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?
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.
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.
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"
Related
I'm a beginner with R.
I tried making a function that selects the first element of a vector.
Then I used the function with lapply on my vector.
However, instead of choosing the first set of student and score, all students pop out..
I understand my function is selecting the first element per group(?) of the vector. But can anyone explain why my function results in the students instead of the first student and score?
Help please. Thanks!
student_score <- c("Philip:70", "Jimmy:80", "Alex:90", "Steve:100")
split_score <- strsplit(student_score, split = ":")
select_first <- function(x) {
x[1]
}
unlist(lapply(split_score, select_first))
#Used unlist() just to make the view of the result simpler
Expected Result: "Philip" "70" /// Actual Result: "Philip" "Jimmy" "Alex" "Steve"
the lapply function is just hiding a for loop around each element of split_score. Here what's inside of it after the strsplit:
strsplit(student_score, split = ":") # gives a list
[[1]]
[1] "Philip" "70"
[[2]]
[1] "Jimmy" "80"
[[3]]
[1] "Alex" "90"
[[4]]
[1] "Steve" "100"
So your call to lapply could be translated like this:
for each element in split_score (each element is the pair name-score)
extract the first (x[1])
So at the first iteration of the loop inside lapply the x is x = c("Philip" "70") and x[1] is "Philip", and so on.
That's why you get the list of names. Note that lapply is just hiding the for loop.
Your lapply is basically doing this:
for (i in 1:4){
split_score[[i]][1]
}
I have a dataframe with a column of URLs from which I want to remove everything after the first question mark. Some URLs have no question mark, and I want these to remain unchanged. In short, I want to strip off all the tracking. This is a sample URL.
https://www.dummy.com/2017/11/29/four-questions-we-have-about-stuff/?utm_source=exacttarget&utm_medium=newsletter&utm_term=dummydotcom-dummycomnewsletter&utm_content=na-readblog-blogpost&utm_campaign=dummy
This is the result I'm looking for.
https://www.dummy.com/2017/11/29/four-questions-we-have-about-stuff/
Assuming your dataframe is called df and it has a column in it named url:
df$url <- sub('\\?.*', '', df$url)
With strsplit:
url <- "https://www.dummy.com/2017/11/29/four-questions-we-have-about-stuff/?utm_source=exacttarget&utm_medium=newsletter&utm_term=dummydotcom-dummycomnewsletter&utm_content=na-readblog-blogpost&utm_campaign=dummy"
result <- strsplit(url, "\\?")[[1]][1]
Output:
> result
[1] "https://www.dummy.com/2017/11/29/four-questions-we-have-about-stuff/"
And here is an example of using it on a vector rather than a single string:
strings <- c("here?string", "another?string", "stringnoquestion", "one?more")
> sapply(strsplit(strings, "\\?"), function(x) x[1])
[1] "here" "another" "stringnoquestion" "one"
strsplit returns a list because it is written to work for vectors as well as singular elements. So in the first example the [[1]] was accessing the first element of the list and then the [1] was accessing the first element of that, the url before the ?.
Here is the first example broken out in to steps:
# Returns a list of length one
> strsplit(url, "\\?")
[[1]]
[1] "https://www.dummy.com/2017/11/29/four-questions-we-have-about-stuff/"
[2] "utm_source=exacttarget&utm_medium=newsletter&utm_term=dummydotcom-dummycomnewsletter&utm_content=na-readblog-blogpost&utm_campaign=dummy"
# Each element of the list is a vector
> strsplit(url, "\\?")[[1]]
[1] "https://www.dummy.com/2017/11/29/four-questions-we-have-about-stuff/"
[2] "utm_source=exacttarget&utm_medium=newsletter&utm_term=dummydotcom-dummycomnewsletter&utm_content=na-readblog-blogpost&utm_campaign=dummy"
# The first element of that vector
> strsplit(url, "\\?")[[1]][1]
[1] "https://www.dummy.com/2017/11/29/four-questions-we-have-about-stuff/"
I am trying to name a nested list. This would be one of the several lists in my nested list:
paths_list[i]
[[1]]
[[1]]$CLASS
[1] "Signal Transduction (Saccharomyces cerevisiae)"
[[1]]$GENES
[1] "YPR165W"
[[1]]$ORGANISM
[1] "Saccharomyces cerevisiae"
Basically what I want to do is to put an ID name for example R-SCE-198203 as the main name for the list (so above $CLASS it should appear the name R-SCE-198203). List paths_list[i] to have the name R-SCE-198203.
