要在R中列出的文本文件

时间:2011-07-06 20:58:36

标签: list r text statistics

我有一个大文本文件,每行中包含可变数量的字段。每行中的第一个条目对应于生物学途径,并且每个后续条目对应于该途径中的基因。前几行可能看起来像这样

path1   gene1 gene2
path2   gene3 gene4 gene5 gene6
path3   gene7 gene8 gene9

我需要将此文件作为列表读入R中,每个元素都是一个字符向量,列表中每个元素的名称是该行的第一个元素,例如:

> pathways <- list(
+     path1=c("gene1","gene2"), 
+     path2=c("gene3","gene4","gene5","gene6"),
+     path3=c("gene7","gene8","gene9")
+ )
> 
> str(pathways)
List of 3
 $ path1: chr [1:2] "gene1" "gene2"
 $ path2: chr [1:4] "gene3" "gene4" "gene5" "gene6"
 $ path3: chr [1:3] "gene7" "gene8" "gene9"
> 
> str(pathways$path1)
 chr [1:2] "gene1" "gene2"
> 
> print(pathways)
$path1
[1] "gene1" "gene2"

$path2
[1] "gene3" "gene4" "gene5" "gene6"

$path3
[1] "gene7" "gene8" "gene9"

...但我需要自动为数千行做这件事。我看到了similar question posted here previously,但我无法弄清楚如何从该线程中做到这一点。

提前致谢。

4 个答案:

答案 0 :(得分:41)

这是一种方法:

# Read in the data
x <- scan("data.txt", what="", sep="\n")
# Separate elements by one or more whitepace
y <- strsplit(x, "[[:space:]]+")
# Extract the first vector element and set it as the list element name
names(y) <- sapply(y, `[[`, 1)
#names(y) <- sapply(y, function(x) x[[1]]) # same as above
# Remove the first vector element from each list element
y <- lapply(y, `[`, -1)
#y <- lapply(y, function(x) x[-1]) # same as above

答案 1 :(得分:6)

一种解决方案是通过read.table()读取数据,但使用fill = TRUE参数填充具有较少“条目”的行,将结果数据帧转换为列表,然后清理“空”元素。

首先,请阅读以下数据:

con <- textConnection("path1   gene1 gene2
path2   gene3 gene4 gene5 gene6
path3   gene7 gene8 gene9
")
dat <- read.table(con, fill = TRUE, stringsAsFactors = FALSE)
close(con)

接下来,我们删除第一列,首先将其保存为稍后列表的名称

nams <- dat[, 1]
dat <- dat[, -1]

将数据框转换为列表。这里我只是在索引1,2,...,n上拆分数据框,其中n是行数:

ldat <- split(dat, seq_len(nrow(dat)))

清理空单元格:

ldat <- lapply(ldat, function(x) x[x != ""])

最后,应用名称

names(ldat) <- nams

,并提供:

> ldat
$path1
[1] "gene1" "gene2"

$path2
[1] "gene3" "gene4" "gene5" "gene6"

$path3
[1] "gene7" "gene8" "gene9"

答案 2 :(得分:3)

基于链接页面的快速解决方案......

inlist <- strsplit(readLines("file.txt"), "[[:space:]]+")
pathways <- lapply(inlist, tail, n = -1)
names(pathways) <- lapply(inlist, head, n = 1)

答案 3 :(得分:3)

还有一个解决方案:

sl <- c("path1 gene1 gene2", "path2 gene1 gene2 gene3") # created by readLines 
f <- function(l, s) {
  v <- strsplit(s, " ")[[1]]
  l[[v[1]]] <- v[2:length(v)]
  return(l)
}
res <- Reduce(f, sl, list())