如何从具有混合元素的列表中提取元素

Question

我在R中有一个列表,其中包含以下元素:

[[812]]
[1] ""             "668"          "12345_s_at" "667"          "4.899777748" 
[6] "49.53333333"  "10.10930207"  "1.598228663"  "5.087437057" 

[[813]]
[1] ""            "376"         "6789_at"  "375"         "4.899655078"
[6] "136.3333333" "27.82508792" "2.20223398"  "5.087437057"

[[814]]
[1] ""             "19265"        "12351_s_at" "19264"        "4.897730912" 
[6] "889.3666667"  "181.5874908"  "1.846451572"  "5.087437057" 

我知道我可以使用类似于list_elem[[814]][3]我想要提取位置814的第三个元素的东西来访问它们.我需要提取所有列表的第三个元素,例如12345_s_at,我想将它们放在一个向量或列表中所以我可以稍后将其元素与另一个列表进行比较.以下是我的代码:

elem<-(c(listdata))
lp<-length(elem)
for (i in 1:lp)
{
    newlist<-c(listdata[[i]][3]) ###maybe to put in a vector
    print(newlist)
 }

当我打印结果时,我得到了第三个元素,但是像这样:

  [1] "1417365_a_at"
  [1] "1416336_s_at"
  [1] "1416044_at"
  [1] "1451201_s_at"

所以我不能用索引来遍历它们newlist[3],因为它会返回NA.我的错误在哪里？

Answer 1

如果要提取每个列表元素的第三个元素,可以执行以下操作:

List <- list(c(1:3), c(4:6), c(7:9))
lapply(List, '[[', 3)  # This returns a list with only the third element
unlist(lapply(List, '[[', 3)) # This returns a vector with the third element

使用您的示例并考虑@GSee评论,您可以执行以下操作:

yourList <- list(c("","668","12345_s_at","667", "4.899777748","49.53333333",
       "10.10930207", "1.598228663","5.087437057"),
     c("","376", "6789_at",  "375",  "4.899655078","136.3333333",
       "27.82508792", "2.20223398",  "5.087437057"),
     c("", "19265", "12351_s_at", "19264", "4.897730912",
       "889.3666667", "181.5874908","1.846451572","5.087437057" ))

sapply(yourList, '[[', 3)
[1] "12345_s_at" "6789_at"    "12351_s_at"

下次您可以使用dput部分数据集提供一些数据,以便我们轻松地重现您的问题.

Answer 2

随着purrr您可以提取元素,并确保数据类型一致性:

library(purrr)

listdata <- list(c("","668","12345_s_at","667", "4.899777748","49.53333333",
       "10.10930207", "1.598228663","5.087437057"),
     c("","376", "6789_at",  "375",  "4.899655078","136.3333333",
       "27.82508792", "2.20223398",  "5.087437057"),
     c("", "19265", "12351_s_at", "19264", "4.897730912",
       "889.3666667", "181.5874908","1.846451572","5.087437057" ))

map_chr(listdata, 3)
## [1] "12345_s_at" "6789_at"    "12351_s_at"

还有其他map_功能可以强制执行类型一致性,map_df()最终可以帮助结束do.call(rbind, …)疯狂.