I have a list l
, which has the following features:
- It has 3 elements
- Each element is a numeric vector of length 5
- Each vector contains numbers from 1 to 5
l = list(a = c(2, 3, 1, 5, 1), b = c(4, 3, 3, 5, 2), c = c(5, 1, 3, 2, 4))
I want to do two things:
First
I want to know how many times each number occurs in the entire list and I want each result in a vector (or any form that can allow me to perform computations with the results later):
Code 1:
> a <- table(sapply(l, "["))
> x <- as.data.frame(a)
> x
Var1 Freq
1 1 3
2 2 3
3 3 4
4 4 2
5 5 3
Is there anyway to do it without using the table()
function. I would like to do it "manually". I try to do it right below.
Code 2: (I know this is not very efficient!)
x <- data.frame(
"1" <- sum(sapply(l, "[")) == 1
"2" <- sum(sapply(l, "[")) == 2
"3" <- sum(sapply(l, "[")) == 3
"4" <- sum(sapply(l, "[")) == 4
"5" <- sum(sapply(l, "[")) == 5)
I tried the following, but I did not work. I actually did not understand the result.
> sapply(l, "[") == 1:5
a b c
[1,] FALSE FALSE FALSE
[2,] FALSE FALSE FALSE
[3,] FALSE TRUE TRUE
[4,] FALSE FALSE FALSE
[5,] FALSE FALSE FALSE
> sum(sapply(l, "[") == 1:5)
[1] 2
Second
Now, I would like to get the number of times each number appears in the list, but now in each element $a
, $b
and $c
. I thought about using the lapply()
but I don't know how exactly. Following is what I tried, but it is inefficient just like Code 2:
lapply(l, function(x) sum(x == 1))
lapply(l, function(x) sum(x == 2))
lapply(l, function(x) sum(x == 3))
lapply(l, function(x) sum(x == 4))
lapply(l, function(x) sum(x == 5))
What I get with these 5 lines of code are 5 lists of 3 elements each containing a single numeric value. For example, the second line of code tells me how many times number 2
appears in each element of l
.
Code 3:
> lapply(l, function(x) sum(x == 2))
$a
[1] 1
$b
[1] 1
$c
[1] 1
What I would like to obtain is a list with three elements containing all the information I am looking for.
Please, use the references "Code 1", "Code 2" and "Code 3" in your answers. Thank you very much.