How to extract elements from a list with mixed elements

Question

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"

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)
 }

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"

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

score 53 · Accepted Answer · edited Oct 29 '17 at 04:41

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

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"

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

Use `sapply` to avoid the `unlist` part. Also, I think `[` would suffice. +1 — GSee, Sep 30 '12 at 14:38
You can also use `pluck` here instead of `[[` which just makes it a bit more readable with an actual verb. — andrew_overflow, Jun 16 '20 at 14:39

score 17 · Answer 2 · answered Aug 17 '16 at 11:33

With purrr you can extract elements and ensure data type consistency:

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"

There are other map_ functions that enforce the type consistency as well and a map_df() which can finally help end the do.call(rbind, …) madness.

score 1 · Answer 3 · answered Aug 18 '15 at 15:13

In case you wanted to use the code you typed in your question, below is the fix:

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"

How to extract elements from a list with mixed elements

3 Answers3

Linked

Related