31

我在 R 中有一个包含以下元素的列表：

[[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list_elem[[814]][3]的方式访问它们，以防我想提取位置 814 的第三个元素。我需要提取所有列表的第三个元素，例如12345_s_at，我想将它们放入向量或列表中所以我可以稍后将它们的元素与另一个列表进行比较。下面是我的代码：

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"

所以我不能用像这样的索引来遍历它们newlist[3]，因为它会返回NA。我的错误在哪里？

4

3 回答 3

52

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到@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，以便我们轻松重现您的问题。

于 2012-09-30T14:36:51.830 回答
16

您可以提取元素并purrr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_功能也可以强制类型一致性，并且map_df()最终可以帮助结束do.call(rbind, …)疯狂。

于 2016-08-17T11:33:42.300 回答
1

如果您想使用您在问题中键入的代码，以下是修复：

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" ))

v <- character() #creates empty character vector
list_len <- length(listdata)
for(i in 1:list_len)
    v <- c(v, listdata[[i]][3]) #fills the vector with list elements (not efficient, but works fine)

print(v)
[1] "12345_s_at" "6789_at"    "12351_s_at"
于 2015-08-18T15:13:26.103 回答