I have the following list
A = list(c(1,2,3,4), c(5,6,7,8), c(4,6,2,3,1), c(6,2,1,7,12, 15, 16, 10))
A
[[1]]
[1] 1 2 3 4
[[2]]
[1] 5 6 7 8
[[3]]
[1] 4 6 2 3 1
[[4]]
[1] 6 2 1 7 12 15 16 10
I want to check if the element 2
is present each list or not. If it exists, then I need to assign 1
to that corresponding list.
Thank you in advance.
@jasbner's comment can be further refined to
1 * sapply(A, `%in%`, x = 2)
# [1] 1 0 1 1
In this case sapply
returns a logical vector, and then multiplication by 1 coerces TRUE
to 1
and FALSE
to 0. Also, as the syntax is x %in% table
, we may avoid defining an anonymous function function(x) 2 %in% x
and instead write as above. Lastly, using sapply
rather than lapply
returns a vector rather than a list, which seems to be what you are after.
这是带替换功能的简单版本!
lapply(A, function(x) ifelse(x==2,1,x))
Here is an option with tidyverse
library(tidyverse)
map_lgl(A, `%in%`, x = 2) %>%
as.integer
#[1] 1 0 1 1
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.