I have a list of vectors (characters). For example:
my_list <- list(c("a", "b", "c"),
c("a", "b", "c", "d"),
c("e", "d"))
For the intersection of all these three vectors, I could use: Reduce(intersect, my_list). But as you can see, there is no common element in all three vectors.
Then, what if I want to find the common element that appears "at least" a certain amount of times in the list? Such as: somefunction(my_list, time=2) would give me c("a", "b", "c", "d") because those elements appear two times.
Thanks.
We can convert this to a data.table and do the group by action to get the elements
library(data.table)
setDT(stack(setNames(my_list, seq_along(my_list))))[,
if(uniqueN(ind)==2) values , values]$values
#[1] "a" "b" "c" "d"
A base R option would be to unlist the 'my_list', find the frequency count with the replicated sequence of 'my_list' using table, get the column sums, check whether it is equal to 2 and use that index to subset the names.
tblCount <- colSums(table(rep(seq_along(my_list), lengths(my_list)), unlist(my_list)))
names(tblCount)[tblCount==2]
#[1] "a" "b" "c" "d"
If you assume that each element will appear no more than once in a vector, you can "unlist" your vectors and count the frequency.
Here, using dplyr functions
library(dplyr)
my_list %>% unlist %>% data_frame(v=.) %>% count(v) %>% filter(n>=2) %>% .[["v"]]
Or base functions
subset(as.data.frame(table(unlist(my_list))), Freq>=2)$Var1
This works:
my_list %>%
purrr::map(~ .) %>%
purrr::reduce(.f = dplyr::intersect, .x = .)
Related
Is there a method in R, to substitute the values of a vector using a dictionary (2 column dataframe with old and new value)
The only method I know is to extract the old value into a dataframe and merge it with, what I call,the dictionary (which is a two column dataframe with old and new values). Afterwards reassign the new value to the original old value. However, it seems when using merge (at least since R v4.1, the order of the x value is not maintained, so I am using join now which keeps the original order of dataframe x intact. I am thinking that there must be an easier way, I just have not found it. Hope this is understandable, I appreciate any help.
cheers Hermann
You could use a named character vector as a dict for replacement by unquoting with !!! inside of dplyr::recode. If you have your "dict" stored as a two-column dataframe, then tidyr::deframe might be handy.
library(tidyverse)
x <- c("a", "b", "c")
dict <- tribble(
~old, ~new,
"a", "d",
"b", "e",
"c", "f"
)
recode(x, !!!deframe(dict))
#> [1] "d" "e" "f"
Created on 2021-06-14 by the reprex package (v1.0.0)
You can use match to substitute the values of a vector using a dictionary:
D$new[match(x, D$old)]
#[1] "d" "e" "f"
You can also use the names to get the new values:
L <- setNames(D$new, D$old)
L[x]
#"d" "e" "f"
Data:
x <- c("a", "b", "c")
D <- data.frame(old = c("a", "b", "c"), new = c("d", "e", "f"))
This question already has answers here:
"Set Difference" between two vectors with duplicate values
(4 answers)
Closed 2 years ago.
a <- c("A", "B", "C", "A", "A", "B")
b <- c("A", "C", "A")
I want to subset a wrt to b such that the following set is obtained:-
("B" "A" "B")
Tradition subsetting results in removal of all the "A"s and "C"s from set a.
It removes duplicates also. I don't want them to be remove. For ex:- Set b has 2 "A"s and 1 "C". So while subsetting a wrt b only two "A"s and one "C" should be removed from set a. And rest all the elements in a should remain even though they might be "A" or "C".
I just want to know if there is a way of doing this in R.
An easy option is to use vsetdiff from package vecsets, i.e.,
vecsets::vsetdiff(a,b)
such that
> vecsets::vsetdiff(a,b)
[1] "B" "A" "B"
Using tibble and dplyr, you can do:
enframe(a) %>%
transmute(name = value) %>%
group_by(name) %>%
mutate(ID = 1:n()) %>%
left_join(enframe(table(b)), by = c("name" = "name")) %>%
filter(ID > value | is.na(value)) %>%
pull(name)
[1] "B" "A" "B"
Here is a way to do this :
#Count occurrences of `a`
a_count <- table(a)
#Count occurrences of `b`
b_count <- table(b)
#Subtract the count present in b from a
a_count[names(b_count)] <- a_count[names(b_count)] - b_count
#Create a new vector of remaining values
rep(names(a_count), a_count)
#[1] "A" "B" "B"
Or:
a <- c("A", "B", "C", "A", "A", "B")
b <- c("A", "C", "A")
greedy_delete <- function(x, rmv) {
for (i in rmv) {
x <- x[-which(x == i)[1]]
}
x
}
greedy_delete(a, b)
#"B" "A" "B"
I have a list, in which each element is a vector of strings, as:
l <- list(c("a", "b"), c("c", "d"))
I want to find the index of the element in l that contains a specific vector of strings, as c("a", "b"). How do I do that? I thought which(l %in% c("a", "b")) should work, but it returns integer(0) instead of 1.
%in% checks presence of elements of the LHS among elements of the RHS. To treat c("a", "b") as a single element of the RHS, it needs to be in a list:
which(l %in% list(c("a", "b")))
Other possibilities are to go element-by-element through l with sapply, such as
which(sapply(l,function(x) all(c("a","b") %in% x)))
# order doesn't matter, other elements allowed
which(sapply(l, identical, c("a", "b"))) # exact match, in order
I know that similar questions have already been asked (e.g. Passing list element names as a variable to functions within lapply or R - iteratively apply a function of a list of variables), but I couldn't manage to find a solution for my problem based on these posts.
