如何从具有混合元素的列表中提取元素

问题描述

我在R中有一个包含以下元素的列表:

I have a list in R with the following elements:

[[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"

我知道可以提取list_elem[[814]][3]之类的内容来访问它们，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，然后将它们放入向量或列表中，以便以后可以将它们的元素与另一个列表进行比较.下面是我的代码:

I know I can access them with something like list_elem[[814]][3] in case that I want to extract the third element of the position 814.I need to extract the third element of all the list, for example 12345_s_at, and I want to put them in a vector or list so I can compare their elements to another list later on. Below is my code:

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:

When I print the results I get the third element, but like this:

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

所以我无法使用像newlist[3]这样的索引遍历它们，因为它返回了NA.我的错误在哪里?

so I cannot traverse them with an index like newlist[3], because it returns NA. Where is my mistake?

推荐答案

如果要提取每个列表元素的第三个元素，则可以执行以下操作:

If you want to extract the third element of each list element you can do:

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注释，您可以执行以下操作:

Using your example and taking into account @GSee comment you can do:

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在部分数据集中提供一些数据，以便我们轻松重现您的问题.

Next time you can provide some data using dput on a portion of your dataset so we can reproduce your problem easily.

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