1, to=length(gene), by=3)) {
    codon <- tolower(paste0(gene[i], gene[i+1], gene[i+2]))
    answer[codon] <- answer[codon] + 1
  }
  return(answer[valid_codons])
}

result <- lapply(genes_chars, codon_count)
print(head(result, 2))
#> [[1]]
#> aaa aac aag aat aca acc acg act aga agc agg agt ata atc atg att caa cac cag cat 
#>   1   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   1 
#> cca ccc ccg cct cga cgc cgg cgt cta ctc ctg ctt gaa gac gag gat gca gcc gcg gct 
#>   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0 
#> gga ggc ggg ggt gta gtc gtg gtt taa tac tag tat tca tcc tcg tct tga tgc tgg tgt 
#>   0   0   1   0   0   0   1   0   0   1   0   0   0   0   0   0   0   0   0   0 
#> tta ttc ttg ttt 
#>   0   0   0   0 
#> 
#> [[2]]
#> aaa aac aag aat aca acc acg act aga agc agg agt ata atc atg att caa cac cag cat 
#>   1   0   0   0   0   1   0   0   0   0   0   0   0   0   0   0   0   0   0   0 
#> cca ccc ccg cct cga cgc cgg cgt cta ctc ctg ctt gaa gac gag gat gca gcc gcg gct 
#>   0   0   1   0   0   0   1   0   0   0   0   0   0   0   0   0   0   0   1   0 
#> gga ggc ggg ggt gta gtc gtg gtt taa tac tag tat tca tcc tcg tct tga tgc tgg tgt 
#>   0   0   1   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   1 
#> tta ttc ttg ttt 
#>   0   0   0   0

我们可以使用length() 和lengths() 检查尺寸是否正确。

unique(lengths(result))
#> [1] 64
length(result)
#> [1] 3800

不过，我认为下面的代码效率更高。

# Split character vectors into groups of three
# Based on https://stackoverflow.com/questions/11619616/how-to-split-a-string-into-substrings-of-a-given-length
splitgenes <- strsplit(genes, "(?<=.{3})", perl = TRUE)

result2 <- t(vapply(splitgenes, function(gene) {
  table(factor(gene, valid_codons))
}, numeric(length(valid_codons))))

# What are the result2 dimensions and content?
dim(result2)
#> [1] 3800   64
result2[1:5, 1:5]
#>      aaa aac aag aat aca
#> [1,]   1   0   0   0   0
#> [2,]   1   0   0   0   0
#> [3,]   0   0   0   0   0
#> [4,]   0   0   0   0   0
#> [5,]   0   1   0   0   0

编辑：

这是 lapply 语句的 for 循环等效项：

result <- list()
for (i in seq_along(genes_chars)) {
  result[[i]] <- codon_count(genes_chars[[i]])
}

但请注意，这样做效率较低。