I want this:
paths_list[i]
[[1]]R-SCE-198203
[[1]]$CLASS
[1] "Signal Transduction (Saccharomyces cerevisiae)"
[[1]]$GENES
[1] "YPR165W"
[[1]]$ORGANISM
[1] "Saccharomyces cerevisiae"
I have searched and the closest I have found was with lapply but you ends up like this:
setNames(lapply(tabs, setNames, varB), varA)
#$varA1
#$varA1$varB1
#[1] "integer"
#
#$varA1$varB2
#[1] "integer"
# ...
I want to avoid the main ID to appear in every element of the list (do not want $varA1 being repeated all the time).
Is that possible?
Thanks in advance
I think your lapply approach is already what you want. The "true" names of the sub-elements do not have the $ signs. The output in the console when you print a full list object shows these signs to help you read the data, however, if you access the individual sub-element via [[]] their names do not have these signs. Maybe the following code example helps. Check the outputs.
a_list <- list("dummy")
names(a_list) <- "dummyname"
a_list <- list(a_list)
names(a_list) <- "name"
a_list
#$name
#$name$dummyname
#[1] "dummy"
names(a_list)
#[1] "name"
a_list[[1]]
#$dummyname
#[1] "dummy"
names(a_list[[1]])
#[1] "dummyname"
I want to read the synonyms from a csv file , where the first word is the "main" word and the rest of the words in the same record are its synonyms
now i basically want to create a list like i would have in R ,
**synonyms <- list(
list(word="ss", syns=c("yy","yyss")),
list(word="ser", syns=c("sert","sertyy","serty"))
)**
This gives me a list as
synonyms
[[1]]
[[1]]$word
[1] "ss"
[[1]]$syns
[1] "yy" "yyss"
[[2]]
[[2]]$word
[1] "ser"
[[2]]$syns
[1] "sert" "sertyy" "serty"
which is essentially a list of lists of "word" and "syns".
how do i go about creating the similar list while reading the word and synonyms from a csv file
any pointers would help !! Thanks
This process should return what you want.
# read in data using readLines
myStuff <- readLines(textConnection(temp))
This will return a character vector with one element per line in the file. Note that textConnection is not necessary for reading in files. Just supply the file path. Now, split each vector element into a vectors using strsplit and return a list.
myList <- strsplit(myStuff, split=" ")
Now, separate the first element from the remaining element for each vector within the list.
result <- lapply(myList, function(x) list(word=x[1], synonyms=x[-1]))
This returns the desired result. We use lapply to move through the list items. For each list item, we return a named list where the first element, named word, corresponds to the first element of the vector that is the list item and the remaining elements of this vector are placed in a second list element called synonyms.
result
[[1]]
[[1]]$word
[1] "ss"
[[1]]$synonyms
[1] "yy" "yyss"
[[2]]
[[2]]$word
[1] "ser"
[[2]]$synonyms
[1] "sert" "sertyy" "serty"
[[3]]
[[3]]$word
[1] "at"
[[3]]$synonyms
[1] "ate" "ater" "ates"
[[4]]
[[4]]$word
[1] "late"
[[4]]$synonyms
[1] "lated" "lates" "latee"
data
temp <-
"ss yy yyss
ser sert sertyy serty
at ate ater ates
late lated lates latee"
I try to get an object's name from the list containing this object. I searched through similar questions and find some suggestions about using the deparse(substitute(object)) formula:
> my.list <- list(model.product, model.i, model.add)
> lapply(my.list, function(model) deparse(substitute(model)))
and the result is:
[[1]]
[1] "X[[1L]]"
[[2]]
[1] "X[[2L]]"
[[3]]
[1] "X[[3L]]"
whereas I want to obtain:
[1] "model.product", "model.i", "model.add"
Thank you in advance for being of some help!
You can write your own list() function so it behaves like data.frame(), i.e., uses the un-evaluated arg names as entry names:
List <- function(...) {
names <- as.list(substitute(list(...)))[-1L]
setNames(list(...), names)
}
my.list <- List(model.product, model.i, model.add)
Then you can just access the names via:
names(my.list)
names(my.list) #..............
Oh wait, you didn't actually create names did you? There is actually no "memory" for the list function. It returns a list with the values of its arguments but not from whence they came, unless you add names to the pairlist given as the argument.
You won't be able to extract the information that way once you've created my.list.
The underlying way R works is that expressions are not evaluated until they're needed; using deparse(substitute()) will only work before the expression has been evaluated. So:
deparse(substitute(list(model.product, model.i, model.add)))
should work, while yours doesn't.
To save stuffing around, you could employ mget to collect your free-floating variables into a list with the names included:
one <- two <- three <- 1
result <- mget(c("one","two","three"))
result
#$one
#[1] 1
#
#$two
#[1] 1
#
#$three
#[1] 1
Then you can follow #DWin's suggestion:
names(result)
#[1] "one" "two" "three"