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"Set Difference" between two vectors with duplicate values
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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"
Related
I want to sort a character variable into two categories in a new variable based on conditions, in conditions are not met i want it to return "other".
If variable x cointains 4 character values "A", "B", "C" & "D" I want to sort them into a 2 categories, 1 and 0, in a new variable y, creating a dummy variable
Ideally I want it to look like this
df <- data.frame(x = c("A", "B", "C" & "D")
y <- if x == "A" | "D" then assign 1 in y
if x == "B" | "C" then assign 0 in y
if x == other then assign NA in y
x y
1 "A" 1
2 "B" 0
3 "C" 0
4 "D" 1
library(dplyr)
df <- df %>% mutate ( y =case_when(
(x %in% df == "A" | "D") ~ 1 ,
(x %in% df == "B" | "C") ~ 1,
x %in% df == ~ NA
))
I got this error message
Error: replacement has 3 rows, data has 2
Here's the proper case_when syntax.
df <- data.frame(x = c("A", "B", "C", "D"))
library(dplyr)
df <- df %>%
mutate(y = case_when(x %in% c("A", "D") ~ 1,
x %in% c("B", "C") ~ 0,
TRUE ~ NA_real_))
df
#> x y
#> 1 A 1
#> 2 B 0
#> 3 C 0
#> 4 D 1
You're combining syntaxes in a way that makes sense in speech but not in code.
Generally you can't use foo == "G" | "H". You need to use foo == "G" | foo == "H", or the handy shorthand foo %in% c("G", "H").
Similarly x %in% df == "A" doesn't make sense x %in% df makes sense. df == "A" makes sense. Putting them together x %in% df == ... does not make sense to R. (Okay, it does make sense to R, but not the same sense it does to you. R will use its Order of Operations which evaluates x %in% df first and gets a result from that, and then checks whether that result == "A", which is not what you want.)
Inside a dplyr function like mutate, you don't need to keep specifying df. You pipe in df and now you just need to use the column x. x %in% df looks like you're testing whether the column x is in the data frame df, which you don't need to do. Instead use x %in% c("A", "D"). Aron's answer shows the full correct syntax, I hope this answer helps you understand why.
suppose I have 3 vectors:
a = c("A", "B", "C")
b = c("D", "E", "F")
c = c("G", "H", "I")
then I have an element:
element = "E"
I want to find which list does my element belongs to. In this case, list b.
It will be appreciated if the solution to this problem is more general because my real data set have more than a hundred lists.
element = "E"
names(our_lists)[sapply(our_lists, `%in%`, x = element)]
# [1] "b"
Data
our_lists <- list(
a = c("A", "B", "C"),
b = c("D", "E", "F"),
c = c("G", "H", "I")
)
Using grep.
element <- "E"
l <- mget(c("a", "b", "c"))
names(l)[grep(element, l)]
# [1] "b"
If you keep the data in individual objects, you need to check for the element in each one individually. Get them in a list.
list_data <- mget(c('a', 'b', 'c'))
names(Filter(any, lapply(list_data, `==`, element)))
#[1] "b"
If all your vectors have the same length then a vectorised idea can be,
c('a', 'b', 'c')[ceiling(which(c(a, b, c) == 'E') / length(a))]
#[1] "b"
You can use dplyr::lst that creates named list from variable names. Then purrr::keep to keep only the vectors that contain your element.
require(tidyverse)
lst(a, b, c) %>%
keep(~ element %in% .x) %>%
names()
output:
[1] "b"
Very new to R. So I am wondering if you can use two different parameters to get the position of both elements from a list. See the below example...
x <- c("A", "B", "A", "A", "B", "B", "C", "C", "A", "A", "B")
y <- c(which(x == "A"))
[1] 1 3 4 9 10
x[y]
[1] "A" "A" "A" "A" "A"
x[y+1]
[1] "B" "A" "B" "A" "B"
But I would like to return the positions of both y and y+1 together in the same list. My current solution is to merge the two above lists by row number and create a dataframe from there. I don't really like that and was wondering if there is another way. Thanks!
I dont know what exactly you want, but this could help:
newY = c(which(x == "A"),which(x == "A")+1)
After that you can sort it with
finaldata <- newY[order(newY)]
Or you do both in one step:
finaldata <- c(which(x == "A"),which(x == "A")+1)[order(c(which(x == "A"),which(x == "A")+1))]
Then you could also delete duplicates if you want to. Please tell me if this is what you wanted.
I have a list of vectors containing strings and I want R to give me another list with all vectors that contain certain strings. MWE:
list1 <- list("a", c("a", "b"), c("a", "b", "c"))
Now, I want a list that contains all vectors with "a" and "b" in it. Thus, the new list should contain two elements, c("a", "b") and c("a", "b", "c").
As list1[grep("a|b", list1)] gives me a list of all vectors containing either "a" or "b", I expected list1[grep("a&b", list1)] to do what I want, but it did not (it returned a list of length 0).
This should work:
test <- list("a", c("a", "b"), c("a", "b", "c"))
test[sapply(test, function(x) sum(c('a', 'b') %in% x) == 2)]
Try purrr::keep
library(purrr)
keep(list1, ~ all(c("a", "b") %in% .))
We can use Filter
Filter(function(x) all(c('a', 'b') %in% x), test)
#[[1]]
#[1] "a" "b"
#[[2]]
#[1] "a" "b" "c"
A solution with grepl:
> list1[grepl("a", list1) & grepl("b", list1)]
[[1]]
[1] "a" "b"
[[2]]
[1] "a" "b" "c"
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 = .)