I have an event dataset (~100 variables, >2000 observations) that contains variables with information on the involved actors. One variable can only contain one actor, so if several actors have been involved in the event, they are spread over several variables (e.g. actor1, actor2, ...). These actors can be classified into two groups ("s" and "nons"). For later use, I need two lists of actors: one that contains all actors of the category "s" and one that contains all actors of "nons". "s" only consists of three actors while "nons" consists of dozens of actors.
# create example data
df <- data.frame(id = c(1:8),
actor1 = c("A", "B", "D", "E", "F", "G", "H", NA),
actor2 = c("A", NA, "B", "C", "E", "I", "D", "G"))
df <-
df %>%
mutate(actor1 = as.character(actor1),
actor2 = as.character(actor2))
Since the script I am about to prepare is supposed to be used on updated versions of the dataset in the future, I would like to automate as much as possible and keep the parts of the script that would need to be adapted as limited as possible. My idea was to create one function per category that extracts the actors of the respective category (e.g. "nons") from one variable (e.g. actor1) in a list and then "loop" this function over the other variables (ideally with the apply family).
I know which category each actor belongs to ("A", "B", and "C" are category "s"), which allows me to define a separation rule as used in the function below (the filter command).
# create function
nons_function <- function(col) {
col_ <- enquo(col)
nons_list <-
df %>%
filter(!is.na(!!col_), !!col_ != "A", !!col_ != "B", !!col_ != "C") %>%
distinct(!!col_) %>%
pull()
nons_list
}
# create list of variables to "loop" over
actorlist <- c("actor1", "actor2")
This results in the following. Instead of two lists of actors I get a list that contains the variable names as character strings.
> lapply(actorlist, nons_function)
[[1]]
[1] "actor1"
[[2]]
[1] "actor2"
What I would like to get is something like the following:
> lapply(actorlist, nons_function)
[[1]]
[1] "D" "E" "F" "G" "H"
[[2]]
[1] "E" "I" "D" "G"
The problem is probably the way I am passing the variable names to my function within lapply. Apparently, my function is not able use a character input as variable names. However, I have not found a way to either adapt my function in a way that allows for character input or to provide my function with a list of variables to loop over in a way it can digest.
Any help appreciated!
EDIT: Initially I had named the actors in a misleading way (actor names indicated which category an actor belongs to), which lead to answers that do not really help in my case. I have changed the actor names from "s1", "s2", "nons1", "nons2" etc to "A", "B", "C" etc now.
here is an option using base r.
for nons-actors:
lapply( df[, 2:3], function(x) grep( "^nons", x, value = TRUE ) )
#$actor1
#[1] "nons1" "nons2" "nons3" "nons4" "nons5"
#
#$actor2
#[1] "nons2" "nons6" "nons1" "nons4"
and for s-actors:
lapply( df[, 2:3], function(x) grep( "^s", x, value = TRUE ) )
# $actor1
# [1] "s1" "s2"
#
# $actor2
# [1] "s1" "s2" "s3"
Here is an option
library(dplyr)
library(stringr)
library(purrr)
map(actorlist, ~ df %>%
select(.x) %>%
filter(!str_detect(!! rlang::sym(.x), "^s\\d+$")) %>%
pull(1))
#[[1]]
#[1] "nons1" "nons2" "nons3" "nons4" "nons5"
#[[2]]
#[1] "nons2" "nons6" "nons1" "nons4"
It can be wrapped as a function as well. Note that the input is string, so instead of enquo, use sym to convert to symbol and then evaluate (!!)
f1 <- function(dat, colNm) {
dat %>%
select(colNm) %>%
filter(!str_detect(!! rlang::sym(colNm), "^s\\d+$")) %>%
pull(1) %>%
unique
}
map(actorlist, f1, dat = df)
NOTE: This can be done more easily, but here we are using similar code from the OP's post
Another option is to use split with grepl in base R and that returns a list of both 'nons' and 's' after removing the NAs
lapply(df[2:3], function(x) {
x1 <- x[!is.na(x)]
split(x1, grepl("nons", x1))})
Check my solution and see if it works for you.
require("dplyr")
# create example data
df <- data.frame(id = c(1:8),
actor1 = c("s1", "s2", "nons1", "nons2", "nons3", "nons4", "nons5", NA),
actor2 = c("s1", NA, "s2", "s3", "nons2", "nons6", "nons1", "nons4"))
df <-
df %>%
mutate(actor1 = as.character(actor1),
actor2 = as.character(actor2))
# Function for getting the category
category_function <- function(col,categ){
if(categ == "non"){
outp = grep("^non",col,value = T)
}else{
outp = grep("^s",col,value = T)
}
return(outp)
}
# Apply the function to all variables whose name starts with "actor"
sapply(df[grep("actor",names(df),value=T)],category_function,categ="non")
sapply(df[grep("actor",names(df),value=T)],category_function,categ="s")
My output was the following:
> sapply(df[grep("actor",names(df),value=T)],category_function,categ="non")
$actor1
[1] "nons1" "nons2" "nons3" "nons4" "nons5"
$actor2
[1] "nons2" "nons6" "nons1" "nons4"
> sapply(df[grep("actor",names(df),value=T)],category_function,categ="s")
$actor1
[1] "s1" "s2"
$actor2
[1] "s1" "s2" "s3"
This is probably pretty straightforward but I'm really stuck: Let's say that I have a vector=c("a", "b", "c","d","e"). How can I concatenate only some specific elements? For example, how do I merge "b" and "c", which will lead me to the vector=c("a","bc","d","e") ?
Thank you
We can do
i1 <- vector %in% c("b", "c")
c(vector[!i1], paste(vector[i1], collapse=""))