I have a question of subsetting a nested list by names.
I have an example list like:
test_list <- list(a = list(A1 = c(1,2,3), A2 = c(4,5,6)),
b = c(7,8,9),
c = list(C1 = c(10,11,12), C2 = list(C21 =c(13,14,15))))
And I want to subset values based on a vector like lnames <-
c('c','C2','C21'). The way I can think of doing this is using:
exp_str <- paste0('test_list','$',paste0(lnames, collapse = '$'))
eval(parse(text = exp_str))
But this seems a little be clunky to me. I am just wondering if there's a functional way to do this like using reduce function.
You can just do
test_list[[lnames]]
# [1] 13 14 15
This is somewhat cryptically described in the ?Extract help page.
[[ can be applied recursively to lists, so that if the single index i is a vector of length p, alist[[i]] is equivalent to alist[[i1]]...[[ip]] providing all but the final indexing results in a list.
Related
lets say I have nested list and I want to convert it into flat data frame in R. look into picture for reference. SO what should I do ?
Nested List
You may be able to use the unlist command,
l.ex <- list(a = list(1:5, LETTERS[1:5]), b = "Z", c = NA)
unlist(l.ex, recursive = FALSE)
unlist(l.ex, recursive = TRUE)
Also look into as.data.frame(do.call(cbind, vectorOfUnlist))
This is the list we can work with:
list1 <- list(L1 = c(a = "abc", b = "bca"),
L2 = c(a = 1, b = 2))
In this case i need a list, that only contains lists that meet the condition, so L1.
This is how i can check it by hand, but i dont know how to go further.
class(list1[[1]][[1]])
class(list1[[1]][[2]])
class(list1[[2]][[1]])
class(list1[[2]][[2]])
Checking for is.character in an sapply and subset the list.
list1[sapply(list1, is.character)]
# $L1
# a b
# "abc" "bca"
Using Filter from base R
Filter(is.character, list1)
I am facing the following challenge:
I have a list of dataframes in R and I'd like to extract some specific information from it. Here is an example:
df_1 <- data.frame(A = c(1,2), B = c(3,4), D = c(5,6))
df_2 <- data.frame(A = c(7,8), B = c(9,10), D = c(11,12))
df_3 <- data.frame(A = c(0,1), B = c(2,3), D = c(4,5))
L <- list(df_1, df_2, df_3)
What I'd like to extract are the values at position (1,1) in each of these dataframes. In the above case this would be: 1, 7, 0.
Is there a way to extract this information easily, probably with one line of code?
As Ronak has suggested , you can use function like lapply and wrap it with unlist for desired output.
unlist(lapply(L,function(x) x[1,1]))
In addition to the *apply methods shown above, you can also do this in a Vectorized manner. Since all the data frames in your list have the same column names, and you want the first element from the first column, i.e. 'A1', then you can simply unlist (which will create a named vector) and grab the values with the name A1.
v1 <- unlist(L)
v1[names(v1) == 'A1']
#A1 A1 A1
# 1 7 0
I have a list similar to this one:
set.seed(1602)
l <- list(data.frame(subst_name = sample(LETTERS[1:10]), perc = runif(10), crop = rep("type1", 10)),
data.frame(subst_name = sample(LETTERS[1:7]), perc = runif(7), crop = rep("type2", 7)),
data.frame(subst_name = sample(LETTERS[1:4]), perc = runif(4), crop = rep("type3", 4)),
NULL,
data.frame(subst_name = sample(LETTERS[1:9]), perc = runif(9), crop = rep("type5", 9)))
Question: How can I extract the subst_name-column of each data.frame and combine them with cbind() (or similar functions) to a new data.frame without messing up the order of each column? Additionally the columns should be named after the corresponding crop type (this is possible 'cause the crop types are unique for each data.frame)
EDIT: The output should look as follows:
Having read the comments I'm aware that within R it doesn't make much sense but for the sake of having alook at the output the data.frame's View option is quite handy.
With the help of this SO-Question I came up with the following sollution. (There's probably room for improvement)
a <- lapply(l, '[[', 1) # extract the first element of the dfs in the list
a <- Filter(function(x) !is.null(unlist(x)), a) # remove NULLs
a <- lapply(a, as.character)
max.length <- max(sapply(a, length))
## Add NA values to list elements
b <- lapply(a, function(v) { c(v, rep(NA, max.length-length(v)))})
e <- as.data.frame(do.call(cbind, d))
names(e) <- unlist(lapply(lapply(lapply(l, '[[', "crop"), '[[', 2), as.character))
It is not really correct to do this with the given example because the number of rows is not the same in each one of the list's data frames . But if you don't care you can do:
nullElements = unlist(sapply(l,is.null))
l = l[!nullElements] #delete useless null elements in list
columns=lapply(l,function(x) return(as.character(x$subst_name)))
newDf = as.data.frame(Reduce(cbind,columns))
If you don't want recycled elements in the columns you can do
for(i in 1:ncol(newDf)){
colLength = nrow(l[[i]])
newDf[(colLength+1):nrow(newDf),i] = NA
}
newDf = newDf[1:max(unlist(sapply(l,nrow))),] #remove possible extra NA rows
Note that I edited my previous code to remove NULL entries from l to simplify things
I have the following named list output from a analysis. The reproducible code is as follows:
list(structure(c(-213.555409754509, -212.033637890131, -212.029474755074,
-211.320398316741, -211.158815833294, -210.470525157849), .Names = c("wasn",
"chappal", "mummyji", "kmph", "flung", "movie")), structure(c(-220.119433774144,
-219.186901747536, -218.743319709963, -218.088361753899, -217.338920075687,
-217.186050877079), .Names = c("crazy", "wired", "skanndtyagi",
"andr", "unveiled", "contraption")))
I want to convert this to a data frame. I have tried unlist to data frame options using reshape2, dplyr and other solutions given for converting a list to a data frame but without much success. The output that I am looking for is something like this:
Col1 Val1 Col2 Val2
1 wasn -213.55 crazy -220.11
2 chappal -212.03 wired -219.18
3 mummyji -212.02 skanndtyagi -218.74
so on and so forth. The actual out put has multiple columns with paired values and runs into many rows. I have tried the following codes already:
do.call(rbind, lapply(df, data.frame, stringsAsFactors = TRUE))
works partially provides all the character values in a column and numeric values in the second.
data.frame(Reduce(rbind, df))
didn't work - provides the names in the first list and numbers from both the lists as tow different rows
colNames <- unique(unlist(lapply(df, names)))
M <- matrix(0, nrow = length(df), ncol = length(colNames),
dimnames = list(names(df), colNames))
matches <- lapply(df, function(x) match(names(x), colNames))
M[cbind(rep(sequence(nrow(M)), sapply(matches, length)),
unlist(matches))] <- unlist(df)
M
didn't work correctly.
Can someone help?
Since the list elements are all of the same length, you should be able to stack them and then combine them by columns.
Try:
do.call(cbind, lapply(myList, stack))
Here's another way:
as.data.frame( c(col = lapply(x, names), val = lapply(x,unname)) )
How it works. lapply returns a list; two lists combined with c make another list; and a list is easily coerced to a data.frame, since the latter is just a list of vectors having the same length.
Better than coercing to a data.frame is just modifying its class, effectively telling the list "you're a data.frame now":
L = c(col = lapply(x, names), val = lapply(x,unname))
library(data.table)
setDF(L)
The result doesn't need to be assigned anywhere with = or <- because L is modified "in place."